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  • Unknown column even thoug it exits

    - by george
    I have SELECT servisler.geo_location, servisler.ADRES_MERKEZ, servisler.ADRES_ILCE, servisler.ADRES_IL, servisler.FIRMA_UNVANI, servisler.ADRES_ISTEL, servisler.YETKILI_ADISOYADI, urun_gruplari.GRUP_ADI FROM servisler INNER JOIN urun_gruplari ON kullanici_cihaz.URUN_GRUP_NO= urun_gruplari.RECNO INNER JOIN kullanici ON kullanici.SERVIS_RECNO = servisler.RECNO INNER JOIN kullanici_cihaz ON kullanici.RECNO = kullanici_cihaz.KUL_RECNO AND kullanici_cihaz.URUN_GRUP_NO = urun_gruplari.RECNO where kullanici.kullanici = 'MAR.EDI.003' but it says [Err] 1054 - Unknown column 'kullanici_cihaz.URUN_GRUP_NO' in 'on clause' enen though the column exits. What is its problem? schema Server version: 5.1.33-community-log

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  • How to use where condition for the for a selected column using subquery?

    - by Holicreature
    I have two columns as company and product. I use the following query to get the products matching particular string... select id,(select name from company where product.cid=company.id) as company,name,selling_price,mrp from product where name like '$qry_string%' But when i need to list products of specific company how can i do? i tried the following but in vein select id,(select name from company where product.cid=company.id) as company,name,selling_price,mrp from product where company like '$qry_string%' Help me

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  • Adding to a multidimensional array in PHP

    - by b. e. hollenbeck
    I have an array being returned from the database that looks like so: $data = array(201 => array('description' => blah, 'hours' => 0), 222 => array('description' => feh, 'hours' => 0); In the next bit of code, I'm using a foreach and checking the for the key in another table. If the next query returns data, I want to update the 'hours' value in that key's array with a new hours value: foreach ($data as $row => $value){ $query = $db->query($sql); if ($result){ $value['hours'] = $result['hours']; } I've tried just about every combination of declarations for the foreach loop, but I keep getting the error that it's a non-object. Surely this is easier than my brain is perceiving it.

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  • Need help in Select query..

    - by Parth
    If I have four rows for a field with different values with other fields similar and then other four rows with same condition, as given below here u can see there different rows for insert with only difference in the "newvalue" and "field" excluding "id" for the table_name=jos_menu, operation=INSERT and live=0 now here what select query should be used to get only single row from the table on every change of table_name...??

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  • AIR: sync gui with data-base?

    - by John Isaacks
    I am going to be building an AIR application that shows a list (about 1-25 rows of data) from a data-base. The data-base is on the web. I want the list to be as accurate as possible, meaning as soon as the data-base data changes, the list displayed in the app should update asap. I do not know of anyway that the air application could be notified when there is a change, I am thinking I am going to have to poll the data-base at certain intervals to keep an up to date list. So my question is, first is there any way to NOT have to keep checking the data-base? or if I do keep have to keep checking the data-base what is a reasonable interval to do that at? Thanks.

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  • VBA How to find last insert id?

    - by Muiter
    I have this code: With shtControleblad Dim strsql_basis As String strsql_basis = "INSERT INTO is_calculatie (offerte_id) VALUES ('" & Sheets("controleblad").Range("D1").Value & "')" rs.Open strsql_basis, oConn, adOpenDynamic, adLockOptimistic Dim last_id As String last_id = "select last_insert_id()" End With The string last_id is not filled. What is wrong? I need to find te last_insert_id so I can use it in an other query.

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  • Submitting a URL into a Form without "http://", with "www.", or with neither

    - by John
    (EDITED) Hello, In the form below, the filed for <div class="urlfield"><input name="url" type="url" id="url" maxlength="500"></div> fine when a URL is submitted that has a "http://" at the beginning of it. However, it doesn't work if a URL is submitted with only a "www." in front of it, or with neither a "http://" nor a "www." How can I make it work in all if the submitted URL has any or none of the following at the beginning of it: http:// www. http://www. Thanks in advance, John Form: echo '<div class="submittitle">Submit an item.</div>'; echo '<form action="http://www...com/.../submit2.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <div class="submissiontitle"><label for="title">Story Title:</label></div> <div class="submissionfield"><input name="title" type="title" id="title" maxlength="1000"></div> <div class="urltitle"><label for="url">Link:</label></div> <div class="urlfield"><input name="url" type="url" id="url" maxlength="500"></div> <div class="submissionbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; submit2.php: <?php if($_SERVER['REQUEST_METHOD'] == "POST"){header('Location: http://www...com/.../submit2.php');} require_once "header.php"; if (isLoggedIn() == true) { $remove_array = array('http://www.', 'http://', 'https://', 'https://www.', 'www.'); $cleanurl = str_replace($remove_array, "", $_POST['url']); $cleanurl = strtolower($cleanurl); $cleanurl = preg_replace('/\/$/','',$cleanurl); $cleanurl = stripslashes($cleanurl); $title = $_POST['title']; $uid = $_POST['uid']; $title = mysql_real_escape_string($title); $title = stripslashes($title); $cleanurl = mysql_real_escape_string($cleanurl); $site1 = 'http://' . $cleanurl; $displayurl = parse_url($site1, PHP_URL_HOST); function isURL($url1 = NULL) { if($url1==NULL) return false; $protocol = '(http://|https://)'; $allowed = '[-a-z0-9]{1,63}'; $regex = "^". $protocol . // must include the protocol '(' . $allowed . '\.)'. // 1 or several sub domains with a max of 63 chars '[a-z]' . '{2,6}'; // followed by a TLD if(eregi($regex, $url1)==true) return true; else return false; } if(isURL($site1)==true) mysql_query("INSERT INTO submission VALUES (NULL, '$uid', '$title', '$cleanurl', '$displayurl', NULL)"); else echo "<p class=\"topicu\">Not a valid URL.</p>\n"; } else { show_loginform(); } if (!isLoggedIn()) { if (isset($_POST['cmdlogin'])) { if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { show_loginform(); } } else { show_userbox(); } require_once "footer.php"; ?>

