Hierarchical connect by and start with and joined?
I have an Employees table and a table of identifiers. The table of identifiers is hierarchical, with parents and children. Each employee has one or more identifiers, but that a unique identifier is considered to be the "primary" or root for each employee identifier. Unfortunately, the employee table can point to one of the children identifier lines and not the root. I need a quick query to reach employees with their most recent ID (root).Here's the code to define the problem.
create table employees (employeeid varchar2(8), fakeNationalID varchar2(9), empname varchar2(30));
insert into employees (employeeid, fakeNationalID, empname) values (1,'001000001','John Smith');
insert into employees (employeeid, fakeNationalID, empname) values (2,'002000002','James Jones');
create table realids (realidkey NUMBER, fakeNationalID VARCHAR2(9) not null,
realNationalID VARCHAR2(9) UNIQUE, parent_realidkey number);
insert into realids (realidkey, fakeNationalID, realNationalID, parent_realidkey) values
(1,'001000001','111111111',3);
insert into realids (realidkey, fakeNationalID, realNationalID, parent_realidkey) values
(2,'002000002','222222222',null);
insert into realids (realidkey, fakeNationalID, realNationalID, parent_realidkey) values
(3,'003000003','333333333',null);
commit;
create or replace function get_parentid (fakeID in VARCHAR2) return varchar2 is
tempid VARCHAR2(9);
begin
select realNationalID into tempid
from (
select realNationalID, fakeNationalID
from realids
start with fakeNationalID = fakeID
connect by nocycle prior parent_realidkey = realidkey
order by level desc)
where rownum = 1;
return tempid;
exception
when NO_DATA_FOUND then
return NULL;
when others then raise;
end;
select get_parentid('001000001') from dual; -- returns 333333333 because its linked to a parent
select get_parentid('002000002') from dual; -- returns 222222222 because there is only one child
select get_parentid('003000003') from dual; -- returns 333333333 because it is the parentIt works, but it is NOT very effective:
select employeeid, get_parentid(fakeNationalID) realid, empname from employees;
employeeid realid empname
---------- ----------- ------------
1 333333333 John Smith
2 222222222 James JonesIt seemed like a good way to do it, but with a sub query.
select e.employeeid, e.fakenationalid, e.empname, sub.realnationalid
from employees,
(select realidkey, fakenationalid, realnationalid, parent_realidkey
from realids r
start with r.fakenationalid = e.fakenationalid
connect by prior r.parent_realidkey = r.realidkey) subAnyone has any ideas on how to get top most parent node of the realids for each row in the employees table? In real life, there are 6 or more employees tables across multiple remote instances of what children in the realids table and how much to the parents. We always want the highest parent of the page realid. Any help would be appreciated.
Hello
Thanks for posting the sample data in a convenient form!
It is always useful to post your version of Oracle, too, especially when it comes with CONNECT BY queries.
What follows is what you asked for in Oracle 10:
WITH got_roots AS
(
SELECT CONNECT_BY_ROOT fakenationalid AS leaf_id
, realnationalid
FROM realids
WHERE CONNECT_BY_ISLEAF = 1
START WITH fakenationalid IN ( SELECT fakenationalid
FROM employees
)
CONNECT BY realidKEY = PRIOR parent_realidkey
)
SELECT e.employeeid
, r.realnationalid
, e.empname
FROM employees e
JOIN got_roots r ON r.leaf_id = e.fakenationalid
;
In any query, call a function defined by the user for each line is going to be slow. Fortunately, Oracle now has the built-in functions and operators that can take the place of get_parentid. The CONNECT_BY_ROOT operator, which was introduced in Oracle 10, is the key to the problem. In Oracle 9, you can get the same results using SYS_CONNECT_BY_PATH.
It is generally faster to CONNECT BY query separately, and then join some other tables you need for results.
You had a good idea in your last query. The problem was that void and employees were equal tables in the FROM clause, and you cannot establish a correlation between equals. You can only correlate a subquery to its Super application. You could make to this general idea work by changing void in a scalar sub-requete, which can be connected to the employees, but I think it would be much less effective than what I posted above.
