Search Results

Search found 17036 results on 682 pages for 'mysql administrator'.

Page 196/682 | < Previous Page | 192 193 194 195 196 197 198 199 200 201 202 203  | Next Page >

• 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?

  Read the article

• 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>'; }

  Read the article

• Problem with MySQL query

  - by Psyche

  This time my setup looks like this: one table with galleries names (gallery_id, gallery_name) and another table with galleries photos (photo_id, photo_gallery_id, photo_name). What I need is to get all the galleries with one random picture for each gallery. Is it possible to do this with a single query?

  Read the article

• MYSQL sum() for distinct rows

  - by makeee

  I'm looking for help using sum() in my SQL query (not posting full query since the scenario is fairly simple). I have COUNT(DISTINCT conversions.id) in my query. I use DISTINCT because I'm doing "group by" for multiple columns and this ensures the same row is not counted more than once. Now I want to add: SUM(conversions.value) as conversion_value The problem is that the "value" for each row is counted more than once (due to the multiple group bys) I basically want to do SUM(conversions.value) for each DISTINCT conversions.id. Is that possible?

  Read the article

• How to prevent negative number in Mysql

  - by Jerry

  Hello guys.. I have data which is starting from 0 in my database. My php will add 1 or -1 to the data depending on the user's input. My problem is that if data is 0 and a user try to subtract 1. The data become 4294967295 which is the maximum value of INT data type. Are there anyways to make the data stays in 0 even when the user asks for -1? Thanks for the reply.. my sql command is like below update board set score=score-1 where team='TeamA' //this would generate 4294967295 if the score is 0.....

  Read the article

• 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?

  Read the article

• 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

  Read the article

• 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++; } ?>

  Read the article

• mysql - How select a akin records?

  - by zeroonea

  I have a table with name varchar address varchar country varchar city varchar ..... to store address of location example: name|address|country HaLong hotel|156 blahblah street|Vietnam Hotel Ha Long|156 blah blah|Vietnam Two rows above is duplicate data. I have a form, when user submit new location. The code need to find akin records to give a message (ex: This location already in db, use it or create new?) How to make a query to get akin record like this?

  Read the article

• 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.

  Read the article

• 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.

  Read the article

• 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>

  Read the article

• 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?

  Read the article

• MYSQL: Udate table column from subquery result

  - by Jhourlad Estrella

  On the Members table are columns "MemberID" and "PointsEarned". I want to update the PointsEarned column from the result of this query: SELECT m.MemberID, m.UserName, ( (SELECT COUNT(*) FROM EventsLog as e WHERE e.MemberID=m.MemberID AND e.EventsTypeID=2)*10 ) + ( (SELECT COUNT(*) FROM EventsLog as e WHERE e.MemberID=m.MemberID AND e.EventsTypeID=3)*3 ) + ( (SELECT COUNT(*) FROM ChatMessages as c WHERE c.MemberID=m.MemberID)*.1 ) as PointsEarned FROM Members as m Can anybody tell me how I should do it with a single query? Thanks!

  Read the article

• 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...

  Read the article

• comparing rows on a mysql table

  - by user311324

  Ok here's the deal I got one table with a bunch of client information. Each client makes up to one purchase a year which is represented by an individual row. there's a column for the year and there's a column the contains a unique identifier for each client. What I need to do is to construct a query that takes last year and this year and shows me which clients were here made a purchase last year but not make a purchase this year. I also need to build a query that shows me which clients did not make a purchase last year and the year before last but did make a purchase this year.

  Read the article

• 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. :)

  Read the article

• 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 ) );

  Read the article

• 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!

  Read the article

• PHP/MySQL - an array filter for bots

  - by Mike

  Hello, I'm making a hit counter. I have a database and I store the IP and $_SERVER['HTTP_USER_AGENT']; of the visitors. Now I need to add a filter, so I can put away the hits, that are made by bots. I found out, that many bots usually keep some common words in the $_SERVER['HTTP_USER_AGENT']; , so I's like to make and array of words, that would keep the bot from displaying in the results. Here is what I have now: while($row = mysql_fetch_array($yesterday, MYSQL_ASSOC)) { <-- Here I need a code, that would run through an array and check, if it containts the keywords and if it doesn't ... just count++; -- } Also if you know any other way of detecting and removing the bots from the results, I'd be verry thankful. Cheers

  Read the article

• MySQL GROUP BY with three tables

  - by Psaniko

  I have the following tables: posts (post_id, content, etc) comments (comment_id, post_id, content, etc) posts_categories (post_category_id, post_id, category_id) and this query: SELECT `p`.*, COUNT(comments.comment_id) AS cmts, posts_categories.*,comments.* FROM `posts` AS `p` LEFT JOIN `posts_categories` ON `p`.post_id = `posts_categories`.post_id LEFT JOIN `comments` ON `p`.post_id = `comments`.post_id GROUP BY `p`.`post_id` There are three comments on post_id=1 and four in total. In posts_categories there are two rows, both assigned to post_id=1. I have four rows in posts. But if I query the statement above I get a result of 6 for COUNT(comments.comment_id) at post_id=1. How is this possible? I guess the mistake is somewhere in the GROUP BY clause but I can't figure out where. Any suggestions?

  Read the article

• 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 ;-)

  Read the article

• Loading a large list of sentences into a MySQL table

  - by Josh K

  I have a large list of phrases / quotes / sentences in a text file. Mixed alphanumeric text, it hasn't been screened for "'s or ''s. Is there an easy way to throw this into a table?

  Read the article

• 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

  Read the article

• 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?

  Read the article

< Previous Page | 192 193 194 195 196 197 198 199 200 201 202 203  | Next Page >