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  • mysql error dont understand what it is saying

    - by sea_1987
    Cannot add or update a child row: a foreign key constraint fails (`mydb`.`job_listing_has_employer_details`, CONSTRAINT `job_listing_has_employer_details_ibfk_2` FOREIGN KEY (`employer_details_id`) REFERENCES `employer_details` (`id`)) INSERT INTO `job_listing_has_employer_details` (`job_listing_id`, `employer_details_id`) VALUES (6, '5') What does this mean? The two ID's I am inserting into the table exsist.

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  • MySQL Joining three tables

    - by text
    I am doing a query with three tables, the problem was one table has many occurrences of id of another. sample data: users: id answers: id:1 user_answer :1 id:1 user_answer :2 id:1 user_answer :3 Questions: id:1 answers :answer description id:2 answers :answer description id:3 answers :answer description How can I get all user information and all answer and its description, I used GROUP by user.id but it only returns only one answer. I want to return something like this list all of users answer: Name Q1 Q2 USERNAME ans1,ans2 ans1,ans2 comma separated description of answer here

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  • Mysql help with view more button.

    - by WAC0020
    I am working on a widget that is a lot like twitters widget where there is a list of postings and a view more button. I can get it to work with using ID variables but I would like to sort by popular posts. Here is my mysq code: $sql = "SELECT id, title, category, icon_normal, status, description, views_monthly FROM posts WHERE views_monthly<=".$lastPost." AND status='1' ORDER BY views_monthly DESC LIMIT 9" So the problem that I am having is it shows the first 9 just fine. When it gets to the point where views_monthly = 0 then it just loads the same 9 post over again. How do it get it to switch to using ID when it reaches Views_monthly = 0 and load fresh posts?

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  • How do I mirror a MySQL database?

    - by user366133
    I'm running two load balanced servers for one website, and I'd like the databases to be synchronized. Queries may be run on either of the two servers because they are both production sites, so the replication can't just work one way. It doesn't have to be in real-time, just fairly accurate so people don't notice a difference when they get switched to a different server.

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  • Possibly simple PHP/MYSQL issue with retrieving and showing data

    - by envoys
    I have been racking my brains over this for a while now. Here is the data i have in the sql data base as an example: ID | TYPE | DATA 1 | TXT | TEST 2 | PHP | php 3 | JS | JAVASCRIPT That is just an example, there are multiple listing for TXT, PHP and JS throughout the table. What I want to do is retrive all the data and display it all into separate drop down/select boxes. Meaning, select box one would list all data with type TXT, select box two would list all data with type PHP and select box 3 would list all data with type JS. The only way I have came about doing this is doing individual sql queries for each different type. I know there is a way to do it all in 1 query and then display it the way I want to but I just can't seem to figure out how and I know its going to drive me nuts when someone helps and I see just how they did it. Thanks for the input.

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  • Why my mysql DISTINCT doesn't work ?

    - by belaz
    Hello, Why the two query below return duplicate member_id and not the third ? i need the second query to work with distinct. Anytime i run a GROUP BY, this query is incredibly slow and the resultset doesn't return the same value as distinct (the value is wrong). SELECT member_id, id FROM ( SELECT * FROM table1 ORDER BY created_at desc ) as u LIMIT 5 +-----------+--------+ | member_id | id | +-----------+--------+ | 11333 | 313095 | | 141831 | 313094 | | 141831 | 313093 | | 12013 | 313092 | | 60821 | 313091 | +-----------+--------+ SELECT distinct member_id, id FROM ( SELECT * FROM table1 ORDER BY created_at desc ) as u LIMIT 5 +-----------+--------+ | member_id | id | +-----------+--------+ | 11333 | 313095 | | 141831 | 313094 | | 141831 | 313093 | | 12013 | 313092 | | 60821 | 313091 | +-----------+--------+ SELECT distinct member_id FROM ( SELECT * FROM table1 ORDER BY created_at desc ) as u LIMIT 5 +-----------+ | member_id | +-----------+ | 11333 | | 141831 | | 12013 | | 60821 | | 64980 | +-----------+ my table sample CREATE TABLE `table1` ( `id` int(11) NOT NULL AUTO_INCREMENT, `member_id` int(11) NOT NULL, `s_type_id` int(11) NOT NULL, `created_at` datetime DEFAULT NULL, PRIMARY KEY (`id`), KEY `s_FI_1` (`member_id`), KEY `s_FI_2` (`s_type_id`) ) ENGINE=InnoDB AUTO_INCREMENT=313096 DEFAULT CHARSET=utf8;

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  • What's the error in my MySQL statement?