Tags: Database
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Join the two trees connect by prior Start With and return only common records?
Oracle 10g Release 2 (10.2)
I have two tables which have structured data. The results, when running queries individually are correct, but I need to join tree a tree two to get only the common records between them.
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SELECT ip_entity_name, entity_code, hier_level, entity_parent
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If I understand correctly, the joints can not work with CONNECT BY / START WITH queries?
A WITH clause is an option?
If possible, I don't want to put a selection in a database to display object and join against other queries.
Thank you.Hello
jtp51 wrote:
Oracle 10g Release 2 (10.2)
...
If I understand correctly, the joints can not work with CONNECT BY / START WITH queries?Before Oracle 9 it was true. Since you're using Oracle 10, you can if you wish; but I'm guessing that you don't want in this case.
A WITH clause is an option?
If possible, I don't want to put a selection in a database to display object and join against other queries.
Yes, a WITH clause that is an option. Viewed online, as Zhxiang has shown, are another option. Either way gives you a regular display effect without creating a database object.
You did not show a sample and the results, so no one can tell if a join is really what you want. Other possibilities include INTERSECT or an IN subquery.
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START WITH and CONNECT BY in Oracle SQL (hierarchical)
Hi, the original table as below
To identify the hierarchical relationship of the data, which are PARENT_ACCOUNT_ID & ACCOUNT_ID, here's the query I used.Customer_ID Account_ID Paying_Account_ID Parent_Account_ID Company_ID 158 158 158 158 0 159 159 158 158 0 160 160 158 158 0 181 181 181 181 0 183 183 183 183 0 24669 24669 24669 24669 0 24671 24671 24671 24669 0 24670 24670 24670 24669 0 3385127 3385127 3385127 24670 0
It is the result of the queryselect lpad(' ', 2*level) || A.ACCOUNT_ID AS LEVEL_LABEL, CONNECT_BY_ISCYCLE "Cycle", LEVEL, A.* from ACCOUNT A START WITH parent_account_id = account_id CONNECT BY NOCYCLE PRIOR A.ACCOUNT_ID = A.PARENT_ACCOUNT_ID AND account_id != parent_account_id ;
My question is how can I changed the query to calculate the values for:Level_Label Level Cycle Customer_ID Account_ID Paying_Account_ID Parent_Account_ID Company_ID 158 1 0 158 158 158 158 0 159 2 0 159 159 158 158 0 160 2 0 160 160 158 158 0 181 1 0 181 181 181 181 0 183 1 0 183 183 183 183 0 24669 1 0 24669 24669 24669 24669 0 24671 2 0 24671 24671 24671 24669 0 24670 2 0 24670 24670 24670 24669 0 3385127 3 0 3385127 3385127 3385127 24670 0
My_Total_PR - number of my accounts to child PR which do not include himself.
Total_PR - Total number of accounts PR in the overall structure
My_Total_NPR - number of my accounts of child NPR which do not include himself.
Total_NPR - Total number of accounts NPR in the overall structure
* PR stand for responsible for payment, for example the responsible account payment 158 158 (Paying_Account_ID), therefore the Total_PR to 158 is 3 (158, 159, 160)
* NPR stand responsible for Non-payment, for example the responsible account payment 159 is 158 (Paying_Account_ID), so the Total_NPR for 159 1
This is the expected result, any advice appreciated. Thank you
Published by: user11432758 on February 14, 2012 01:00Level_Label Level Cycle My_Total_PR Total_PR My_Total_NPR Total_NPR Paying_Account 158 1 0 2 3 0 0 158 159 2 0 0 0 0 1 158 160 2 0 0 0 0 1 158 181 1 0 0 1 0 0 181 183 1 0 0 1 0 0 183 24669 1 0 0 1 3 3 24669 24671 2 0 0 1 0 0 24671 24670 2 0 0 1 1 1 24670 3385127 3 0 0 1 0 0 3385127
Published by: user11432758 on February 14, 2012 07:05Hello
user11432758 wrote:
Hi here is the statement DDL, thank youCREATE TABLE "SYSTEM"."ACCOUNT" ...Do not create your own objects in the diagram of the SYSTEM or any scheme that comes with the database. Create a separate schema and place your items. You'll have fewer security problems, and the migration to a new database will be easier.
