Search Results

Search found 63924 results on 2557 pages for 'mysql error 1050'.

Page 29/2557 | < Previous Page | 25 26 27 28 29 30 31 32 33 34 35 36  | Next Page >

  • [MYSQL] Select users who own both a dog and a cat

    - by matte
    Hi, I have this sample table: CREATE TABLE `dummy` ( `id` int(11) NOT NULL AUTO_INCREMENT, `userId` int(11) NOT NULL, `pet` varchar(50) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ; INSERT INTO `dummy` (`id`, `userId`, `pet`) VALUES(1, 1, 'dog'); INSERT INTO `dummy` (`id`, `userId`, `pet`) VALUES(2, 1, 'cat'); INSERT INTO `dummy` (`id`, `userId`, `pet`) VALUES(3, 2, 'dog'); INSERT INTO `dummy` (`id`, `userId`, `pet`) VALUES(4, 2, 'cat'); INSERT INTO `dummy` (`id`, `userId`, `pet`) VALUES(5, 3, 'cat'); INSERT INTO `dummy` (`id`, `userId`, `pet`) VALUES(6, 4, 'dog'); How can I write the statements below in mysql: Retrieve all users who own both a dog and a cat Retrieve all users who own a dog or a cat Retrieve all users who own only a cat Retrieve all users who doesn't own a cat Thanks!

    Read the article

  • using check contraint in MySQL for controlling string length not working

    - by ptrn
    Dear stackoverflow, I'm tumbled with a problem! I've set up my first check constraint using MySQL, but unfortunately I'm having a problem. When inserting a row that should fail the test, the row is inserted anyway. The structure: CREATE TABLE user ( id INT UNSIGNED NOT NULL AUTO_INCREMENT, uname VARCHAR(10) NOT NULL, fname VARCHAR(50) NOT NULL, lname VARCHAR(50) NOT NULL, mail VARCHAR(50) NOT NULL, PRIMARY KEY (id), CHECK (LENGTH(fname) > 3) ); The insert statement: INSERT INTO user VALUES (null, 'user', 'Fname', 'Lname', '[email protected]'); I'm pretty sure I'm missing something basic here.

    Read the article

  • MySQL - Sort on a calculated value based on two dates

    - by Petter Magnusson
    I have the following problem that needs to be solved in a MySQL query: Fields info - textfield date1 - a date field date2 - a date field offset1 - a text field with a number in the first two positions, example "10-High" offset2 - a text field with a number in the first two positions, example "10-High" I need to sort the records by the calculated "sortvalue" based on the current date (today): If today=date2 then sortvalue=offset1*10+offset2*5+1000 else sortvalue=offset1*10+offset2*5 I have quite good understanding of basic SQL with joins etc, but this I am not even sure if its possible...if it helps I could perhaps live with a single formula giving the same sort of effect as the IFs do....ie. before date1 = low value, after date2 = high value... Rgds PM

    Read the article

  • Searching a table MySQL & PHP.

    - by S1syphus
    I want to be able to search through a MySQL table using values from a search string, from the url and display the results as an XML output. I think I have got the formatting and declaring the variables from the search string down. The issue I have is searching the entire table, I've looked over SO for previous answers, and they all seem to have to declare each column in the table to search through. So for example my database layout is as follows: **filesindex** -filename -creation -length -wall -playlocation First of all would the following be appropriate: $query = "SELECT * FROM filesindex WHERE filename LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE creation LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE length LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE wall LIKE '".$searchterm."%' UNION SELECT * FROM filesindex WHERE location LIKE '".$searchterm."%'"; Or ideally, is there an easier way that involves less hardcoding to search a table. Any ideas? Thanks

    Read the article

  • Reversing column values in mysql command line

    - by user94154
    I have a table posts with the column published, which is either 0 (unpublished) or 1 (published). Say I want to make all the published posts into unpublished posts and all the unpublished posts into published posts. I know that running UPDATE posts SET published = 1 WHERE published = 0; UPDATE posts SET published = 0 WHERE published = 1; will end up turning all my posts into published posts. How can I run these queries in the mysql command line so that it truly "reverse" the values, as opposed to the mistake outlined above? Thanks

    Read the article

  • MySQL: Insert row on table2 if row in table1 exists

    - by Andrew M
    I'm trying to set up a MySQL query that will insert a row into table2 if a row in table1 exist already, otherwise it will just insert the row into table1. I need to find a way to adapt the following query into inserting a row into table2 with the existing row's id. INSERT INTO table1 (host, path) VALUES ('youtube.com', '/watch') IF NOT EXISTS ( SELECT * FROM table1 WHERE host='youtube.com' AND path='/watch' LIMIT 1); Something kind of like this: INSERT ... IF NOT EXISTS(..) ELSE INSERT INTO table2 (table1_id) VALUES(row.id); Except I don't know the syntax for this.