    - by Jim
    The following SQL statement has a syntax error according to phpMyAdmin, but I can't spot what it is. Any ideas? CREATE TABLE allocations( `student_uid` INT unsigned NOT NULL DEFAULT 0, `active` INT unsigned NOT NULL DEFAULT 1, `name` VARCHAR( 255 ) NOT NULL DEFAULT '', `internal_id` VARCHAR( 255 ) DEFAULT '', `tutor_uid` INT NOT NULL DEFAULT 0, `allocater_uid` INT unsigned NOT NULL DEFAULT 0, `time_created` INT NOT NULL DEFAULT 0, `remote_time` FLOAT NOT NULL DEFAULT 0, `next_lesson` VARCHAR NOT NULL DEFAULT -1, PRIMARY KEY ( student_uid ) );

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  • MySQL table with similar column info - HELP!!!

    - by George Garman
    I have a DB with a table that is named "victim". The form that dumps the info into the table has room for two victims and therefore there is vic1_fname, vic1_lname, vic2_fname, vic2_lname, etc.. (business name, person first, person last, address, city, state, zip) a "1" and "2" of each. Now I want to search the DB and locate listed victims. This is what I have so far: $result = mysql_query( "SELECT victim.* FROM victim WHERE vic1_business_name OR vic2_business_name LIKE '%$search_vic_business_name%' AND vic1_fname OR vic2_fname LIKE '%$search_vic_fname%' AND vic1_lname OR vic2_lname LIKE '%$search_vic_lname%' AND vic1_address OR vic2_address LIKE '%$search_vic_address%' AND vic1_city OR vic2_city LIKE '%$search_vic_city%' AND vic1_state OR vic2_state LIKE '%$search_vic_state%' AND vic1_dob OR vic2_dob LIKE '%$search_vic_dob%' "); <table width="960" style="border: groove;" border=".5"> <tr><th colspan=10>You search results are listed below:</th></tr> <tr> <th>Case Number</th> <th>Business Name</th> <th>First Name</th> <th>Last Name</th> <th>DOB / Age</th> <th>Address</th> <th>City</th> <th>State</th> </tr> <?php while($row = mysql_fetch_array($result)) { ?> <tr> <td align="center"><?php print $row['vic_business_name']; ?></td> <td align="center"><?php print $row['vic_fname']; ?></td> <td align="center"><?php print $row['vic_lname']; ?></td> <td align="center"><?php print $row['vic_dob']; ?></td> <td align="center"><?php print $row['vic_adress']; ?></td> <td align="center"><?php print $row['vic_city']; ?></td> <td align="center"><?php print $row['vic_state']; ?></td> </tr> <?php } ?> </table> The info did not display in the table until I changed the table to this: <tr> <td align="center"><?php print $row['vic1_business_name']; ?></td> <td align="center"><?php print $row['vic1_fname']; ?></td> <td align="center"><?php print $row['vic1_lname']; ?></td> <td align="center"><?php print $row['vic1_dob']; ?></td> <td align="center"><?php print $row['vic1_adress']; ?></td> <td align="center"><?php print $row['vic1_city']; ?></td> <td align="center"><?php print $row['vic1_state']; ?></td> </tr> <tr> <td align="center"><?php print $row['vic2_business_name']; ?></td> <td align="center"><?php print $row['vic2_fname']; ?></td> <td align="center"><?php print $row['vic2_lname']; ?></td> <td align="center"><?php print $row['vic2_dob']; ?></td> <td align="center"><?php print $row['vic2_adress']; ?></td> <td align="center"><?php print $row['vic2_city']; ?></td> <td align="center"><?php print $row['vic2_state']; ?></td> </tr> Now it displays both rows, even if its empty. It doesn't matter if the victim was listed originally as vic1 or vic2, i just want to know if they are a victim. I hope this makes sense. I can't get it to display the way I want, line-by-line, irregardless of whether you are vic1 or vic2.

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  • PHP mySQL query's and PHP Variables

    - by jon
    I'm trying to make an OO Login system for a project I'm working on, and am having trouble with inserting variables into the query strings. In the code below, if I replace "$TBL_NAME" with the actual table name it works. Why isn't $TBL_NAME translating to the value of $TBL_NAME? class UserDB { private $TBL_NAME = "users"; public static function CheckLogin($username, $password) { Database::Connect(); $username = stripslashes($username); $password = stripslashes($password); $username = mysql_real_escape_string($username); $password = mysql_real_escape_string($password); $sql="SELECT uid FROM $TBL_NAME WHERE username='$username' AND password='$password' "; $result =mysql_query($sql); $count=mysql_num_rows($result); if ($count==1) return true; else return false; } The Query is returning false.

