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  • SQL with codition on calculated value

    - by user619893
    I have a table with products, their amount and their price. I need to select all entries where the average price per article is between a range. My query so far: SELECT productid,AVG(SUM(price)/SUM(amount)) AS avg FROM stock WHERE avg=$from AND avg<=$to GROUP BY productid If do this, it tells me avg doesnt exist. Also i obviously need to group by because the sum and average need to be per wine

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  • PHP - How to display other values, when a query is limited by 3?

    - by Dodi300
    Hello. Can anyone tell me how to display the other values, when a query is limited my 3. In this question I asked how to order and limit values, but now I want to show the others in another query. How would I go about doing this? Here's the code I used before: $query = "SELECT gmd FROM account ORDER BY gmd DESC LIMIT 3"; $result = mysql_query($query); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { } Thanks!

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  • Tracking Votes and only allowing 1 vote per member

    - by MikeAdams
    What I'm trying to do is count the votes when someone votes on a "page". I think I lost myself trying to figure out how to track when a member votes or not. I can't seem to get the code to tell when a member has voted. //Generate code ID $useXID = intval($_GET['id']); $useXrank = $_GET['rank']; //if($useXrank!=null && $useXID!=null) { $rankcheck = mysql_query('SELECT member_id,code_id FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'" AND WHERE code_id="'.$useXID.'"'); if(!mysql_fetch_array($rankcheck) && $useXrank=="up"){ $rankset = mysql_query('SELECT * FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'"'); $ranksetfetch = mysql_fetch_array($rankset); $rankit = htmlentities($ranksetfetch['ranking']); $rankit+="1"; mysql_query("INSERT INTO code_votes (member_id,code_id) VALUES ('$_MEMBERINFO_ID','$useXID')") or die(mysql_error()); mysql_query("UPDATE code SET ranking = '".$rankit."' WHERE ID = '".$useXID."'"); } elseif(!mysql_fetch_array($rankcheck) && $useXrank=="down"){ $rankset = mysql_query('SELECT * FROM code_votes WHERE member_id="'.$_MEMBERINFO_ID.'"'); $ranksetfetch = mysql_fetch_array($rankset); $rankit = htmlentities($ranksetfetch['ranking']); $rankit-="1"; mysql_query("INSERT INTO code_votes (member_id,code_id) VALUES ('$_MEMBERINFO_ID','$useXID')") or die(mysql_error()); mysql_query("UPDATE code SET ranking = '".$rankit."' WHERE ID = '".$useXID."'"); } // hide vote links since already voted elseif(mysql_fetch_array($rankcheck)){$voted="true";} //}

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  • Database Formatting for Album Tracks

    - by Sev
    I would like to store album's track names in a single field in a database. The number of tracks are arbitrary for each album. Each album is one record in the table. Each track must be linked to a specific URL which also should be stored in the database somewhere. Is it possible to do this by storing them in a single field, or is a relational table for the track names/urls the only way to go?

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  • Should I change $_REQUEST to $_POST

    - by Scarface
    Hey guys quick question, I have a checkbox system where a list of items can be checked and deleted on the click of a button. I currently use request and it does the job but I was wondering if $_REQUEST was some sort of security risk or improper. If anyone has any advice I would appreciate it. Should I change to $_POST? If so, what is the best way to go about it? foreach ($_REQUEST as $key=>$value) { if (substr($key,0,3)==="img") { $id = substr($key,3); if(isset($_REQUEST['Delete'])) { $sql = 'SELECT file_name,username FROM images WHERE id=?'; $stmt = $conn->prepare($sql); $result=$stmt->execute(array($id)); while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) { $image=$row['file_name']; $user=$row['username']; $myFile = "$user/images/$image"; unlink($myFile); } <input id=\"img".$id."\" name=\"img".$id."\" type=\"checkbox\">

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  • Is this possible with sql?

    - by Andrew
    Is it possible to do something like this: INSERT INTO table(col1, col2) VALUES(something_from_another_table, value); With something_from_another_table being a SQL command? Like, is there something I can do that's equivelant to: INSERT INTO table(col1, col2) VALUES((SELECT value FROM table2 WHERE id = 3), value);

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  • What information about a user is available via PHP?