    Read the article

  • insert array to mysql db function

    - by ganjan
    Hi. I have an array where the keys represent each column in my database. Now I want a function that makes a mysql update query. Something like $db['money'] = $money_input + $money_db; $db['location'] = $location $query = 'UPDATE tbl_user SET '; for($x = 0; $x < count($db); $x++ ){ $query .= $db something ".=." $db something } $query .= "WHERE username=".$username." ";

    Read the article

  • MySQL Query like not returning correct results

    - by Herr Kaleun
    Hello friends, i've a MySQL query that should return some rows that have the letters Ö or Ü in it but it actually does not. The query code is this: $this->db->like('title', $text ); It's PHP CodeIgniter active query. Lets assume we have 2 rows. 1. Büm 2. Bom if i search for Bü, the 1. row has to be returned but it does not. When i search for Bo the second row gets returned successfully and when i search for B both rows are returned. How could i fix this? What may be the underlieng cause? Thanks for reading.

    Read the article

  • MySQL script to delete data in chunks until everything lower then id has been deleted

    - by Chriswede
    I need an MySQL Skript which does the following: delete chunks of the database until it has deleted all link_id's greater then 10000 exmaple: x = 10000 DELETE FROM pligg_links WHERE link_id > x and link_id < x+10000 x = x + 10000 ... So it would delete DELETE FROM pligg_links WHERE link_id > 10000 and link_id < 20000 then DELETE FROM pligg_links WHERE link_id > 20000 and link_id < 30000 until all id's less then 10000 have been removed I need this because the database is very very big (more then a gig) thank in advance

    Read the article

  • MySQL: filling empty fields with zeroes when using GROUP BY

    - by SaltLake
    I've got MySQL table CREATE TABLE cms_webstat ( ID int NOT NULL auto_increment PRIMARY KEY, TIMESTAMP_X timestamp DEFAULT CURRENT_TIMESTAMP, # ... some other fields ... ) which contains statistics about site visitors. For getting visits per hour I use SELECT hour(TIMESTAMP_X) as HOUR , count(*) AS HOUR_STAT FROM cms_webstat GROUP BY HOUR ORDER BY HOUR DESC which gives me | HOUR | HOUR_STAT | | 24 | 15 | | 23 | 12 | | 22 | 9 | | 20 | 3 | | 18 | 2 | | 15 | 1 | | 12 | 3 | | 9 | 1 | | 3 | 5 | | 2 | 7 | | 1 | 9 | | 0 | 12 | And I'd like to get following: | HOUR | HOUR_STAT | | 24 | 15 | | 23 | 12 | | 22 | 9 | | 21 | 0 | | 20 | 3 | | 19 | 0 | | 18 | 2 | | 17 | 0 | | 16 | 0 | | 15 | 1 | | 14 | 0 | | 13 | 0 | | 12 | 3 | | 11 | 0 | | 10 | 0 | | 9 | 1 | | 8 | 0 | | 7 | 0 | | 6 | 0 | | 5 | 0 | | 4 | 0 | | 3 | 5 | | 2 | 7 | | 1 | 9 | | 0 | 12 | How should I modify the query to get such result? Thanks.

    Read the article

  • MYSQL Inserting rows that reference main rows.

    - by Andrew M
    I'm transferring my access logs into a database. I've got two tables: urlRequests id : int(10) host : varchar(100) path: varchar(300) unique index (host, path) urlAccesses id : int(10) request : int(10) <-- reference to urlRequests row ip : int(4) query : varchar(300) time : timestamp I need to insert a row into urlAccesses for every page load, but first a row in urlRequests has to exist with the requested host and path so that urlAccesses's row can reference it. I know I can do it this way: A. check if a row exists in urlRequests B. insert a row in urlRequests if it needs it C. insert a row into urlAccesses with the urlRequests's row id referenced That's three queries for every page load if the urlRequests row doesn't exist. I'm very new to MySQL, so I'm guessing that there's a way to go about this that would be faster and use less queries.

