SQL query - SEARCH
I use 4.0.27 - version of SQL Standard and I can't seem to execute search commands. I tried contains freetext, and a few others I found here and there. If someone could help me on some search commands, maybe I'm the wrong syntax or something. For the moment, I have this:SELECT *.
QUIZ
WHEN FREETEXT (TITLE, "SQL");
The query does not work, I don't know if it was a problem with the clause of research I would get a result of 0 isn't it, thought it might be related to my SQL version put so into it at the beginning.
Thanks for your help guys!
slowpoke115 wrote:
> I use 4.0.27 - version of SQL Standard
There is no such thing. Do you mean MySQL 4.0.27?
> I tried contains freetext, and a few others that I found here and
> it.
Different databases use slightly different SQL dialects, so you must
to use the correct syntax. If you mean MySQL, none of the foregoing be
work.
> SELECT *.
> QUIZ
> WHERE FREETEXT (TITLE, "SQL");
In MySQL, this would be written like this:
SELECT * QUIZ
WHERE TITLE LIKE '% SQL % '.
On non-windows servers, MySQL is case sensitive with respect to the database,
the names of table and column, so make sure that you use the
combination of capital and small letters (is the table QUIZ, quiz, or quiz)?
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Published by: SeenuGuddu on October 15, 2009 21:55
Published by: SeenuGuddu on October 15, 2009 21:55
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Best regards
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Published by: Bilal on December 27, 2012 09:52
Published by: Bilal on December 27, 2012 10:07Bilal, thanks for the nice problem! Another possibility to double check is to write a small PL/SQL function that returns 1 if a duplicate id is found, then equate to 0: "NUMBER of RETURN of Duplicate_Token_Found (p_str_main in VARCHAR2, p_str_trial VARCHAR2). It should analyze the second string and could use p_str_main LIKE '%', | l_id | ', %' for each id. In any case, the query complete (without that) is given below:
Solution with names SQL> WITH rsf_itm (con_id, max_weight, nxt_id, lev, tot_weight, tot_profit, path, root_id, lev_1_id) AS ( 2 SELECT c.id, 3 c.max_weight, 4 i.id, 5 0, 6 i.item_weight, 7 i.item_profit, 8 ',' || i.id || ',', 9 i.id, 10 0 11 FROM items i 12 CROSS JOIN containers c 13 UNION ALL 14 SELECT r.con_id, 15 r.max_weight, 16 i.id, 17 r.lev + 1, 18 r.tot_weight + i.item_weight, 19 r.tot_profit + i.item_profit, 20 r.path || i.id || ',', 21 r.root_id, 22 CASE WHEN r.lev = 0 THEN i.id ELSE r.nxt_id END 23 FROM rsf_itm r 24 JOIN items i 25 ON i.id > r.nxt_id 26 AND r.tot_weight + i.item_weight <= r.max_weight 27 ORDER BY 1, 2 28 ) SEARCH DEPTH FIRST BY nxt_id SET line_no 29 , rsf_con (nxt_con_id, nxt_line_no, con_path, itm_path, tot_weight, tot_profit, lev) AS ( 30 SELECT con_id, 31 line_no, 32 To_Char(con_id), 33 ':' || con_id || '-' || (lev + 1) || ':' || path, 34 tot_weight, 35 tot_profit, 36 0 37 FROM rsf_itm 38 UNION ALL 39 SELECT r_i.con_id, 40 r_i.line_no, 41 r_c.con_path || ',' || r_i.con_id, 42 r_c.itm_path || ':' || r_i.con_id || '-' || (r_i.lev + 1) || ':' || r_i.path, 43 r_c.tot_weight + r_i.tot_weight, 44 r_c.tot_profit + r_i.tot_profit, 45 r_c.lev + 1 46 FROM rsf_con r_c 47 JOIN