How to find the first Sunday and the second Saturday of each month

Hi all

Thank you

Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - 64 bit Production

994122 wrote:

Hello

I need to pass the months parameter how to do this? like Jan, Feb etc... (one of those)

Do you have a procedure?

Should you output only for the months you passed in the parameter?

The easiest way is to set the parameter as date. When you go such as p_date as DATE ' 2014-10-01', then you can

PROCEDURE two_dates)

p_date IN DATE

p_first_sunday DATE

p_second_saturday DATE

)

BEGIN

p_first_sunday: = NEXT_DAY (TRUNC (p_date, 'MM') - 1, TO_CHAR (DATE ' 2014-10-12', 'DAY'));

p_second_saturday: = NEXT_DAY (TRUNC (p_date, 'MM') - 1, TO_CHAR (DATE ' 2014-10-11', 'DAY')) + 7;

END two_dates;

Or you describe what you need.

Tags: Database

Similar Questions

How to get the second Monday of each month in a given date range?

In Oracle forms, how to get the second Monday of each month in a given date range?
I tried below using the query WITH the Clause, but it seems that WITH Clause does not work in Oracle forms. So is there another way to do this in Oracle forms?
WITH month_range AS
(
SELECT TO_DATE ('Dec 2013', 'Mon YYYY') AS first_month
, TO_DATE ('Mar 2014', 'Mon YYYY') AS last_month
OF the double
)
SELECT NEXT_DAY (6 + ADD_MONTHS (first_month
, LEVEL - 1
)
, 'MONDAY '.
) AS second_monday
OF month_range
CONNECTION OF LEVEL < = 1 + MONTHS_BETWEEN (last_month, first_month)
;
Thanks in advance.

Good fishing, when the first day of the month is Thursday... So I changed the query accordingly... Try the below

SELECT CASE WHEN TO_CHAR (ADD_MONTHS (TRUNC(startdate,'MM'),(LEVEL-1)), 'DY') = 'game '.

THEN NEXT_DAY (ADD_MONTHS (TRUNC(startdate,'MM'),(LEVEL-1)), 'THU')

Of OTHER NEXT_DAY (ADD_MONTHS (TRUNC(startdate,'MM'),(LEVEL-1)), 'Game') + 7

END AS second_day

FROM (SELECT SYSDATE startdate,

SYSDATE + 300 enddate

THE DOUBLE)

CONNECT BY LEVEL<=>

To find the third Saturday of a month
Hi all

Y at - it a simple way to find the third Saturday of each month
other than two queries below.

< CODE >

-The first code (No nested query)
SELECT MAX (DECODE (LENGTH (TO_CHAR (LEVEL)), 1))
, TO_DATE (TO_CHAR (SYSDATE, 'YYYYMM') |) '0' || TO_CHAR (LEVEL), 'YYYYMMDD')
, TO_DATE (TO_CHAR (SYSDATE, 'YYYYMM') |) TO_CHAR (LEVEL), 'YYYYMMDD'))) THIRD_SAT
OF THE DOUBLE
WHERE DECODE (LENGTH (TO_CHAR (LEVEL)), 1).
, TO_CHAR (TO_DATE (TO_CHAR (SYSDATE, 'YYYYMM') |)) '0' || TO_CHAR (LEVEL), 'YYYYMMDD'), "DY")
, TO_CHAR (TO_DATE (TO_CHAR (SYSDATE, 'YYYYMM') |)) TO_CHAR (LEVEL), 'YYYYMMDD'), "DY")
) = 'SAT '.
CONNECT BY LEVEL < 22