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  • MYSQL: COUNT with GROUP BY, LEFT JOIN and WHERE clause doesn't return zero values

    - by Paul Norman
    Hi guys, thanks in advance for any help on this topic! I'm sure this has a very simply answer, but I can't seem to find it (not sure what to search on!). A standard count / group by query may look like this: SELECT COUNT(`t2`.`name`) FROM `table_1` `t1` LEFT JOIN `table_2` `t2` ON `t1`.`key_id` = `t2`.`key_id` GROUP BY `t1`.`any_col` and this works as expected, returning 0 if no rows are found. So does: SELECT COUNT(`t2`.`name`) FROM `table_1` `t1` LEFT JOIN `table_2` `t2` ON `t1`.`key_id` = `t2`.`key_id` WHERE `t1`.`another_column` = 123 However: SELECT COUNT(`t2`.`name`) FROM `table_1` `t1` LEFT JOIN `table_2` `t2` ON `t1`.`key_id` = `t2`.`key_id` WHERE `t1`.`another_column` = 123 GROUP BY `t1`.`any_col` only works if there is at least one row in table_1 and fails miserably returning an empty result set if there are zero rows. I would really like this to return 0! Anyone enlighten me on this? Beer can be provided in exchange if you are in London ;-)

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  • Install mySQL data using ftp?

    - by Jane
    I am trying to install magento (open source e-commerce platform) sample data on my webhost. I have uploaded the file magento_sample_data.sql via ftp, and setup a new database and have assigned it to a user. How do I get the sample data into my empty database?

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  • MySQL: Query Cacheing (How do I use memcache?)

    - by Rachel
    I have an query like: SELECT id as OfferId FROM offers WHERE concat(partycode, connectioncode) = ? AND CURDATE() BETWEEN offer_start_date AND offer_end_date AND id IN ("121211, 123341,151512,5145626 "); Now I want to cache the results of this query using memcache and so my question is How can I cache an query using memcache. I am currently using CURDATE() which cannot be used if we want to implement caching and so how can I get current date functionality without using CURDATE() function ?

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  • Mysql - Join matches and non-matches

    - by jwzk
    This is related to my other question: http://stackoverflow.com/questions/2579249/managing-foreign-keys I am trying to join the table of matches and non-matches. So I have a list of interests, a list of users, and a list of user interests. I want the query to return all interests, whether the user has the interest or not (should be null in that case), only where the user = x. Every time I get the query working its only matching interests that the user specifically has, instead of all interests whether they have it or not.

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  • mysql_num_rows(): supplied argument is not a valid MySQL result resource

    - by php-b-grader
    I am getting this error when I pass an invalid SQL string... I spent the last hour trying to find the problem assuming - It's not my SQL it must be the db handle... ANyway, I've now figured out that it was bad SQL... What I want to do is test the result of the mysql_query() for a valid resultset. I am simply using empty($result)... Is this the most effective test? Is there a more widely accepted method of testing a resultset for a valid result?

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  • Mysql question about UPDATE

    - by Beck
    UPDATE counter_reports SET `counter`=`counter`+1,`date`=? WHERE report_id IN( (SELECT report_id FROM counter_reports WHERE report_name="emails_sent" AND `year`=1 ORDER BY report_id DESC LIMIT 1), (SELECT report_id FROM counter_reports WHERE report_name="emails_sent" AND `month`=1 ORDER BY report_id DESC LIMIT 1), (SELECT report_id FROM counter_reports WHERE report_name="emails_sent" AND `week`=1 ORDER BY report_id DESC LIMIT 1), (SELECT report_id FROM counter_reports WHERE report_name="emails_sent" AND `day`=1 ORDER BY report_id DESC LIMIT 1) ) Is there any alternative for such sql? I need to update(increment by 1) last counter reports for day,week,month and year. If I'm adding manually, sql works fine, but with subqueries it fails to launch. Thanks. :)

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  • MySQL select column length in php

    - by Patrick
    Hello! How do i get the actual max length of a specified column in php? For instance, this table: id - int(11) name - string(20) I want in php to select the maximum number of characters that a field can have, like SELECT length(name) from table1 and it should then return 20 (since its the maximum number of characters for that field).

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  • MySQL using function IN with variable

    - by misanov
    I have problem querying table with variable in IN function. SELECT s.date, (SELECT GROUP_CONCAT(value) FROM value WHERE id_value IN(s.ref_values) ) AS vals FROM stats s ORDER BY s.date DESC LIMIT 1 Where s.ref_values is '12,22,54,15'. I get only one return for first number (12). When I insert that value directly in IN(12,22,54,15) it finds all 4. So, there must be problem with using variable in IN. What am I doing wrong?