rsf_itm r_i 48 ON r_i.con_id > r_c.nxt_con_id 49 WHERE r_c.itm_path NOT LIKE '%,' || r_i.root_id || ',%' 50 AND r_c.itm_path NOT LIKE '%,' || r_i.lev_1_id || ',%' 51 AND r_c.itm_path NOT LIKE '%,' || r_i.nxt_id || ',%' 52 ) 53 , paths_ranked AS ( 54 SELECT itm_path || ':' itm_path, tot_weight, tot_profit, lev + 1 n_cons, 55 Rank () OVER (ORDER BY tot_profit DESC) rnk, 56 Row_Number () OVER (ORDER BY tot_profit DESC) sol_id 57 FROM rsf_con 58 ), best_paths AS ( 59 SELECT itm_path, tot_weight, tot_profit, n_cons, sol_id 60 FROM paths_ranked 61 WHERE rnk = 1 62 ), row_gen AS ( 63 SELECT LEVEL lev 64 FROM DUAL 65 CONNECT BY LEVEL <= (SELECT Count(*) FROM items) 66 ), con_v AS ( 67 SELECT b.itm_path, r.lev con_ind, b.sol_id, b.tot_weight, b.tot_profit, 68 Substr (b.itm_path, Instr (b.itm_path, ':', 1, 2*r.lev - 1) + 1, 69 Instr (b.itm_path, ':', 1, 2*r.lev) - Instr (b.itm_path, ':', 1, 2*r.lev - 1) - 1) 70 con_nit_id, 71 Substr (b.itm_path, Instr (b.itm_path, ':', 1, 2*r.lev) + 1, 72 Instr (b.itm_path, ':', 1, 2*r.lev + 1) - Instr (b.itm_path, ':', 1, 2*r.lev) - 1) 73 itm_str 74 FROM best_paths b 75 JOIN row_gen r 76 ON r.lev <= b.n_cons 77 ), con_split AS ( 78 SELECT itm_path, con_ind, sol_id, tot_weight, tot_profit, 79 Substr (con_nit_id, 1, Instr (con_nit_id, '-', 1) - 1) con_id, 80 Substr (con_nit_id, Instr (con_nit_id, '-', 1) + 1) n_items, 81 itm_str 82 FROM con_v 83 ), itm_v AS ( 84 SELECT c.itm_path, c.con_ind, c.sol_id, c.con_id, c.tot_weight, c.tot_profit, 85 Substr (c.itm_str, Instr (c.itm_str, ',', 1, r.lev) + 1, 86 Instr (c.itm_str, ',', 1, r.lev + 1) - Instr (c.itm_str, ',', 1, r.lev) - 1) 87 itm_id 88 FROM con_split c 89 JOIN row_gen r 90 ON r.lev <= c.n_items 91 ) 92 SELECT v.sol_id, 93 v.tot_weight s_wt, v.tot_profit s_pr, c.id c_id, c.name c_name, c.max_weight m_wt, 94 Sum (i.item_weight) OVER (PARTITION BY v.sol_id, c.id) c_wt, 95 i.id i_id, i.name i_name, i.item_weight i_wt, i.item_profit i_pr 96 FROM itm_v v 97 JOIN containers c 98 ON c.id = To_Number (v.con_id) 99 JOIN items i 100 ON i.id = To_Number (v.itm_id) 101 ORDER BY sol_id, con_id, itm_id 102 / SOL_ID S_WT S_PR C_ID C_NAME M_WT C_WT I_ID I_NAME I_WT I_PR ---------- ---- ---- ----- --------------- ---- ---- ----- ---------- ---- ---- 1 255 255 1 SEECS UG Block 100 100 1 BIT-10 35 35 2 BIT-11 40 40 6 BICSE-7 25 25 2 IAEC Building 70 70 4 BSCS-3 40 40 7 BESE-3 30 30 3 RIMMS Building 90 85 3 BSCS-2 35 35 5 BEE-4 50 50 2 255 255 1 SEECS UG Block 100 95 4 BSCS-3 40 40 6 BICSE-7 25 25 7 BESE-3 30 30 2 IAEC Building 70 70 1 BIT-10 35 35 3 BSCS-2 35 35 3 RIMMS Building 90 90 2 BIT-11 40 40 5 BEE-4 50 50 3 255 255 1 SEECS UG Block 100 100 3 BSCS-2 35 35 4 BSCS-3 40 40 6 BICSE-7 25 25 2 IAEC Building 70 65 1 BIT-10 35 35 7 BESE-3 30 30 3 RIMMS Building 90 90 2 BIT-11 40 40 5 BEE-4 50 50 4 255 255 1 SEECS UG Block 100 100 3 BSCS-2 35 35 4 BSCS-3 40 40 6 BICSE-7 25 25 2 IAEC Building 70 70 2 BIT-11 40 40 7 BESE-3 30 30 3 RIMMS Building 90 85 1 BIT-10 35 35 5 BEE-4 50 50 5 255 255 1 SEECS UG Block 100 95 2 BIT-11 40 40 6 BICSE-7 25 25 7 BESE-3 30 30 2 IAEC Building 70 70 1 BIT-10 35 35 3 BSCS-2 35 35 3 RIMMS Building 90 90 4 BSCS-3 40 40 5 BEE-4 50 50 6 255 255 1 SEECS UG Block 100 100 2 BIT-11 40 40 3 BSCS-2 35 35 6 BICSE-7 25 25 2 IAEC