Here's a way to can get the aggregates you want:
WITH got_descendants AS ( SELECT CONNECT_BY_ROOT account_id AS ancestor_id , paying_account_id , LEVEL AS lvl FROM account CONNECT BY NOCYCLE PRIOR account_id = parent_account_id AND account_id != parent_account_id ) SELECT ancestor_id , COUNT (CASE WHEN lvl > 1 AND ancestor_id = paying_account_id THEN 1 END) AS my_total_pr , COUNT (CASE WHEN ancestor_id = paying_account_id THEN 1 END) AS total_pr , COUNT (CASE WHEN lvl > 1 AND ancestor_id != paying_account_id THEN 1 END) AS my_total_npr , COUNT (CASE WHEN ancestor_id != paying_account_id THEN 1 END) AS total_npr FROM got_descendants GROUP BY ancestor_id ;Output:
` MY_ MY_ TOTAL TOTAL TOTAL TOTAL ANCESTOR_ID _PR _PR _NPR _NPR ----------- ----- ----- ----- ----- 158 2 3 0 0 159 0 0 0 1 160 0 0 0 1 181 0 1 0 0 183 0 1 0 0 24669 0 1 3 3 24670 0 1 1 1 24671 0 1 0 0 3385217 0 1 0 0This gives the correct numbers, but how can bring us in an order that reflects the hierarchy, with the columns (for example lvl) that come from the hierarchy?
A solution would be to make two CONNECT BY queries; a service without START WITH clause (like the one above) who collects the aggregates and the other with a START WITH clause (as your original query), which is in the right order and columns such as level_label and level. We could join result sets and get exactly what we want. I'll leave that as an exercise.Here is another way, which gets good results with only one CONNECTION PER request:
WITH got_descendants AS ( SELECT CONNECT_BY_ROOT account_id AS ancestor_id , paying_account_id , account_id , LEVEL AS lvl , CONNECT_BY_ISCYCLE AS cycle , CASE WHEN CONNECT_BY_ROOT account_id = CONNECT_BY_ROOT parent_account_id THEN ROWNUM END AS r_num FROM account CONNECT BY NOCYCLE PRIOR account_id = parent_account_id AND account_id != parent_account_id ORDER SIBLINGS BY account_id -- Optional ) , got_o_num AS ( SELECT got_descendants.* , MIN (r_num) OVER (PARTITION BY account_id) AS o_num , MAX (lvl) OVER (PARTITION BY account_id) AS max_lvl FROM got_descendants ) SELECT LPAD ( ' ' , 2 * (MIN (max_lvl) - 1) ) || ancestor_id AS level_label , MIN (max_lvl) AS "Level" , MIN (cycle) AS "Cycle" , COUNT (CASE WHEN lvl > 1 AND ancestor_id = paying_account_id THEN 1 END) AS my_total_pr , COUNT (CASE WHEN ancestor_id = paying_account_id THEN 1 END) AS total_pr , COUNT (CASE WHEN lvl > 1 AND ancestor_id != paying_account_id THEN 1 END) AS my_total_npr , COUNT (CASE WHEN ancestor_id != paying_account_id THEN 1 END) AS total_npr , MIN (paying_account_id) AS paying_account FROM got_o_num GROUP BY ancestor_id ORDER BY MIN (o_num) ;Output:
` MY_ MY_ TOTAL TOTAL TOTAL TOTAL PAYING_ LEVEL_LABEL Level Cycle _PR _PR _NPR _NPR ACCOUNT --------------- ----- ----- ----- ----- ----- ----- -------- 158 1 0 2 3 0 0 158 159 2 0 0 0 0 1 158 160 2 0 0 0 0 1 158 181 1 0 0 1 0 0 181 183 1 0 0 1 0 0 183 24669 1 0 0 1 3 3 24669 24670 2 0 0 1 1 1 24670 3385217 3 0 0 1 0 0 3385217 24671 2 0 0 1 0 0 24671That's exactly what you asked for, except that you have posted the line with level_label =' 24671' before the line with level_label = "24671 '. You may not care about who comes first, but if it's important, explains why these lines should be in descending order of account_id, while "159 and 160" are in ascending order. You will need change the ORDERBY brothers and SŒURS clause accordingly.