    - by Camran
    This is about a classifieds website, where anyone may post classifieds. I have a security database which I intend to fill with information about the user who posts the classifieds. I intend to record information such as IP, name, tel, email, classified_text, classified_title etc etc. The reason for all this is that sometimes people become victims of fraud (fake classifieds etc). So I wonder, what information is possible to get from the poster which may help in tracking him/her down? IP is a given, but what else could be useful? And I would much like examples of how it would be useful also, as well as the code for it please, like $_SERVER['REMOTE_ADDR']. And btw, I use PHP and have Sql as a database. Thanks

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  • retrieving same column twice from a table

    - by GJ
    hello all i hav a table named address which has id, title and parent_id fields. in title column the name of regions and districts are inserted. the regions have parent_id zero and parent_id of the districts are id of the regions. i want a query which display regions in one column and its respective districts in another column. hope u guys understand what i mean.. thank u all.

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  • Empty files generated from running `mysqldump` using PHP

    - by alex
    I keep getting empty files generated from running $command = 'mysqldump --opt -h localhost -u username -p \'password\' dbname > \'backup 2009-04-15 09-57-13.sql\''; command($command); Anyone know what might be causing this? My password has strange characters in it, but works fine with connecting to the db. I've ran exec($command, $return) and outputted the $return array and it is finding the command. I've also ran it with mysqldump > file.sql and the file contains Usage: mysqldump [OPTIONS] database [tables] OR mysqldump [OPTIONS] --databases [OPTIONS] DB1 [DB2 DB3...] OR mysqldump [OPTIONS] --all-databases [OPTIONS] For more options, use mysqldump --help So it would seem like the command is working.

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  • reading into table: comma values and quotes SQL

    - by every_answer_gets_a_point
    i have a string like this something = "something, something1, "something2, something else", something3" i need it to be read into a table like this: field1 = "something" field2= "something2" field3 = "something2, something else" field4 = "something3" please notice that the double quotes in the something string signified that the string inside the quotes is to be placed in one field anyone know how to do this with an insert into statement or some other way? the answer can be purely sql or can be vba with sql. thanks!

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  • Insert into select and update in single query

    - by Ossi
    I have 4 tables: tempTBL, linksTBL and categoryTBL, extra on my tempTBL I have: ID, name, url, cat, isinserted columns on my linksTBL I have: ID, name, alias columns on my categoryTBL I have: cl_id, link_id,cat_id on my extraTBL I have: id, link_id, value How do I do a single query to select from tempTBL all items where isinsrted = 0 then insert them to linksTBL and for each record inserted, pickup ID (which is primary) and then insert that ID to categoryTBL with cat_id = 88. after that insert extraTBL ID for link_id and url for value. I know this is so confusing, put I'll post this anyhow... This is what I have so far: INSERT IGNORE INTO linksTBL (link_id,link_name,alias) VALUES(NULL,'tex2','hello'); # generate ID by inserting NULL INSERT INTO categoryTBL (link_id,cat_id) VALUES(LAST_INSERT_ID(),'88'); # use ID in second table I would like to add here somewhere that it only selects items where isinserted = 0 and iserts those records, and onse inserted, will change isinserted to 1, so when next time it runs, it will not add them again.

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  • Improve SQL query performance

    - by Anax
    I have three tables where I store actual person data (person), teams (team) and entries (athlete). The schema of the three tables is: In each team there might be two or more athletes. I'm trying to create a query to produce the most frequent pairs, meaning people who play in teams of two. I came up with the following query: SELECT p1.surname, p1.name, p2.surname, p2.name, COUNT(*) AS freq FROM person p1, athlete a1, person p2, athlete a2 WHERE p1.id = a1.person_id AND p2.id = a2.person_id AND a1.team_id = a2.team_id AND a1.team_id IN ( SELECT id FROM team, athlete WHERE team.id = athlete.team_id GROUP BY team.id HAVING COUNT(*) = 2 ) GROUP BY p1.id ORDER BY freq DESC Obviously this is a resource consuming query. Is there a way to improve it?