Building 70 65 1 BIT-10 35 35 7 BESE-3 30 30 3 RIMMS Building 90 90 4 BSCS-3 40 40 5 BEE-4 50 50 7 255 255 1 SEECS UG Block 100 100 2 BIT-11 40 40 3 BSCS-2 35 35 6 BICSE-7 25 25 2 IAEC Building 70 70 4 BSCS-3 40 40 7 BESE-3 30 30 3 RIMMS Building 90 85 1 BIT-10 35 35 5 BEE-4 50 50 8 255 255 1 SEECS UG Block 100 100 1 BIT-10 35 35 4 BSCS-3 40 40 6 BICSE-7 25 25 2 IAEC Building 70 70 2 BIT-11 40 40 7 BESE-3 30 30 3 RIMMS Building 90 85 3 BSCS-2 35 35 5 BEE-4 50 50 9 255 255 1 SEECS UG Block 100 100 1 BIT-10 35 35 4 BSCS-3 40 40 6 BICSE-7 25 25 2 IAEC Building 70 65 3 BSCS-2 35 35 7 BESE-3 30 30 3 RIMMS Building 90 90 2 BIT-11 40 40 5 BEE-4 50 50 10 255 255 1 SEECS UG Block 100 100 1 BIT-10 35 35 3 BSCS-2 35 35 7 BESE-3 30 30 2 IAEC Building 70 65 2 BIT-11 40 40 6 BICSE-7 25 25 3 RIMMS Building 90 90 4 BSCS-3 40 40 5 BEE-4 50 50 11 255 255 1 SEECS UG Block 100 100 1 BIT-10 35 35 3 BSCS-2 35 35 7 BESE-3 30 30 2 IAEC Building 70 65 4 BSCS-3 40 40 6 BICSE-7 25 25 3 RIMMS Building 90 90 2 BIT-11 40 40 5 BEE-4 50 50 12 255 255 1 SEECS UG Block 100 95 1 BIT-10 35 35 3 BSCS-2 35 35 6 BICSE-7 25 25 2 IAEC Building 70 70 2 BIT-11 40 40 7 BESE-3 30 30 3 RIMMS Building 90 90 4 BSCS-3 40 40 5 BEE-4 50 50 13 255 255 1 SEECS UG Block 100 95 1 BIT-10 35 35 3 BSCS-2 35 35 6 BICSE-7 25 25 2 IAEC Building 70 70 4 BSCS-3 40 40 7 BESE-3 30 30 3 RIMMS Building 90 90 2 BIT-11 40 40 5 BEE-4 50 50 14 255 255 1 SEECS UG Block 100 100 1 BIT-10 35 35 2 BIT-11 40 40 6 BICSE-7 25 25 2 IAEC Building 70 65 3 BSCS-2 35 35 7 BESE-3 30 30 3 RIMMS Building 90 90 4 BSCS-3 40 40 5 BEE-4 50 50 98 rows selected. Elapsed: 00:00:01.42Published by: BrendanP on January 20, 2013 11:25
I found the need to deduplicate regular expression:AND RegExp_Instr (r_c.itm_path | r_i.path, ',(\d+),.*?,\1,') = 0)
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Hi all
I have two record values described.
Select 1 as seq_no, "test1" as double data_set
Union
Select 1, '12test' from dual
Union
Select 2, 'abcd' from dual
Union
Select 2, 'ilm' from dual
Union
Select 2, '12test444' from dual
Here, I need to extract the o/p from the DataSet above below
1, 12test
1, test1
2, 12test444
Select 1 as seq_no, "test1" as double data_set
Union
Select 1, '12test' from dual
Union
Select 2, 'abcd' from dual
Union
Select 2, 'ilm' from dual
Here, I need to extract the o/p from the DataSet above below
1, 12test
1, test1
2, abcd
2, ilm
The logic behind both above the o/p is
If the search string 'test' is present in any data set (data set is defined by the same seq_no values), and then print the entire line containing the search string. As in the first data set of record value with seq_val 1 and 2 have two test string 'test', o/p is all the lines contating the search string 'test '.
If the search string 'test' is not present in any data set (data set is defined by the same values seq_no) then print all lines not containing not the chain sought. As in the second set of data record value with seq_val 2 at once is not test string 'test', o/p is all lines not contating the search string 'test' in the test with 1as and all seq_val data the lines contating the search string 'test' in a set of data with seq_val 2
Hope I could explain my logic o/p and I need to do in a sql query.