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START WITH and CONNECT BY PRIOR
Hello
Database: Oracle 11g
1.
SELECT ename, empno, mgr
WCP
START WITH empno = 7839
CONNECT BY PRIOR MGR = EMPNO
Result set:
EMPNO, ENAME MGR
-------- ---------- ----------
KING 7839
2.
SELECT empno.ename, Bishop
WCP
START WITH mgr = 7839
CONNECT BY PRIOR MGR = EMPNO
Result set:
EMPNO, ENAME MGR
-------- ---------- ----------
7566 JONES 7839
7698 BLAKE 7839
7782 CLARK 7839
KING 7839
KING 7839
KING 7839
My questions are:
Q1. What is actually happening in the result defines two queries. I'm not able to grasp the difference when I use START WITH empno = 7839 and START WITH mgr =. 7839
Q2. What is the difference between
CONNECTION BY MGR PRIOR = EMPNO and
CONNECT BY PRIOR EMPNO = MGR?
can someone please help me here?
Thank you
Hello
A CONNECT BY query looks like an operation UNION ALL of the data of different levels, numbered 1, 2, 3 and more.
Level 1 is filled with the START WITH clause.
If there are data on level N, then N + 1 level is filled, using the CONNECT BY clause, which generally refers to something on the N level via the PRIOR operator. Another way to put it is that level N + 1 is filled by a self-join with lines that have already chosen the level N.
If there is no data on the level of N, the query stops.
Let's see how this applies to your queries.
Level being such an important concept in CONNECT BY queries, you might want to see in all your CONNECT BY queries all test and debug the.
1 query, including the level are:
SELECT ename, empno, mgr
LEVEL
FROM scott.emp
START WITH empno = 7839
Empno = mgr PRIOR CONNECTION
;
You will notice that I have re-arranged the CONNECT BY clause. I find it a little more clear medium. Of course, it never changes the results just if you say "x = y" or "y = x.
The results, including the level, are:
LEVEL OF ARCHBISHOP EMPNO, ENAME
---------- ---------- ---------- ----------
7839 KING 1
What happened to produce these results?
First level 1 has been met using the START WITH clause. Level 1 is identical to the results of
SELECT ename, empno, mgr
AS LEVEL 1
FROM scott.emp
WHERE empno = 7839 - same as your START WITH clause
;
It happens to be only 1 row in the table scott.emp who met the empno = 7839 condition, and we show a few columns of this line.
That's all that need the level 1. Something has been on level 1, so we're trying now to complete level 2, using the CONNECT BY condition.
Any line that is included in the level 2 meets the empno = mgr PREREQUISITE condition, where the PREVIOUS operator refers to a line of level 1. In this case, there is only 1 row at level 1, this line gets to have a NULL mgr. Given that PRIOR mgr is NULL in this case, the condition to connect BY
EmpNo = mgr BEFORE equals
EmpNo = NULL and who obviously won't be true for any line, so nothing is added to level 2, and ends the query.
Now let's look at application 2. I'll add another column of debugging, called path, which I'll describe later:
SELECT ename, empno, mgr
LEVEL
, SYS_CONNECT_BY_PATH (empno, "/") AS path
FROM scott.emp
START WITH mgr = 7839
CONNECT BY PRIOR Mgr = empno
LEVEL CONTROL
path
;
Output:
EMPNO, ENAME MGR LEVEL PATH
---------- ---------- ---------- ---------- ---------------
7566 7839 1 7566 JONES
7698 7839 1 7698 BLAKE
7782 7839 1 7782 CLARK
7839 KING 2/7566/7839
7839 KING 2/7698/7839
7839 KING 2/7782/7839
Again, we'll study how people got 1 level. It happens to be 3 scott.emp lines that meet this condition START WITH, so there are 3 lines in the game at level 1.