--deuxieme code (a nested query)
MAX SELECT THIRD_SAT (ALL_DATES)
OF (MAX SELECTION (DECODE (LENGTH (TO_CHAR (LEVEL)), 1)))
, TO_DATE (TO_CHAR (SYSDATE, 'YYYYMM') |) '0' || TO_CHAR (LEVEL), 'YYYYMMDD')
, TO_DATE (TO_CHAR (SYSDATE, 'YYYYMM') |) TO_CHAR (LEVEL), 'YYYYMMDD'))) ALL_DATES
OF THE DOUBLE
CONNECT BY LEVEL < 22
)
WHERE TO_CHAR (ALL_DATES, 'DY') = 'SAT '.
< code >

Thank you and best regards,
CDPRASAD
Hello
Tubby wrote:
```
ME_XE?select next_day(trunc(sysdate, 'MM'), 'SATURDAY') + 14 from dual;

NEXT_DAY(TRUNC(SYSDA
--------------------
21-AUG-2010 00 00:00

1 row selected.

Elapsed: 00:00:00.00
```
Be careful! If the month happens to start Saturday, which returns the 4th Saturday, not the 3rd:
```
select  next_day ( trunc ( DATE '2010-05-15'
                , 'MM'
                )
          , 'SATURDAY'
          ) + 14
from      dual;

NEXT_DAY(
---------
22-MAY-10
```
That's because NEXT_DAY always returns a DATE after (not the same as) its 1st argument.
Do this instead:
```
select  next_day ( trunc ( DATE '2010-05-15'
                , 'MM'
                ) + 13
          , 'SATURDAY'
          )           AS third_sat
from      dual;

THIRD_SAT
---------
15-MAY-10
```
or, if you need something that is independent of the NLS:
```
select  next_day ( trunc ( SYSDATE
                , 'MM'
                ) + 13
          , TO_CHAR ( DATE '2010-07-31'     -- You can use any Saturday here
                 , 'Day'
                 )
          )           AS third_sat
from      dual;

THIRD_SAT
---------
21-AUG-10
```

How to find the child level for each table in a relational model?

Earthlings,

I need your help, and I know that, "Yes, we can change." Change this thread to a question answered.

So: How to find the child level for each table in a relational model?

I have a database of relacional (9.2), all right?

.
     O /* This is a child who makes N references to each of the follow N parent tables (here: three), and so on. */
    /↑\ Fks
   O"O O" <-- level 2 for first table (circle)
  /↑\ Fks
"o"o"o" <-- level 1 for middle table (circle)
   ↑ Fk
  "º"

Tips:
-Each circle represents a table;
-Red no tables have foreign key
-the picture on the front line of tree, for example, a level 3, but when 3 becomes N? How is N? That is the question.

I started to think about the following:

First of all, I need to know how to take the kids:

select distinct child.table_name child
  from all_cons_columns father
  join all_cons_columns child
 using (owner, position)
  join (select child.owner,
               child.constraint_name fk,
               child.table_name child,
               child.r_constraint_name pk,
               father.table_name father
          from all_constraints father, all_constraints child
         where child.r_owner = father.owner
           and child.r_constraint_name = father.constraint_name
           and father.constraint_type in ('P', 'U')
           and child.constraint_type = 'R'
           and child.owner = 'OWNER') aux
 using (owner)
 where child.constraint_name = aux.fk
   and child.table_name = aux.child
   and father.constraint_name = aux.pk
   and father.table_name = aux.father;

Thought...
We will share!

Thanks in advance,
Philips

Published by: BluShadow on April 1st, 2011 15:08
formatting of code and hierarchy for readbility

Have you looked to see if there is a cycle in the graph of dependence? Is there a table that has a foreign key to B and B has a back of A foreign key?

SQL> create table my_emp (
  2    emp_id number primary key,
  3    emp_name varchar2(10),
  4    manager_id number
  5  );

Table created.

SQL> ed
Wrote file afiedt.buf

  1  create table my_mgr (
  2    manager_id number primary key,
  3    employee_id number references my_emp( emp_id ),
  4    purchasing_authority number
  5* )
SQL> /

Table created.

SQL> alter table my_emp
  2    add constraint fk_emp_mgr foreign key( manager_id )
  3         references my_mgr( manager_id );

Table altered.

