Query. Urgent!
Hi all, I have installed MySql 4.1 in my system. when I tried to run Mysql , I always gets this error. Could not start MySql service on a local computer Error 1067: the process terminated unexpectedly. Any help is highly appreciated. Regards, Renish -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Query...Urgent
Hi all, I have installed MySql 4.1 in my system. when I tried to run Mysql , I always gets this error. Could not start MySql service on a local computer Error 1067: the process terminated unexpectedly. Any help is highly appreciated. Regards, Renish -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
A difficult query- urgent for me
Dear Friends, I have a problm, try to solve that. Actually there is a table with columns a and b . So i want if a contains a particular word than a's value should return else 'b' value should return. And there must be one and only one column returning. I have mysql 4.x and i think the logic will be like. I know i am wrong by syntax but ... SELECT IF(a REGEXP CONCAT('word' , '$'),a,b) from table_name where a = 'anything'; I think i am clear if not do please ask me questions. I shall be very grateful if any one of you can give me a solutions. -- Regards Abhishek jain. mail2web - Check your email from the web at http://mail2web.com/ . -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: A difficult query- urgent for me
On 1/15/06, [EMAIL PROTECTED] wrote: Actually there is a table with columns a and b . So i want if a contains a particular word than a's value should return else 'b' value should return. SELECT CASE WHEN a = 'Good' THEN a ELSE b END FROM table Jochem
Re: A difficult query- urgent for me
An example of the _data_ would be very helpful here. For example, let's say this is your data: Col_ACol_B ----- aceexpert doghound hungryravenous If you are searching for the word ace, you should find it. You then want the SQL to return ace, right? If you are searching for the word puppy, you won't find it in the first column. Which word from the second column do you want to return??? There is no obvious reason to prefer any of the different values in the second column when the search word does not appear in the first column. Or do you only have a single row in this table? If so, I'm not sure why you want to create a table just to contain these two values; it might be justified, depending on what you are doing, but it seems unlikely. Can you clarify what you are trying to accomplish? Otherwise, it's going to be hard to help you. Also, which version of MySQL are you on, 4.0.x or 4.1.x? It might make a big difference to the answer I would give since 4.0.x does not support subqueries while 4.1.x does. Rhino - Original Message - From: [EMAIL PROTECTED] To: mysql@lists.mysql.com Sent: Sunday, January 15, 2006 10:21 AM Subject: A difficult query- urgent for me Dear Friends, I have a problm, try to solve that. Actually there is a table with columns a and b . So i want if a contains a particular word than a's value should return else 'b' value should return. And there must be one and only one column returning. I have mysql 4.x and i think the logic will be like. I know i am wrong by syntax but ... SELECT IF(a REGEXP CONCAT('word' , '$'),a,b) from table_name where a = 'anything'; I think i am clear if not do please ask me questions. I shall be very grateful if any one of you can give me a solutions. -- Regards Abhishek jain. mail2web - Check your email from the web at http://mail2web.com/ . -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED] -- No virus found in this incoming message. Checked by AVG Free Edition. Version: 7.1.371 / Virus Database: 267.14.18/230 - Release Date: 14/01/2006 -- No virus found in this outgoing message. Checked by AVG Free Edition. Version: 7.1.371 / Virus Database: 267.14.18/230 - Release Date: 14/01/2006 -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
mysql query urgent!!!
i, I have to write simple mysql script. I need folowing things: Number of people logged in previous 0 to 24 hours Number of people logged in previous 24.5 to 48 hours Number of people logged in previous 48.5 to 72 hours Number of people logged in previous 72.5 hours to 7 days Number of people logged in previous 7.5 to 14 days Number of people logged in previous 14.5 days onwards My tables is as follows: mysql select * from User limit 1; +--+---+-+--+ | partner_name | user_name | last_login | mailbox_size | +--+---+-+--+ | foo.net | edward| 2001-06-02 09:37:41 | 229099 | +--+---+-+--+ 1 row in set (0.62 sec) I also want to save the output to comma delimited file. Cheers kapil _ Do You Yahoo!? Get your free @yahoo.com address at http://mail.yahoo.com - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
Re: mysql query urgent!!!
i, I have to write simple mysql script. I need folowing things: Number of people logged in previous 0 to 24 hours Number of people logged in previous 24.5 to 48 hours Number of people logged in previous 48.5 to 72 hours Number of people logged in previous 72.5 hours to 7 days Number of people logged in previous 7.5 to 14 days Number of people logged in previous 14.5 days onwards My tables is as follows: mysql select * from User limit 1; +--+---+-+--+ | partner_name | user_name | last_login | mailbox_size | +--+---+-+--+ | foo.net | edward| 2001-06-02 09:37:41 | 229099 | +--+---+-+--+ 1 row in set (0.62 sec) I also want to save the output to comma delimited file. Cheers kapil Sir, redefine the last_login column as a timestamp and try the following; CREATE TEMPORARY TABLE time_groups SELECT If( Unix_timestamp(CURRENT_TIMESTAMP) - Unix_timestamp(last_login) = 86400 , '01 day' , If( Unix_timestamp(CURRENT_TIMESTAMP) - Unix_timestamp(last_login) = 172800 , '02 days' , If( Unix_timestamp(CURRENT_TIMESTAMP) - Unix_timestamp(last_login) = 259200 , '03 days' , If( Unix_timestamp(CURRENT_TIMESTAMP) - Unix_timestamp(last_login) = 604800 , '07 days' , If( Unix_timestamp(CURRENT_TIMESTAMP) - Unix_timestamp(last_login) = 1209600 , '14 days' , '14 days' ) ) ) ) ) AS time_group FROM User; SELECT time_group, Count(*) FROM time_groups GROUP BY time_group; It works on my machine. The time_group values are chosen so that they will display in the proper order. Bob Hall Know thyself? Absurd direction! Bubbles bear no introspection. -Khushhal Khan Khatak MySQL list magic words: sql query database - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php