    Read the article

  • Mysql on duplicate key update + sub query

    - by jwzk
    Using the answer from this question: http://stackoverflow.com/questions/662877/need-mysql-insert-select-query-for-tables-with-millions-of-records new_table * date * record_id (pk) * data_field INSERT INTO new_table (date,record_id,data_field) SELECT date, record_id, data_field FROM old_table ON DUPLICATE KEY UPDATE date=old_table.data, data_field=old_table.data_field; I need this to work with a group by and join.. so to edit: INSERT INTO new_table (date,record_id,data_field,value) SELECT date, record_id, data_field, SUM(other_table.value) as value FROM old_table JOIN other_table USING(record_id) ON DUPLICATE KEY UPDATE date=old_table.data, data_field=old_table.data_field, value = value; I can't seem to get the value updated. If I specify old_table.value I get a not defined in field list error.

    Read the article

  • Storing HTML in MySQL using Java

    - by mpcabd
    Hello there again, So, I'm working on a project now where I should store webpages inside a database, I'm using crawler4j to crawl and Proxool along with MySQL Java Connector to connect to my database. When I tested the application I got: com.mysql.jdbc.MysqlDataTruncation: Data truncation: Data too long for column 'HTMLData'. The HTMLData column wasTEXT. When I changed the HTMLData column to LONGTEXT the error was gone, but I'm afraid it might get back in the future. Any idea on how to do that perfectly so I don't worry about that error (or any other similar error) in the future? Thanks :)

    Read the article

  • MySQL Query still executing after a day..?

    - by Matt Jarvis
    Hi - I'm trying to isolate duplicates in a 500MB database and have tried two ways to do it. One creating a new table and grouping: CREATE TABLE test_table as SELECT * FROM items WHERE 1 GROUP BY title; But it's been running for an hour and in MySQL Admin it says the status is Locked. The other way I tried was to delete duplicates with this: DELETE bad_rows.* from items as bad_rows inner join ( select post_title, MIN(id) as min_id from items group by title having count(*) 1 ) as good_rows on good_rows.post_title = bad_rows.post_title; ..and this has been running for 24hours now, Admin telling me it's Sending data... Do you think either or these queries are actually still running? How can I find out if it's hung? (with Apple OS X 10.5.7)

    Read the article

  • MySQL temp table issue

    - by AmyD
    Hi folks! I'm trying to use temp tables to speed up my MySQL 4.1.22-standard database and what seems like a simple operation is causing me all kinds of issues. My code is below.... CREATE TEMPORARY TABLE nonDerivativeTransaction_temp (accession_number varchar(30), transactionDateValue date)) TYPE=HEAP; INSERT INTO nonDerivativeTransaction_temp VALUES( SELECT accession_number, transactionDateValue FROM nonDerivativeTransaction WHERE transactionDateValue = "2010-06-15"); SELECT * FROM nonDerivativeTransaction_temp; The original table (nonDerivativeTransaction) has two fields, accession_number (varchar(30)) and transactionDateValue (date). Apparently I am getting an issue with the first two statements but I can't seem to nail down what it is. Any help would be appreciated. Amy D.

    Read the article

  • Match two mysql cols on alpha chars (ignoring numbers in same field)

    - by Steve
    I was wondering if you know of a way I could filter a mysql query to only show the ‘alpha’ characters from a specific field So something like SELECT col1, col2, **alpha_chars_only(col3)** FROM table I am not looking to update only select. I have been looking at some regex but without much luck most of what turned up was searching for fields that only contain ‘alpha’ chars. In a much watered down context... I have col1 which contains abc and col two contains abc123 and I want to match them on alpha chars only. There can be any number of letters or numbers. Any help very much wel come

    Read the article

  • Mysql Query - Order By Not Working

    - by jwzk
    I'm running Mysql 5.0.77 and I'm pretty sure this query should work? SELECT * FROM purchases WHERE time_purchased BETWEEN '2010-04-15 00:00:00' AND '2010-04-18 23:59:59' ORDER BY time_purchased ASC, order_total DESC time_purchased is DATETIME, and an index. order_total is DECIMAL(10,2), and not an index. I want to order all purchases by the date (least to greatest), and then by the order total (greatest to least). So I would output similar to: 2010-04-15 $100 2010-04-15 $80 2010-04-15 $20 2010-04-16 $170 2010-04-16 $45 2010-04-16 $15 2010-04-17 $274 .. and so on. The output I am getting from that query has the dates in order correctly, but it doesn't appear to sort the order total column at all. Thoughts? Thanks.