SQL> ed
Wrote file afiedt.buf

  1   select level lvl,
  2          child_table_name,
  3          sys_connect_by_path( child_table_name, '/' ) path
  4     from (select parent.table_name      parent_table_name,
  5                  parent.constraint_name parent_constraint_name,
  6                  child.table_name        child_table_name,
  7                  child.constraint_name   child_constraint_name
  8             from user_constraints parent,
  9                  user_constraints child
 10            where child.constraint_type = 'R'
 11              and parent.constraint_type = 'P'
 12              and child.r_constraint_name = parent.constraint_name
 13           union all
 14           select null,
 15                  null,
 16                  table_name,
 17                  constraint_name
 18             from user_constraints
 19            where constraint_type = 'P')
 20    start with child_table_name = 'MY_EMP'
 21*  connect by prior child_table_name = parent_table_name
SQL> /
ERROR:
ORA-01436: CONNECT BY loop in user data

If you have a cycle, you have some problems.

(1) it is a NOCYCLE keyword does not cause the error, but that probably requires an Oracle version which is not so far off support. I don't think it was available at the time 9.2 but I don't have anything old enough to test on

SQL> ed
Wrote file afiedt.buf

  1   select level lvl,
  2          child_table_name,
  3          sys_connect_by_path( child_table_name, '/' ) path
  4     from (select parent.table_name      parent_table_name,
  5                  parent.constraint_name parent_constraint_name,
  6                  child.table_name        child_table_name,
  7                  child.constraint_name   child_constraint_name
  8             from user_constraints parent,
  9                  user_constraints child
 10            where child.constraint_type = 'R'
 11              and parent.constraint_type = 'P'
 12              and child.r_constraint_name = parent.constraint_name
 13           union all
 14           select null,
 15                  null,
 16                  table_name,
 17                  constraint_name
 18             from user_constraints
 19            where constraint_type = 'P')
 20    start with child_table_name = 'MY_EMP'
 21*  connect by nocycle prior child_table_name = parent_table_name
SQL> /

       LVL CHILD_TABLE_NAME               PATH
---------- ------------------------------ --------------------
         1 MY_EMP                         /MY_EMP
         2 MY_MGR                         /MY_EMP/MY_MGR
         1 MY_EMP                         /MY_EMP
         2 MY_MGR                         /MY_EMP/MY_MGR

(2) If you try to write on a table and all of its constraints in a file and do it in a valid order, the entire solution is probably wrong. It is impossible, for example, to generate the DDL for MY_EMP and MY_DEPT such as all instructions for a table come first, and all the instructions for the other are generated second. So even if NOCYCLE to avoid the error, you would end up with an invalid DDL script. If that's the problem, I would rethink the approach.

-Generate the DDL for all tables without constraint
-Can generate the DDL for all primary key constraints
-Can generate the DDL for all unique key constraints
-Can generate the DDL for all foreign key constraints

This is not solidarity all the DOF for a given in the file object. But the SQL will be radically simpler writing - there will be no need to even look at the dependency graph.

Justin

How to find the second largest in a pl/sql table

Hello friends,

I want to find the first and second maximum items in a pl/sql table.

Here's the code...

DECLARE
Max_earnings_type TYPE TABLE IS NUMBER;
max_earnings_tab max_earnings_type: = max_earnings_type();
number of v_count: = 0;
number of v_max_earnings;

Can someone give me how to find the maximum first max and second in the type of the given table.

appreciate your help.

Thank you/kumar

Published by: kumar73 on October 21, 2010 09:42

kumar73 wrote:

When I tried to implement your logic in my application, I get the following error...