Waiting for your answers.
Thanks in advance.Hello
Here's one way:
WITH got_rnk AS ( SELECT seq_no, data_set , DENSE_RANK () OVER ( PARTITION BY seq_no ORDER BY CASE WHEN INSTR (data_set, 'test') > 0 THEN 1 ELSE 2 END ) AS rnk FROM table_x ) SELECT seq_no, data_set FROM got_rnk WHERE rnk = 1 ;The CASE expression returns 1 if data_set contains 'test', and it returns 2 if it is not. We will call this number x.
You are not really interested in x itself. On the contrary, you want all the lines that have the lowest value of x between the lines with the same seq_no. In other words, if a row with a given_seq_num x = 1, then you want all the lines with this seq_no and x = 1, but it the lowest x for a seq_no is 2, then you want the lines with x = 2 instead.
DENSE_RANK returns the number 1, if the riow data is the more low (or tied for the lowest) in its partition, so if him are all rows with x = 1 in the group, rnk = 1 corresponds to x = 1. But if all lines with the same seq_no have x = 2, then rnk = 1 corresponds to x = 2. -
Find the name of the view based on a sql query
Can anyone suggest me how to find the name of the view based on a sql query? When I try to the following select statement:
but I get this error:select view_name from user_views where text like '%SELECT * from TABLE%';
But as I notice that the TEXT column is really LONG to type... Are there any other table system that store information about the text using each view?SQL Error: ORA-00932: inconsistent datatypes: expected NUMBER got LONG 00932. 00000 - "inconsistent datatypes: expected %s got %s"See this example, using DBMS_METADATA. GET_DDL:
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How to use the multiple selection list values in sql query
Hi all
In the search form, I have a multiselect llist tell (P3_STATUS) and I want to have a query using this element and fetch documents
How to do a sql query based on the selection of the value of this element.
Please could someone help me on this?SELECT "W"."START_DT" "Start Date", "W"."CAMPAIGN_CODE" "Campaign Name", "W"."MKT_CHANNEL" "Channel", "W"."MKT_SUB_CHANNEL" "Sub Channel", "W"."PROMO_CODE" "Promo Code", "W"."TRACKING_CODE" "Tracking Code", "W"."TFN" "TFN", "W"."STATUS" "Status", "W"."CAMPAIGN_CODE" "Edit" FROM "WC_MKT_CAMPAIGN_DS" "W" WHERE (MKT_CHANNEL = decode(:P1_CHANNEL,'%null%',MKT_CHANNEL,NULL,MKT_CHANNEL,:P1_CHANNEL)) AND (MKT_SUB_CHANNEL= decode(:P1_SUB_CHANNEL,'%null%',MKT_SUB_CHANNEL,NULL,MKT_SUB_CHANNEL,:P1_SUB_CHANNEL)) AND *STATUS = decode(.................*
Thanks in advance
Robert LTry to change your selection of status to
AND INSTR(':'||:P3_STATUS||':',':'||STATUS||':') > 0 -
Failed to parse the SQL query by user
Hi all
in my application, I have a piece of text with a button "submit". In this article, I type a name and a report after the element region show me the result (s). This works for all my users (> 2000) perfectly, but that users become an error in the area of the report:
Failed to parse the SQL query:
ORA-01403: no data found
We try this with the same searchstring on the same computer/browser. If I connected the result is ok, if the logged on user, the error message appears. If I try this on the user's computer with me connected, ok result. If the user try this on another pc, error results.
I have a production and a developer workspace. In the developer workspace the user can try this perfectly without errors. That in the space of productive work, clear the error.
The SQL-Select in the ist verry simple reprirt:
Select id
name
raum
table
where instr (upper (name), upper (:P60_SEARCH)) > 0
However, all users can use this search box with report perfectly, only this one user has the error. There is no restrictions on this point or a State.
Can someone help me?Hi Carsten,
I don't think that there is a way to do it.
Could you please mark the correct and useful answers in this thread? Otherwise I'll never get near Andy ;) :)
Denes Kubicek
------------------------------------------------------------------------------
http://deneskubicek.blogspot.com/
http://www.Opal-consulting.de/training
http://Apex.Oracle.com/pls/OTN/f?p=31517:1
------------------------------------------------------------------------------ -
Hi all
I think I'm really stupid but I have problems of SQL query.
