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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_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: 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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  • convert mysql code to codeigniter

    - by Jethro Tamares Doble
    How can i convert this code into an acceptable codeigniter code: mysql_select_db($database_connection_ched, $connection_ched); $query_Institutions = "SELECT * FROM tb_institutional_profile ORDER BY tb_institutional_profile.institution_name ASC"; $Institutions = mysql_query($query_Institutions, $connection_ched) or die(mysql_error()); $row_Institutions = mysql_fetch_assoc($Institutions); $totalRows_Institutions = mysql_num_rows($Institutions); <td width="192"><select name="institution_id"> <?php do { <option value="<?php echo $row_Institutions['institution_id']?>" ><?php echo $row_Institutions['institution_name']?></option> <?php } while ($row_Institutions = mysql_fetch_assoc($Institutions)); ?> </select></td>

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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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  • mysql: managing memory usage

    - by every_answer_gets_a_point
    i am doing a delete with a LIKE statement my keybuffer is 25m, the sort buffer size is 256k the delete has been taking over 2 hours should i increase memory usage? there are about 50 megs of data in the table from which i am deleting, thats about 500,000 rows is there anything else i can do on the adminsitration size to speed up this delete?

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  • MySql Not Like Regexp?

    - by KnockKnockWhosThere
    I'm trying to find rows where the first character is not a digit. I have this: SELECT DISTINCT(action) FROM actions WHERE qkey = 140 AND action NOT REGEXP '^[:digit:]$'; But, I'm not sure how to make sure it checks just the first character...

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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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  • 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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  • 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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  • Sum of a summation in mysql

    - by dames
    I have the following query, in the top select statement (sum(l.app_ln_amnt)/count(l.app_ln_amnt)) works well but in the union I want to find the total of (sum(l.app_ln_amnt)/count(l.app_ln_amnt)) query from the top select statement However my solution seems to be off I need some help please select (sum(l.app_ln_amnt)/count(l.app_ln_amnt)), from receipt_history l UNION select SUM(sum(l.app_ln_amnt)/count(l.app_ln_amnt)), from receipt_history l

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  • mysql bug database error

    - by user963499
    Why is that when I search a row in my database, it ended up 0 results? In fact, there are rows that meet my search criterion when I view them manually, but the search button doesnt work as it is. Heres how you can see there are rows that have '0000-00-00 00:00:00' in them, but when I used the search feature, it ended up like this: take note that Im entering it in the right field which is 'AcctStopTime'. TIA

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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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  • 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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  • Problem with multi-table MySQL query

    - by mahle
    I have 3 tables. Here is the relevant information needed for each. items prod_id order_id item_qty orders order_id order_date order_status acct_id accounts acct_id is_wholesale items is linked to order by the order_id and orders is linked to accounts via acct_id I need to sum item_qty for all items where prod_id=464 and the order stats is not 5 and where the is_wholesale is 0 and the order_date is between two dates. Im struggling with this and would appreciate any help. Here is what I have but it's not working correctly: SELECT SUM(items.item_qty) as qty FROM items LEFT JOIN orders ON orders.order_id = items.order_id LEFT JOIN accounts on orders.acct_id = accounts.acct_id WHERE items.prod_id =451 AND orders.order_date >= '$from_date' AND orders.order_date <= '$to_date' AND orders.order_status <>5 AND accounts.is_wholesale=0; Again, any help would be greatly appreciated!

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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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  • onUpdate in MySQL means?

    - by ajsie
    i know that if you create a foreign key on a field (parent_id) in a child table that refer to a parent table's primary key (id), then if this parent table is deleted the child class will be deleted as well if you set onDelete to cascade when creating the foreign key in the child class. but what happens if i set it to onUpdate = cascade?

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  • mysql query trouble

    - by Bharanikumar
    Hi , in my database i have phone numbers with country code , which look somthing like 0044-123456 0044-123456 0014-123456 0014-123456 0024-123456 0024-123456 0034-123456 0044-123456 0044-123456 0024-123456 0034-123456 084-123456 084-123456 i want to total up the numbers by country, something like this output 0044 (2) 0024 (2) 0034 (1) 084 (2) 064 (5) Is it possible to do this with a SQL query?

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  • Mysql select most frequent and sort alphabetically

    - by user2605793
    I am trying to select the most common 100 names from a table then display the list showing the names and count. I want the user to be able to re-sort the list alphabetically rather than based on count. I thought the following code would do it. It works for the default sort by count but fails on the sort alphabetically. The line "$count = mysql_num_rows($table);" gives an error: mysql_num_rows() expects parameter 1 to be resource, boolean given. Any help would be greatly appreciated. // Get most popular surnames echo '<h2>Most Common Surnames</h2>'; if ($sort == "") { // default sort by count echo '<a href="http://mysite/names.php?id='.$id.'&sort=name">Sort by name</a><br>'; $query = "SELECT family_name, COUNT(*) as count FROM namefile WHERE record_key = $id GROUP BY family_name ORDER BY count DESC LIMIT 100"; } else { // sort alphabetically echo '<a href="http://mysite/names.php?id='.$id.'">Sort by count</a><br>'; $query = "SELECT * FROM ( SELECT family_name, COUNT(*) as count FROM namefile WHERE record_key = $id GROUP BY family_name ORDER BY count DESC LIMIT 100) AS alpha ORDER BY family_name"; } $table = mysql_query($query); $count = mysql_num_rows($table); $tot = 0; $i = 0; echo '<table><tr>'; while ($tot < $count2) { $rec2 = mysql_fetch_array($table2); echo '<td>'.$rec2[0].'</td><td>'.$rec2[1].'</td><td width="40">&nbsp;</td><td>'; if ($i++ == 6) { echo '</tr><tr>'; $i = 0; } $tot++; } echo '</tr></table><br>'; UPDATE: I needed to add "AS alpha" to give the outer select a unique name. (alpha is just a random name I made up.) It now works perfectly. Code updated for the benefit of any others who need something similar.

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  • Mysql SELECT with an OR across 2 columns

    - by Haroldo
    I'm creating a 'similar items' link table. i have a 2 column table. both columns contains product ids. The table is showing that these items are similar. However ids in the left column are more valuable. Say i want to select similar items to product '125b'. i only want 3 similar items to 125b. If there are any instances of 125b in col1 I would prefer these to finding 125b in col2. so i need a select statement along the lines of SELECT * FROM similar_items WHERE col_1={$id} OR col_2={$id} ORDER BY column(?) LIMIT 3 i do not want to do 2 separate queries ( ie query 2 if count(query1) <3 )

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