Given that the data on the level 1, the test of the query to complete level 2, referring to some PRIOR line on level 1. Any line that meets the condition to connect BY, with a line any level 1 in the PREVIOUS line, will appear at level 2.
Let's look at the line at level 1 where ename = 'JONES '. Are there lines in sccott.emp that met the empno = mgr PREREQUISITE condition, where mgr PREREQUISITE is the column of Archbishop of the line with "JONES"? Yes, there are one, the line with ename = 'KING', so that the rank is included at level 2.
Let's look at the line at level 1 where ename = 'BLAKE '. Are there lines in sccott.emp that met the empno = mgr PREREQUISITE condition, where mgr PREREQUISITE is the column of Archbishop of the line with "BLAKE"? Yes, there are one, the line with ename = 'KING', so that the rank is included at level 2.
Let's look at the line at level 1 where ename = 'CLARK '. Are there lines in sccott.emp that met the empno = mgr PREREQUISITE condition, where mgr PREREQUISITE is the column of Archbishop of the line with 'CLARK '? Yes, there are one, the line with ename = 'KING', so that the rank is included at level 2.
There are thus 3 rows at level 2. They happen to all be on the same line of the table emp; It is correct. Remember, CONNECT BY is like a UNION ALL (not just a UNION). It is a UNION of
lines that are at level 1, because him meets the condition to BEGIN WITH, and
lines that are at level 2 because puts it CONNECT BY condition regarding the 'JONES', and
lines that are at level 2, because they meet the condition to connect BY regarding the "BLAKE", and
lines that are at level 2, because they meet the condition to connect BY regarding the "CLARK".
SYS_CONNECT_BY_PATH can enlighten us on that. SYS_CONNECT_BY_PATH (empno, ' / ') shows the empno of each level that caused this line appears in the result set. It's a delimited list /, where the nth element (i.e. the empno after bar oblique nth) is the empno who found the N level.
Since there were data at level 2, the quert now trying to complete level 3.
Is there all the rows in the table that satisfy the CONNECT BY condition (mgr PRIOR = empno) with respect to any line level 2? No, Bishop is be NULL on all lines of level 2, so no line can satisfy this condition CONNECT BY, no lines are added at level 3, and ends the query.
I hope that answers the question:
Q1. What is actually happening in the result defines two queries. I'm not able to grasp the difference when I use START WITH empno = 7839 and START WITH mgr =. 7839
I'll try to not be so detailed answering
Q2. What is the difference between
CONNECTION BY MGR PRIOR = EMPNO and
CONNECT BY PRIOR EMPNO = MGR?
These 2 CONNECT BY conditions are different where you put the PRIOR operator. The operator PRIOR to switching as it change the direction, upward or downward, which move you through the tree you get from level to level.
Bishop PRÉALABLE = empno means the employee on level N + 1 will be the same as the Manager of level N. This means that higher level numbers will be the most senior people in the hierarchy. This is called a query from the bottom up. (Both of the queries that you have posted this same CONNECT BY exact state; both are requests from bottom to top).
Mgr = empno PREREQUISITE or equivalent
PRIOR empno = mgr means exactly the opposite. When you move from level N to level N + 1 in the query, you will move to an older person, to a junior position in the hierarchy. This is called a query from top to bottom. The employee level N will be the Manager of wover is a level N + 1.
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Understanding "CONNECT BY" and "START WITH".
OK, I'm trying to update the rows with values determined by lines joined in a recursive relationship of unknown depth. I am told that "CONNECT BY" and "START WITH" can be useful in this, but I don't see how to get the value I'm looking for.
In my case, there are 3 values in my table.