    Read the article

  • Getting mysql syntax error and cant find source

    - by eGGzy
    I have function that updates log table. function wslog($userID, $log, $where) { safe_query("INSERT INTO ".PREFIX."log ( time, userID, log, where ) values( '".time()."', '".$userID."', '".$log."', '".$where."' ) "); } And I have this php code: wslog($userID, 'server|'.mysql_insert_id().'', 'servers'); But I keep getting syntax error: Query failed: errorno=1064 error=You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'where ) values( '1269208030', '1', 'server|14', 'servers' )' at line 1 query=INSERT INTO ws_DII_log ( time, userID, log, where ) values( '1269208030', '1', 'server|14', 'servers' )

    Read the article

  • MySQL query problem

    - by SaltLake
    I've got MySQL table CREATE TABLE stat ( ID int NOT NULL auto_increment PRIMARY KEY, TIMESTAMP_X timestamp DEFAULT CURRENT_TIMESTAMP, # ... some other fields ... ) which contains statistics about site visitors. For getting visits per hour I use SELECT hour(TIMESTAMP_X) as HOUR , count(*) AS HOUR_STAT FROM cms_webstat GROUP BY HOUR ORDER BY HOUR DESC which gives me | HOUR | HOUR_STAT | | 24 | 15 | | 23 | 12 | | 22 | 9 | | 20 | 3 | | 18 | 2 | | 15 | 1 | | 12 | 3 | | 9 | 1 | | 3 | 5 | | 2 | 7 | | 1 | 9 | | 0 | 12 | And I'd like to get following: | HOUR | HOUR_STAT | | 24 | 15 | | 23 | 12 | | 22 | 9 | | 21 | 0 | | 20 | 3 | | 19 | 0 | | 18 | 2 | | 17 | 0 | | 16 | 0 | | 15 | 1 | | 14 | 0 | | 13 | 0 | | 12 | 3 | | 11 | 0 | | 10 | 0 | | 9 | 1 | | 8 | 0 | | 7 | 0 | | 6 | 0 | | 5 | 0 | | 4 | 0 | | 3 | 5 | | 2 | 7 | | 1 | 9 | | 0 | 12 | How should I modify the query to get such result? Thanks.

    Read the article

  • PHP MySQL Insert Data

    - by happyCoding25
    Hello, Im trying to insert data into a table in MySQL. I found/modified some code from w3Schools and still couldn't get it working. Heres what I have so far: <?php $rusername=$_POST['username']; $rname=$_POST['name']; $remail=$_POST['emailadr']; $rpassword=$_POST['pass']; $rconfirmpassword=$_POST['cpass']; if ($rpassword==$rconfirmpassword) { $con = mysql_connect("host","user","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("mydbname ", $con); } mysql_query("INSERT INTO members (id, username, password) VALUES ('4', $rusername, $rpassword)"); ?> Did I mistype something? To my understanding "members" is the name of the table. If anyone knows whats wrong I appreciate the help. Thanks

    Read the article

  • Optimize MySQL database query

    - by rajeeesh
    I had a commenting application in my web site. The comments will store in a MySQL table . table structure as follows id | Comment | user | created_date ------------------------------------------------------ 12 | comment he | 1245 | 2012-03-30 12:15:00 ------------------------------------------------------ I need to run a query for listing all the comments after a specific time. ie .. a query like this SELECT * FROM comments WHERE created_date > "2012-03-29 12:15:00" ORDER BY created_date DESC Its working fine.. My question is if I got a 1-2 lakh entry in this table is this query is sufficient for the purpose ? or this query will take time to execute ? In most cases I have to show last 2 days data + periodically ( interval of 10 mins ) checking for updates with ajax from this table ... Please help Thanks

    Read the article

  • mysql_close(): supplied argument is not a valid MySQL-Link resource

    - by maxedison
    I'm trying to get the hang of using custom session handlers to store session data in a MySQL database. However, I keep getting the following warning: mysql_close(): supplied argument is not a valid MySQL-Link resource Here's the code I'm using, which I got from here: function _open(){ global $_sess_db; $_sess_db = mysql_connect("localhost", "root", "******"); if ($_sess_db) { return mysql_select_db('style', $_sess_db); } return false; } function _close(){ global $_sess_db; return mysql_close($_sess_db); //error happens here } The full text of the error message ultimately points to the final "return mysql_close($_sess_db);" line. I can confirm that the mysql_connect info does in fact work, and I do have the rest of the session handler functions defined as well. And in case it helps, I get these errors immediately upon page load, without actually calling any of the session handler functions, and without having any current sessions open.