I have a table with many lines of operations which all refer to different stores, for example:
Store... _Sales_... _Month_
Reading... 200 k... April
Leeds... 50k................ April
Manchester... 70k................ May
Reading... 100 k... May
I need to arrive at the average sales for the month combined as a total. That is the average income of store for a given period is 200 k + 50 k + 70 k + 100 k / * 3 * (because there are only 3 unique shops) - I hope this makes sense!
So, basically, I'm doing both a query of the SUM (and company store) and then out of the average of all the stores together. Is this possible?
Thank youHello
This query returns 140 which seems to be the correct value:
with data as ( select 'Reading' Store , 200 sales from dual union select 'Leeds' , 50 from dual union select 'Manchester' , 70 from dual union select 'Reading' , 100 from dual ) select sum(sales) , count( distinct store ) , sum(sales)/count(distinct store) from dataThere are several options:
1. you can get two IR on one page (using IFRAMEs - search for that word on the Forum)...
2. you create another region with the above query and that position just below the report
3. in the foot of the region call a process of PL/SQL (using AJAX) that calculates the value using the query and print it (via htp.p)Greetings,
Roelhttp://roelhartman.blogspot.com/
http://www.bloggingaboutoracle.org/
http://www.Logica.com/You can assign this answer to your question in marking it as useful or Correct ;-)
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How to measure the performance of the sql query?
Hi Experts,
How to measure the cost of performance, efficiency and CPU of an sql query?
What are all the measures available to a sql query?
How to identify the optimal query writing?
I use Oracle 9i...
It'll be useful for me to write the effective query...
Thanks and greetingsPSRAM wrote:
Could you tell me how to activate the PLUSTRACE role?First put on when you do a search on PLUSTRACE: http://forums.oracle.com/forums/search.jspa?threadID=&q=plustrace&objID=f75&dateRange=all&numResults=15&rankBy=10001
Kind regards
Rob. -
Œuvres SQL query to MS SQL Server 2008, but not when you use the database kit
I have this SQL query:
DECLARE TABLE (@DataTypeTable)
Name varchar (128).
TypeID INT)-Add comma delimeted type data in the temporary table names
INSERT INTO @DataTypeTable (name)
SELECT * from WhatWeShouldDoRead.func_Split (@DataTypeTrimmed, ',')SELECT the name OF @DataTypeTable
That takes a comma delimited by the string and returns the string as a table. It works correctly in Microsoft SQL Server Management Studio. When I run this as a stored procedure I return nothing. There are no errors, SQL or otherwise. I checked that I am connected to the correct database and the stored procedure is responsible without changing any error chain which is reported of this stored procedure (that code is not shown in the example above). Has anyone seen this problem before, or have experience with SQL/Labview interfaces to tell me what I am doing wrong?
Thanks in advance.
-
HI HI... I'm a student doing a project related to labview. My task is to create a vi, type a user name and password to continue the whole VI.
As I am a newbie to SQL query language, can anyone help me this?... This isn't like the VI with password lock
There is a connection of the user called button in my main façade... u by clicking on it, a pop-up window will come out asking you a user name and a password. If the user name and the password is correct, then you can proceed. The problem is that I'm stuck with database...
Help me pls!with respect,
Ray
Hello
You have two cases:
(1) connect to the database with string (link a string of connection information), and then type something like this:
Driver is SQL Native Client;. Server = IP. Add.re.SS; UID = username; PW = *** ; Database = MyDatabase (depends on your database)(2) use a UDL file (you can configure it to connect to your database, with specific format). Remember that the connection is successful with test button.
There is a UDL file that you can edit here: C:\Program NIUninstaller Instruments\LabVIEW 2010\examples\database\Labview.udl
Edit: The connection dropped, you can set the path to an mdb file, and I think you can give the path of your accdb file.
Kind regards
-
I am trying to execute the following SQl query, SELECT * failure WHERE ID = '123 ' AND RepairAction =' '; using the DB tools run Query.vi. This query never find record in my database. My database contains a record where the registered ID contains the value '123' and the RepairAction field is an empty string. If I remove the declaration 'AND RepairAction' ';' my query text, the record is found. I think my problem is that I do not use the correct syntax to describe and an empty string. I tried the following: "," ",""," "and NULL as empty and none of these work string arguments.
I was hoping someone might be able to tell me what the correct syntax is an empty string or if there is another approach that I take.
Thanks in advance for your help,
Jim
Jim,
Just to be sure, have you used 'is' instead of '=' in this command?
This makes all the difference in this command.
Cerati
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