ID
ID of the parent
Invoice
On some lines, the Bill is null. For records, you get the ID of the invoice by searching for the invoice of the parent folder. I'm trying to update the table so that all THE rows in the table have an ID of invoice.
Here is an example of table and the lines.
What I would do is update the inv_num column in the table so that a select statement * would look like this...CREATE TABLE DISTRIBUTION ( ID INT, INV_NUM INT, PARENT_ID INT ) INSERT INTO DISTRIBUTION 1, 111, NULL; INSERT INTO DISTRIBUTION 2, 112, NULL; INSERT INTO DISTRIBUTION 3, NULL, 2; INSERT INTO DISTRIBUTION 4, 113, NULL; INSERT INTO DISTRIBUTION 5, NULL, 4; INSERT INTO DISTRIBUTION 6, NULL, 5; INSERT INTO DISTRIBUTION 7, NULL, 6;
You can provide any help would be greatly appreciated.ID INV_NUM PARENT_ID ----- ------------- --------------- 1 111 null 2 112 null 3 112 2 4 113 null 5 113 4 6 113 5 7 113 6Hello
Thank you post the CREATE TABLE and INSERT instructions, but please make sure that they work.
None of the INSERT statements; I think you meant something like the statements shown after the query.Here's a way to get the desired results:
UPDATE distribution m SET inv_num = ( SELECT inv_num FROM distribution WHERE CONNECT_BY_ISLEAF = 1 START WITH id = m.id CONNECT BY id = PRIOR parent_id AND PRIOR inv_num IS NULL ) WHERE inv_num IS NULL ;This statement is Bottom-Up of subqueries, where we START WITH the lines that need to update, and process to the top of the tree, until you get to an ancestor who was an inv_num.
In your sample data, only the roots (the lines that have no parents) have inv_num. In this case, it might be a little easier (but only a little) to make a request from top to bottom , where we START WITH the roots and low process in the tree to find their subordinates.
If we add some data examples where a nonroot has inv_num:INSERT INTO DISTRIBUTION (id, inv_num, parent_id) VALUES ( 91, 910, 1); INSERT INTO DISTRIBUTION (id, inv_num, parent_id) VALUES ( 92, NULL, 91);What results would you like?
Using the UPDATE statement above, id = 92 would get his inv_nuym of the closest ancestor (in the case of thios, parent) who had an inv_num:. ID INV_NUM PARENT_ID ---------- ---------- ---------- 1 111 91 910 1 92 910 91 2 112 3 112 2 4 113 5 113 4 6 113 5 7 113 6Either the row with id = 92 gets inv_num = 910, no 111.
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This seems to be a simple problem, but I think that I have enough intelligence SQL to solve.
I have a table called DE_DOMAIN. The columns of interest are:
DOMAIN_ID - PK
NAME
PARENT_ID - FK (can be NULL), poining to CF of the parent
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Regardless of the name of the parent, it works fine:
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OF DE_DOMAIN
WHERE SUPERDOMAINE = 2673
Start by PARENT_ID IS NULL
Connect DOMAIN_ID PARENT_ID = prior
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Thank you very much.Hello
You can use the FIRST operator in the SELECT clause.
I don't have a version of your table, so I'll use scott.emp to illustrate:SELECT empno , ename , mgr , PRIOR ename AS mgr_name FROM scott.emp START WITH mgr IS NULL CONNECT BY mgr = PRIOR empno ORDER SIBLINGS BY ename ;Output:
. EMPNO ENAME MGR MGR_NAME ---------- ---------- ---------- ---------- 7839 KING 7698 BLAKE 7839 KING 7499 ALLEN 7698 BLAKE 7900 JAMES 7698 BLAKE 7654 MARTIN 7698 BLAKE 7844 TURNER 7698 BLAKE 7521 WARD 7698 BLAKE 7782 CLARK 7839 KING 7934 MILLER 7782 CLARK 7566 JONES 7839 KING 7902 FORD 7566 JONES 7369 SMITH 7902 FORD 7788 SCOTT 7566 JONES 7876 ADAMS 7788 SCOTT -
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Switch issues. Help, please.