PL/SQL: digital or value error: NULL index key value table

What happens if the PL/SQL table has NULL values. Question is how you want to handle NULL values. You want to ignore nulls as GROUP BY do? If you want to consider NULL values, you can say if you want to order the NULLS FIRST or NULLS LAST. I guess that logical GROUP BY:

DECLARE
    TYPE max_earnings_type IS TABLE OF NUMBER;
    TYPE max_earnings_sorted_type IS TABLE OF NUMBER
      INDEX BY BINARY_INTEGER;
    max_earnings_tab        max_earnings_type;
    max_earnings_tab_sorted max_earnings_sorted_type;
BEGIN
    SELECT  sal + comm
      BULK COLLECT
      INTO  max_earnings_tab
      FROM  emp;
    FOR v_i in 1..max_earnings_tab.count LOOP
      IF max_earnings_tab(v_i) IS NOT NULL
        THEN
          max_earnings_tab_sorted(max_earnings_tab(v_i)) := 1;
      END IF;
    END LOOP;
    DBMS_OUTPUT.PUT_LINE('MAX value in PL/SQL table is ' || nvl(to_char(max_earnings_tab_sorted.last),'NULL'));
    DBMS_OUTPUT.PUT_LINE('Second MAX value in PL/SQL table is ' || nvl(to_char(max_earnings_tab_sorted.prior(max_earnings_tab_sorted.last)),'NULL'));
END;
/
MAX value in PL/SQL table is 2650
Second MAX value in PL/SQL table is 1900

PL/SQL procedure successfully completed.

SQL> SELECT  sal + comm
  2    FROM  emp;

  SAL+COMM
----------

      1900
      1750

      2650

      1500

  SAL+COMM
----------

14 rows selected.

SQL>

SY.

How to find the rank lowest for each student using a Select query?
Hey I'm having a little trouble here

I have this table

Student - Grade

John - 8
Richard - 9
Louis - 9
Francis - 5
John - 13
Richard - 10
Peter - 12

Shades may vary from 0 to 20.

Name - varchar (50)
Grade - wide

I'm trying to generate a SQL search to search for each rank lowest for each student.

So far, I did:

Select s.name, s.grade
s student
where s.grade = (select min (grade) of student)

The result of this search returns me only the rank lowest of all ranks posted above, which is Francis - 5.

I want to find the lowest rank, but not only for Francis, but for all students in the table.

How do I do that?

Thank you in advance.
Ok

Now we head into analytical SQL:
```
with student as (select 'John' name, 8 grade, 'fail' result from dual union all
                 select 'John' name, 13 grade, 'pass' result from dual union all
                 select 'Francis',     5 grade,  'fail' from dual union all
                 select 'Peter', 12, 'pass' from dual)
-- End of your test data
SELECT   name,
         MAX (result) KEEP (DENSE_RANK FIRST ORDER BY grade) result,
         MIN (grade) min_grade
FROM     student
GROUP BY name
```
Please note that I passed ;)

Concerning
Peter

To learn more:
http://download.Oracle.com/docs/CD/B19306_01/server.102/b14200/functions056.htm#SQLRF00641

How to find the value duplicate of each column.
I have it here are four columns,
How can I find the duplicate of each columns value.

with All_files like)
Select ' 1000 'like BILL, "2000" AS DELIVERYNOTE, CANDELINVOICE ' 3000', '4000' CANDELIVERYNOTE of all union double
Select ' 5000 ', ' 6000', ' 7000 ', ' 8000' Union double all the
Select '9000 ', '1000', '1100',' 1200' from dual union all
Select ' 1200 ', ' 3400', ' 6700 ', ' 8790' Union double all the
Select ' 1000 ', ' 2000', ' 3000 ', ' 9000' Union double all the
Select '1230', '2340', ' 3450 ', ' 4560' double
)
SELECT * from All_files

Output should be as shown below.

1000 2000 3000 4000
9000 1000 1100 1200
1200 3400 6700 8790
1000 2000 3000 9000

Required to check uniqueness columns.