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  • How can I remove this query from within a loop?

    - by Chris
    I am currently designing a forum as a personal project. One of the recurring issues I've come across is database queries in loops. I've managed to avoid doing that so far by using table joins or caching of data in arrays for later use. Right now though I've come across a situation where I'm not sure how I can write the code in such a way that I can use either of those methods easily. However I'd still prefer to do at most 2 queries for this operation rather than 1 + 1 per group of forums, which so far has resulted in 5 per page. So while 5 isn't a huge number (though it will increase for each forum group I add) it's the principle that's important to me here, I do NOT want to write queries in loops What I'm doing is displaying forum index groupings (eg admin forums, user forums etc) and then each forum within that group on a single page index, it's the combination of both in one page that's causing me issue. If it had just been a single group per page, I'd use a table join and problem solved. But if I use a table join here, although I can potentially get all the data I need it'll be in one mass of results and it needs displaying properly. Here's the code (I've removed some of the html for clarity) <?php $sql= "select * from forum_groups"; //query 1 $result1 = $database->query($sql); while($group = mysql_fetch_assoc($result1)) //first loop {?> <table class="threads"> <tr> <td class="forumgroupheader"> <?php echo $group['group_name']; ?> </td> </tr> <tr> <td class="forumgroupheader2"> <?php echo $group['group_desc']; ?> </td> </tr> </table> <table> <tr> <th class="thforum"> Forum Name</th> <th class="thforum"> Forum Decsription</th> <th class="thforum"> Last Post </th> <tr> <?php $group_id = $group['id']; $sql = "SELECT forums.id, forums.forum_group_id, forums.forum_name, forums.forum_desc, forums.visible_rank, forums.locked, forums.lock_rank, forums.topics, forums.posts, forums.last_post, forums.last_post_id, users.username FROM forums LEFT JOIN users on forums.last_post_id=users.id WHERE forum_group_id='{$group_id}'"; //query 2 $result2 = $database->query($sql); while($forum = mysql_fetch_assoc($result2)) //second loop {?> So how can I either a) write the SQL in such a way as to remove the second query from inside the loop or b) combine the results in an array either way I need to be able to access the data as an when so I can format it properly for the page output, ie within the loops still.

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  • checkbox checked with php form post?

    - by Patrick
    how do I check a checkbox? I've tried 1, On, Yes that doesn't work. putting the worked "checked" alone works but then how do I check with php after form post of the checkbox is checked? <input type="checkbox" class="inputcheckbox" id="newmsg" name=chk[newmsg2] value="1" />

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  • differentiating results of sql right join

    - by Sourabh
    Hi I have a below SQL query SELECT `User`.`username` , Permalink.perma_link_id, Permalink.locale, Permalink.title, DATEDIFF( CURDATE( ) , Permalink.created ) AS dtdiff, `TargetSegment`.segment_text, TargetSegment.source_segment_id ,TargetSegment.perma_link_id ,TargetSegment.created ,TargetSegment.updated, DATEDIFF( CURDATE( ) , TargetSegment.updated ) AS datediff FROM `users` AS `User` RIGHT JOIN perma_links AS `PermaLink` ON ( `PermaLink`.`username` = `User`.`username` ) RIGHT JOIN target_segments AS `TargetSegment` ON ( `TargetSegment`.`username` = `User`.`username` ) RIGHT JOIN source_segments AS `SourceSegment` ON ( `SourceSegment`.`source_detail_id` = `PermaLink`.`source_detail_id` ) LEFT JOIN source_details AS `SourceDetail` ON ( `SourceSegment`.`source_detail_id` = `SourceDetail`.`id` ) WHERE `TargetSegment`.`username` = "xxxx" AND `TargetSegment`.`segment_text` <> "" AND `Permalink`.`perma_link_id` = `TargetSegment`.`perma_link_id` AND `TargetSegment`.`source_segment_id` = `SourceSegment`.`id` AND `Permalink`.`source_detail_id` = `SourceDetail`.`id` ORDER BY `TargetSegment`.`updated` DESC LIMIT 0 , 10 This SQL is fetching correct results for me.I want to identify from which table each row if from , to be specific which result is due to PermaLink table and which is from TargetSegment table. is this achievable ?