How is it that a computer which works very well with a direct Ethernet connection won't work with a switch, however any other computer using the same port, cable, ect, and so on, can? We already checked the firewall as a question, that it was not, and now, we are puzzled.Well, I know that the issue is long-term... believe me, it's a long, but as they say, the devil is in the details. Anywho, the question is, my grandfather has a HP laptop, which for two years has been able to run through a Linksys switch to his router and connect its printers to his laptop. About two months ago it suddenly doesn't work like that. Now, it works fine if it plugs directly into the router/Ethernet. My uncle suggested that the switch was bad, but after testing the same port, cable and all, the switch worked well for him. ' GRAMPS really needs this answered, but someone else, it is called can not understand and now I, in turn, ask you all for what you can offer. Here are the ideas that we have already discredited.
-Switch bad: as above, is that this particular laptop, even under identical conditions, the works of my uncle very well.
-Bad NIC: debunked through the fact we connected to the computer directly to the router and it worked fine.
-Bad configuration of the firewall: we have disabled the firewall (please do not notice, he wasn't the smartest idea, we already know) and even if she recorded the switch exists, the internet is always triggered when you are connected with the switch.Thanks in advance for any assistance that you can provide and Merry Christmas to you all.
It would be useful to consult the results of the ipconfig/all command both when it is connected to the switch and when it is connected directly to the router. In addition, what is the model of the switch?
To save the manual copy and the new hits of the ipconfig/all command output-
First connect through switch.
Open a command prompt window (start > run > cmd > OK)
Type the following lines in the black command prompt window and press ENTER after each lineecho "Connected via the button" > "% UserProfile%\Desktop\ipinfo.txt".
ipconfig/all > "% UserProfile%\Desktop\ipinfo.txt".Now connect directly to the router, type the following lines in the command prompt window and press ENTER after each line
echo "Connected to the router" > "% UserProfile%\Desktop\ipinfo.txt".
ipconfig/all > "% UserProfile%\Desktop\ipinfo.txt".
Notepad '% UserProfile%\Desktop\ipinfo.txt '.
outputCopy the contents of the Notepad window in your response (if you use the laptop when it is connected directly to the router) or close the Notepad window, and then copy ipinfo.txt of the laptop to the Gramps in a USB FlashDrive you can connect to any computer allows you to answer.
You can delete ipinfo.txt on the desktop when you are finished.
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Vista, loss of internet connection on cold start and restarts.
I use windows Vista, recently (after only 4 years of satisfactory used) I lose the internet connection on cold starting and restarting and duty use windows diagnostics to establish the connection. I bought a new Modem Sagem, and cable ethernet and filters, still the problems persists. ISP tells me that the problem is with Vista, can anyone help please?
The game may 17, 2012 03:43:10 + 0000, BIGKEV2257 wrote:
I am not satisfied by McAfee, three days and three nights, waiting for their technicians to ring me, I get automated emails asking if I'm happy and saying that they can not communicate with me, seems since Mr. McAfee sold his company, that it went to the pack, and I understand not why Microsoft sell McAfee with new computers when mcAfee are not appropriate technical support.
Two points:
1. in my opinion and that of many others of us here, McAfee is one of
the poorest choice of security these days software here. I strongly
recommend you take and replace with eSet NOD32 or
Kaspersky, if you are willing to pay for it. If you want a free
antivirus, I recommend one (do not run more than one antivirus
program) of three of the following:
Avira AntiVir
Avast
Microsoft Security Essentials
You also need anti-spyware software (even if you run a program such as)
Microsoft Security Essentials, with integrated anti-spyware capability
It). I recommend that you download and install MalwareBytes (freeware)
AntiMalware
2 Microsoft isn't selling McAfee with new computers. Microsoft
is not even do or do not sell computers. If your computer comes with
McAfee, it's because the manufacturer of your computer (Dell, HP,
Gateway, or anyone else) has chosen to sell like that.Ken Blake, Microsoft MVP
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Can not get to start remote access connection manager and the connections don't work Internet
Original title: wired & wireless connections does not.