Thank you.
Hello

If you are not too concerned about performance, this should give you the desired result:
```
SELECT distinct INVOICE, DELIVERYNOTE, CANDELINVOICE, CANDELIVERYNOTE
FROM (
  SELECT a.*,
         count(*) over(partition by t.column_value) cnt
  FROM All_files a,
       table(
         sys.odcivarchar2list( a.INVOICE
                             , a.DELIVERYNOTE
                             , a.CANDELINVOICE
                             , a.CANDELIVERYNOTE ) ) t
)
WHERE cnt > 1
;
```

How to find the redo generated by each SQL statement?
Database version is 10.2.0.4. OS is AIX.
We must again excessive generation during a specific time. We want to identify what a statement contributes to excessive recovery.
ADDM shows this as a first observation

Waiting on event "log file sync" during the execution of operations of COMMIT and ROLLBACK consumed at the time of the important data.

Is it possible to know which statement sql generates how many redo in bytes for a moment?

Thank you
Delphine
Is it possible to know which statement sql generates how many redo in bytes for a moment?

It will not help you.
For example, you can see a lot of UPDATE, DELETE, INSERT from the apps statements. Then? What you do with them? If you can modify the application to eliminate the useless LMD - it would be great of course. But it is unlikely.

When the DML statements also generate Redo, does not cause them too much expectations "log file sync", because the writer log buffer empty buffer as soon as it fulfills more than 30% after a certain time or on Commit / Rollback.

Waiting on event "log file sync" during the execution of operations of COMMIT and ROLLBACK consumed at the time of the important data.

Right. This is because the writer written newspaper here synchronous, reliable, he waits until the HDD writes data down and meets its written OK.
And hard drives are not the fastest things in the computer. They are mechanical - that's why.

I would recommend:
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How to find the value max and min for each column in a table 2d?

How to find the value max and min for each column in a table 2d?

For example, in the table max/min for the first three columns would be 45/23, 14/10, 80/67.

Thank you

Chuck,

With color on your bars, you should have enough experience to understand this.

You're a loop in the table already. Now you just need a function like table Max and min. loop. And you may need to transpose the table 2D.

How to find the right server to use windowsmail to send and receive e-mails?

How to find the right server to use windowsmail to send and receive e-mails?

Who is your current e-mail server? (Which is after the @ in your address).

Windows Mail: Setting up an account of end-to-end
http://Windows.Microsoft.com/en-us/Windows-Vista/Windows-mail-setting-up-an-account-from-start-to-finish

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How to find the book, THE, Op units, Org (All) names and ID?
With regard to:
Mr. Shahzad Saleem

Please run the following query

SELECT gl.NAME ledger_name, hou.NAME operating_unit_name,

xep.NAME legal_entity_name

Gl_ledgers gl, hr_operating_units hou, xle_entity_profiles xep

WHERE gl.ledger_id = hou.set_of_books_id

AND hou.default_legal_context_id = xep.legal_entity_id

How to find my WEP key and the name of my network?

I need to reset my wireless printer to scan. It asks for my WEP key and I thought it was one that is not correct. How to find the right WEP key and my correct network name? I have a laptop HP Pavilion dm4 with HP scanner/printe wireless

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Frank

How to find the age of a child and snapshot number snapshot for a virtual machine.

Hi all

How to find the age of a snapshot for a virtual machine child count and snapshot.

This code snippet should help... Assuming that you have the following entries:

VM (VC:VirtualMachine)

snapshot (VC:VirtualMachineSnapshot)