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  • PHP & MySQL query value question.

    - by space
    How can I use the first query's id value $row['id'] again after I run a second query inside the while loop statement? To show you what I mean here is a sample code below of what I'm trying to do. I hope I explained it right. Here is the code. $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT users.* FROM users WHERE user_id = 4"); if (!$dbc) { // There was an error...do something about it here... print mysqli_error($mysqli); } while($row = mysqli_fetch_assoc($dbc)) { echo '<div>User: ' . $row['id'] . '</div>'; echo '<div>Link To User' . $row['id'] . '</div>'; $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc2 = mysqli_query($mysqli,"SELECT COUNT(cid) as num FROM comments WHERE comments_id = $row[id]"); if (!$dbc2) { // There was an error...do something about it here... print mysqli_error($mysqli); } else { while($row = mysqli_fetch_array($dbc2)){ $num = $row['num']; } } echo '<div>User ' . $row['id'] . ' Comments# ' . $num . '</div>'; }

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  • Check if the current date is between two dates + mysql select query

    - by kj7
    I have following table : id dateStart dateEnd active 1 2012-11-12 2012-12-31 0 2 2012-11-12 2012-12-31 0 I want to compare todays date in between dateStart and dateEnd. Following is my query for this : $todaysDate="2012-26-11"; $db = Zend_Registry::get("db"); $result = $db->fetchAll("SELECT * FROM `table` WHERE active=0 AND {$todaysDate} between dateStart and dateEnd"); return $result; But its not working. Any solution. Thanks in advance.

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  • mysql codeigniter active record m:m deletion

    - by sea_1987
    Hi There, I have a table 2 tables that have a m:m relationship, what I can wanting is that when I delete a row from one of the tables I want the row in the joining table to be deleted as well, my sql is as follow, Table 1 CREATE TABLE IF NOT EXISTS `job_feed` ( `id` int(11) NOT NULL AUTO_INCREMENT, `body` text NOT NULL, `date_posted` int(10) NOT NULL, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=3 ; Table 2 CREATE TABLE IF NOT EXISTS `job_feed_has_employer_details` ( `job_feed_id` int(11) NOT NULL, `employer_details_id` int(11) NOT NULL, PRIMARY KEY (`job_feed_id`,`employer_details_id`), KEY `fk_job_feed_has_employer_details_job_feed1` (`job_feed_id`), KEY `fk_job_feed_has_employer_details_employer_details1` (`employer_details_id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1; So what I am wanting to do is, if the a row is deleted from table1 and has an id of 1 I want the row in table to that also has that idea as part of the relationship also. I want to do this in keeping with codeigniters active record class I currently have this, public function deleteJobFeed($feed_id) { $this->db->where('id', $feed_id) ->delete('job_feed'); return $feed_id; }

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  • Export data from mysql table row into an array format

    - by user1804952
    I am trying to output data from table : artists row : artist into this format. Artist Names can have special characters and there are over 16k of them. It needs to be written to a file. called anything artist.php for example $Artist = array( "Name from database", "Name from database", "Name from database", "Name from database", "Name from database" ); ok sorry for not explaining. do this for ajax auto complete.. so i need to create a file with this array in it. here is the exact script http://www.brandspankingnew.net/specials/ajax_autosuggest/ajax_autosuggest_autocomplete.html

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  • PHP/MySQL Problem

    - by Scott
    Why does this only print the sites specific content under the first site, and doesn't do it for the other 2? <?php echo 'NPSIN Data will be here soon!'; // connect to DB $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = 'root'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to DB'); $dbname = 'npsin'; mysql_select_db($dbname); // get number of sites $query = 'select count(*) from sites'; $result = mysql_query($query) or die ('Query failed: ' . mysql_error()); $resultArray = mysql_fetch_array($result); $numSites = $resultArray[0]; echo "<br><br>"; // get all sites $query = 'select site_name from sites'; $result = mysql_query($query); // get site content $query2 = 'select content_name, site_id from content'; $result2 = mysql_query($query2); // get site files // print info $count = 1; while ($row = mysql_fetch_array($result, MYSQL_NUM)) { echo "Site $count: "; echo "$row[0]"; echo "<br>"; $contentCount = 1; while ($row2 = mysql_fetch_array($result2, MYSQL_NUM)) { $id = $row2[1]; if ($id == ($count - 1)) { echo "Content $contentCount: "; echo "$row2[0]"; echo "<br>"; } $contentCount++; } $count++; } ?>

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