I can not get the remote access connection manager to start and so no internet connection is not working, also I can't open the system restore to go back on this machine. What is this?
I am running a Dell Studio 1735 PP31L w model number / Edition Vista Home premium.
Hi Rick,
1. what happens when you try to start the remote access connection manager? You receive messages or error codes?
2. you receive error codes or restore messages when you perform the system?
You can check the status of the following services and make sure that the services are started.
a. Click Start and type Services in start search and press ENTER.
b. in the services with the right button on the phone and then click Properties.
c. under the general tab, select automatic next to startup type.
d. under the general tab, click Start under the service status and then click apply and then click OK.
e. Repeat steps c & d to the remote access connection manager and Remote Access Auto Connection Manager service.
Hope this information is useful.
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Vista has started with a normal window then changes to one with zoom 150 and no navigation
After my computer did an automatic update, the screen changed. It starts with a normal screen but then goes to a 150% or more with no navigation or any level of zoom in the status bar. So cannot use Control Panel controls to adjust the screen resolution, want to do? Do not have original install disks that the computer came preloaded.
Hello
If you could get in Control Panel, you can use the ENTER key to OK to change settings
and read this;
There are a number of things to try:
try going to your graphic card manufacturers site or computer and are looking for the driver download section
Search your computer or graphics card model number based on what you have and download and install the latest graphics drivers for vista
then try to make the screen of solution of problems
http://Windows.Microsoft.com/en-us/Windows-Vista/change-screen-resolution
Change the screen resolution
__________________________________________________________
or try a restore of the system before this happened
http://www.windowsvistauserguide.com/system_restore.htm
If necessary do in safe mode
Windows Vista
Using the F8 method:
- Restart your computer.
- When the computer starts, you will see your computer hardware are listed. When you see this information begins to tap theF8 key repeatedly until you are presented with theBoot Options Advanced Windows Vista.
- Select the Safe Mode option with the arrow keys.
- Then press enter on your keyboard to start mode without failure of Vista.
- To start Windows, you'll be a typical logon screen. Connect to your computer and Vista goes into safe mode.
- Do whatever tasks you need and when you are done, reboot to return to normal mode.
________________________________________________________
and change how to get updates for you to choose what you install to stop it happening again
You may need to install one at a time to find the problem, we
Make sure that you do not use Windows Update to install the drivers of 3rd material part
Find them directly in the hardware manufacturer
and when you see the problem update right click on it - UAC prompt - then hide it
http://www.bleepingcomputer.com/tutorials/tutorial140.html
Download updates but let me choose whether to install them - if you select this option, Windows will download the updates on your computer, but not install them automatically. If you want to install updates, then you must install them manually. You should only select this option if you have a reason to not install updates automatically. Only advanced users should use this option.
Check for updates but let me choose whether to download and install them - if you select this option, you'll be alerted when there are new updates available for download and install. You can then choose to download and install the updates that you want. This option should really be reserved for people who know exactly which updates they need, or those who have little access to the Internet.
______________________________________________________
Here are the different ways to reinstall Vista
Contact the manufacturer of the laptop computer and ask them to send you to vista recovery disks to reinstall the operating system back as it was when you bought it
they do it for a nominal cost of $
____________________________________________________________
also ask them if you have a recovery partition on your hard drive to get back to the way you bought
you would normally press F8, F9, F10 or F11 or Alt + F10 or 0 to start to start the recovery process according to the manufacturer
Ask them of the exact key sequence
__________________________________________________________
or borrow a vista microsoft dvd; not a HP, Acer recovery disk etc
Make sure that you borrow the correct 32-bit or 64-bit microsoft dvd to your computer
they contain all versions of vista
This is the product key that determines which version of vista is installed
http://www.theeldergeek.com/Vista/vista_clean_installation.html
____________________________________________________________
How to replace Microsoft software or hardware, order service packs and replace product manuals
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