System.log("====== Snapshot details for VM: "+snapshot.config.name+" ======");
var snapshotTrees = vm.snapshot.rootSnapshotList;
System.log("Snapshot Tree Count: "+snapshotTrees.length);
for each (snapshotTree in snapshotTrees){
    showSnapshotInfo(snapshotTree, snapshot);
}

function showSnapshotInfo(snapshotTree, selectedSnapshot){
    if(selectedSnapshot == snapshotTree.snapshot){
        System.log("====================");
        System.log("Name: "+snapshotTree.name);
        System.log("Id: "+snapshotTree.id);
        System.log("Description: "+snapshotTree.description);
        System.log("createTime: "+snapshotTree.createTime);
        System.log("quiesced: "+snapshotTree.quiesced);
        System.log("replaySupported: "+snapshotTree.replaySupported);
        System.log("backupManifest: "+snapshotTree.backupManifest);
        System.log("snapshot id: "+snapshotTree.snapshot.id);
        System.log("state: "+snapshotTree.state.value);
        var childSnapshotList = snapshotTree.childSnapshotList;
        if(childSnapshotList != null){
            System.log("childSnapshotList length: "+snapshotTree.childSnapshotList.length);
        }
        System.log("");
    }
    if(snapshotTree.childSnapshotList != null){
        System.log("Children found: "+snapshotTree.childSnapshotList.length);
        for each (childSnapshotTree in snapshotTree.childSnapshotList){
            showSnapshotInfo(childSnapshotTree, selectedSnapshot);
        }
    }
}

During the passage of one of my virtual machines, here is an example of output I get:

[2014-06-12 15:19:21.493] [I] Snapshot Tree Count: 1
[2014-06-12 15:19:21.493] [I] ====================
[2014-06-12 15:19:21.494] [I] Name: before upgrade to 5.0.1
[2014-06-12 15:19:21.494] [I] Id: 31
[2014-06-12 15:19:21.494] [I] Description:
[2014-06-12 15:19:21.495] [I] createTime: Mon Mar 19 2012 13:48:57 GMT-0400 (EDT)
[2014-06-12 15:19:21.495] [I] quiesced: false
[2014-06-12 15:19:21.495] [I] replaySupported: null
[2014-06-12 15:19:21.495] [I] backupManifest: null
[2014-06-12 15:19:21.496] [I] snapshot id: snapshot-83
[2014-06-12 15:19:21.496] [I] state: poweredOff
[2014-06-12 15:19:21.496] [I] childSnapshotList length: 1
[2014-06-12 15:19:21.496] [I]
[2014-06-12 15:19:21.497] [I] ====================
[2014-06-12 15:19:21.497] [I] Name: Before 5.1 upgrade
[2014-06-12 15:19:21.497] [I] Id: 36
[2014-06-12 15:19:21.497] [I] Description:
[2014-06-12 15:19:21.498] [I] createTime: Fri Sep 21 2012 21:59:58 GMT-0400 (EDT)
[2014-06-12 15:19:21.498] [I] quiesced: false
[2014-06-12 15:19:21.498] [I] replaySupported: null
[2014-06-12 15:19:21.498] [I] backupManifest: null
[2014-06-12 15:19:21.499] [I] snapshot id: snapshot-84
[2014-06-12 15:19:21.499] [I] state: poweredOff
[2014-06-12 15:19:21.499] [I] childSnapshotList length: 1
[2014-06-12 15:19:21.499] [I]
[2014-06-12 15:19:21.500] [I] ====================
[2014-06-12 15:19:21.500] [I] Name: Before Named Instance
[2014-06-12 15:19:21.500] [I] Id: 37
[2014-06-12 15:19:21.501] [I] Description:
[2014-06-12 15:19:21.501] [I] createTime: Mon Nov 11 2013 13:40:46 GMT-0500 (EST)
[2014-06-12 15:19:21.501] [I] quiesced: false
[2014-06-12 15:19:21.501] [I] replaySupported: null
[2014-06-12 15:19:21.502] [I] backupManifest: null
[2014-06-12 15:19:21.502] [I] snapshot id: snapshot-236
[2014-06-12 15:19:21.502] [I] state: poweredOff
[2014-06-12 15:19:21.502] [I]

How to find the difference between standard edition and standard edition one
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Hi expert,

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How to find the first Sunday and the second Saturday of each month

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