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  • How to first get different related values from diferent SQL tables (PHP)

    - by Ole Jak
    I am triig to fill options list. I have 2 tables USERS and STREAMS I vant to get all streams and get names of users assigned to that streams. Users consists of username and id Streams consists of id, userID, streamID I try such code: <?php global $connection; $query = "SELECT * FROM streams "; $streams_set = mysql_query($query, $connection); confirm_query($streams_set); $streams_count = mysql_num_rows($streams_set); while ($row = mysql_fetch_array($streams_set)){ $userid = $row['userID']; global $connection; $query2 = "SELECT email, username "; $query2 .= "FROM users "; $query2 .= "WHERE id = '{$userid}' "; $qs = mysql_query($query2, $connection); confirm_query($qs); $found_user = mysql_fetch_array($qs); echo ' <option value="'.$row['streamID'].'">'.$row['userID'].$found_user.'</option> '; } ?> But it does not return USER names from DB=( So what shall I do to this code to see usernames as "options" text?

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  • wp+sql+image not goin in the folder

    - by happy
    this is my code for uploading image in database but image are going to the desird forlder...but when i m tryin to retrieve the images to diaplay,,they are not displayed..anyone help me...... $category=$_POST['category']; $uploadDir = 'D:/xampp/htdocs/js/wordpress/wp-content/plugins/img/imagess/ '; $fileName = $_FILES['Photo']['name']; $tmpName = $_FILES['Photo']['tmp_name']; $fileSize = $_FILES['Photo']['size']; $fileType = $_FILES['Photo']['type']; $filePath = $uploadDir . $fileName; $result = move_uploaded_file($tmpName,$filePath); if (!$result) { echo "Error uploading file"; exit; } if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); $filePath = addslashes($filePath); } global $wpdb; //$insert=$wpdb->insert('images',array('image_name'=>$filePath,'cat_name'=>$category),array('%b','%s')); $insert=$wpdb->insert('images',array('image_name'=>$filePath,'cat_name'=>$category)); $wpdb->insert('categories',array('cat_name'=>$category)); echo "Successfully Submitted";

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  • Set AUTO_INCREMENT value programmatically

    - by Tim
    So this works... ALTER TABLE variation AUTO_INCREMENT = 10; But I want to do this; ALTER TABLE variation AUTO_INCREMENT = (SELECT MAX(id)+1 FROM old_db.varaition); but that doesnt work, and neither does; SELECT MAX(id)+1 INTO @old_auto_inc FROM old_db.variation ALTER TABLE variation AUTO_INCREMENT = @old_auto_inc; So does anyone know how to do this? ( I'm trying to ensure that AUTO_INCREMENT keys dont collide between an old and a new site and need to do this automatically. So I can just run a script when the new db goes live )

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  • Turning user ID into name (seperate tables) in PHP

    - by mobile
    I am currently trying to display the username of people who i am following, the problem is that during the following process, only the ID of me and the person i'm following is stored. I've got it to the point where the ID's are displayed but i'd like to show the names hyperlinked. $p_id is the profile ID. Here's what I've got: $following = mysql_query("SELECT `follower`, `followed` FROM user_follow WHERE follower=$p_id"); I am following: <?php while($apple = mysql_fetch_array($following)){ echo '<a href="'.$apple['followed'].'">+'.$apple['followed'].'</a> '; }?> The usernames are in a different table "users" under the field "username" - I need them to match up with the ID's that are currently displayed, and be displayed. Any help appreciated, thanks guys

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  • Is this SQL is valid?

    - by Beck
    UPDATE polls_options SET `votes`=`votes`+1, `percent`=ROUND((`votes`+1) / (SELECT voters FROM polls WHERE poll_id=? LIMIT 1) * 100,1) WHERE option_id=? AND poll_id=? Don't have table data yet, to test it properly. :) And by the way, in what type % integers should be stored in database? Thanks for the help!

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