    Read the article

  • Can't Connect To Local Mysql Using IP Address, but CAN connect from remote server

    - by user1782041
    Here's an interesting one that does not seem to fall into any of the mysql connection issues I've read about or searched for: On an Ubuntu 12.04 box I had some system updates waiting to install, and I took care of that this evening. After the install, I started seeing some errors in my syslog complaining about a particular php script that could no longer connect to the mysql instance on the box. Here is the specific error: PHP Warning: mysql_connect(): Can't connect to MySQL server on '192.168.0.40' (4) Now, the server's IP address is 192.168.0.40, and I've checked to make sure that I have mysql listening on 0.0.0.0 so that I can connect using either "localhost" or "192.168.0.40". Here's where things get odd: From the local machine, if I try the following: mysql -uroot -p -h192.168.0.40 I get this error: ERROR 2003 (HY000): Can't connect to MySQL server on '192.168.0.40' (110) I've checked, and error 110 indicates an OS timeout, and error 2003 is the mysql generic "can't connect" error. This indicates that it is not permissions with the user. However, if I do the same thing from a remote machine (say, from 192.168.0.30), I log right in with no problems. Futher, other scripts on the local machine that connect to mysql using "localhost" for the host rather than "192.168.0.40" connect with no problems. Also, I can connect via the mysql socket with no problems both from the command line and php scripts. So, this feels like a networking issue of some kind on the local box, but there are no iptables rules on this box (it is firewalled externally) and I can't figure out what else may be causing this. This problematic script worked perfectly prior to the latest system update. For now, I'll simply change the script to connect via localhost, but I'd really like to know why it broke for 2 reasons: There may be other scripts that connect using 192.168.0.40 that don't run very often which are now broken. Auditing them all will take more time than I feel like devoting at the moment. I'm curious, and want to know why it broke so I can fix it correctly. Any help?

    Read the article

  • mysql command line not working

    - by Sandeepan Nath
    I have mysql running in my fedora system. I have xampp setup on the system and php projects present in the webspace are working fine. PhpMyAdmin is working fine. echoing phpinfo() in a PHP script also shows mysql enabled. But running mysql connect command mysql -u[username] -p[password] Gives this - bash: mysql: command not found How do I fix that? Any pointers? I guess I need to do some pointing (define some path in some file) so that my system knows that mysql is installed. What exactly do I have to do? Additional Details This system was someone else's and he is not available here. May be PHP/Mysql was setup already in the system. I just freshly extracted xampp for linux into /opt/lampp/ and have put all the above mentioned things (PHP projects and PhpMyAdmin) there. After doing that I had a socket problem (PhpMyAdmin was not working and showing this)- #2002 - The server is not responding (or the local MySQL server's socket is not correctly configured) I restarted lampp using ./lampp restart but problem remained. Then after turning on system today, I started lampp and everything worked just fine. No project issues anymore only command line Mysql not working

    Read the article

  • MySQL partition "full"?

    - by gdea73
    I have a server that runs Debian 6.2, with Apache, PHP5, and MySQL. Well, I hadn't done anything with MySQL at all so far, just Apache and PHP; I must have installed it (mysql-server) at some point along the line, and I decided to login to the database for the first time a couple days ago as I was considering using the database for a future website project. I noticed that the "root" user had a password, and I didn't recall having set one. My usual root password was incorrect. So I attempted to reset the password. sudo service mysql stop (stopped successfully) sudo /usr/bin/mysqld_safe --skip-grant-tables --skip-networking & started successfully, from what I can tell. However, mysql itself returns "Can't connect to local MySQL server through socket '/var/run/mysqld/mysqld,sock' (2)", and additionally sudo service mysql start returns "/etc/init.d/mysql: ERROR: The partition with /var/lib/mysql is too full! ... failed!" df -h tells me that / is 26% used, a 20GB partition, and /home, roughly 900GB, has only 5% usage. On a potentially related note, I've been experiencing random hangs since I noticed this problem, my tty2 randomly froze several times while idle, and the entire system is suddenly unstable. gnome-terminal also does not open. (Gnome-terminal apparently works now, disregard that part, but the server is still being somewhat unstable, I randomly lost connection when I was SSHed into it from my laptop, twice now.)

    Read the article

< Previous Page | 25 26 27 28 29 30 31 32 33 34 35 36  | Next Page >