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  • Create fulltext index on a VIEW

    - by kylex
    Is it possible to create a full text index on a VIEW? If so, given two columns column1 and column2 on a VIEW, what is the SQL to get this done? The reason I'd like to do this is I have two very large tables, where I need to do a FULLTEXT search of a single column on each table and combine the results. The results need to be ordered as a single unit. Suggestions? EDIT: This was my attempt at creating a UNION and ordering by each statements scoring. (SELECT a_name AS name, MATCH(a_name) AGAINST('$keyword') as ascore FROM a WHERE MATCH a_name AGAINST('$keyword')) UNION (SELECT s_name AS name,MATCH(s_name) AGAINST('$keyword') as sscore FROM s WHERE MATCH s_name AGAINST('$keyword')) ORDER BY (ascore + sscore) ASC sscore was not recognized.

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  • problem in counting children category

    - by moustafa
    I have this table: fourn_category (id , sub) I am using this code to count: function CountSub($id){ $root = array($id); $query = mysql_query("SELECT id FROM fourn_category WHERE sub = '$id'"); while( $row = mysql_fetch_array( $query, MYSQL_ASSOC ) ){ array_push($root,$row['id']); CountSub($row['id']); } return implode(",",$root); } It returns the category id as 1,2,3,4,5 to using it to count the sub by IN() But the problem is that it counts this: category 1 category 2 category 3 category 4 category 5 Category 1 has 1 child not 4. Why? How can I get all children's trees?

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  • ob_start() is partially capturing data

    - by AAA
    I am using the following code: PHP: // Generate Guid function NewGuid() { $s = strtoupper(uniqid(rand(),true)); $guidText = substr($s,0,8) . '-' . substr($s,8,4) . '-' . substr($s,12,4). '-' . substr($s,16,4). '-' . substr($s,20); return $guidText; } // End Generate Guid $Guid = NewGuid(); $alphabet = '123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ'; function base_encode($num, $alphabet) { $base_count = strlen($alphabet); $encoded = ''; while ($num >= $base_count) { $div = $num/$base_count; $mod = ($num-($base_count*intval($div))); $encoded = $alphabet[$mod] . $encoded; $num = intval($div); } if ($num) $encoded = $alphabet[$num] . $encoded; return $encoded; } function base_decode($num, $alphabet) { $decoded = 0; $multi = 1; while (strlen($num) > 0) { $digit = $num[strlen($num)-1]; $decoded += $multi * strpos($alphabet, $digit); $multi = $multi * strlen($alphabet); $num = substr($num, 0, -1); } return $decoded; } ob_start(); echo base_encode($Guid, $alphabet); //should output: bUKpk $theid = ob_get_contents(); ob_get_clean(); The problem: When i echo $theid, it shows the complete entry, but as it is being inserted into the database, only the first entry in the sequence gets inserted, for example for the entry buKPK, only 'b' is being inserted not the rest.

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  • Display a ranking grid for game : optimization of left outer join and find a player

    - by Jerome C.
    Hello, I want to do a ranking grid. I have a table with different values indexed by a key: Table SimpleValue : key varchar, value int, playerId int I have a player which have several SimpleValue. Table Player: id int, nickname varchar Now imagine these records: SimpleValue: Key value playerId for 1 1 int 2 1 agi 2 1 lvl 5 1 for 6 2 int 3 2 agi 1 2 lvl 4 2 Player: id nickname 1 Bob 2 John I want to display a rank of these players on various SimpleValue. Something like: nickname for lvl Bob 1 5 John 6 4 For the moment I generate an sql query based on which SimpleValue key you want to display and on which SimpleValue key you want to order players. eg: I want to display 'lvl' and 'for' of each player and order them on the 'lvl' The generated query is: SELECT p.nickname as nickname, v1.value as lvl, v2.value as for FROM Player p LEFT OUTER JOIN SimpleValue v1 ON p.id=v1.playerId and v1.key = 'lvl' LEFT OUTER JOIN SimpleValue v2 ON p.id=v2.playerId and v2.key = 'for' ORDER BY v1.value This query runs perfectly. BUT if I want to display 10 different values, it generates 10 'left outer join'. Is there a way to simplify this query ? I've got a second question: Is there a way to display a portion of this ranking. Imagine I've 1000 players and I want to display TOP 10, I use the LIMIT keyword. Now I want to display the rank of the player Bob which is 326/1000 and I want to display 5 rank player above and below (so from 321 to 331 position). How can I achieve it ? thanks.

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  • Avoiding Nested Queries

    - by Midhat
    How Important is it to avoid nested queries. I have always learnt to avoid them like a plague. But they are the most natural thing to me. When I am designing a query, the first thing I write is a nested query. Then I convert it to joins, which sometimes takes a lot of time to get right. And rarely gives a big performance improvement (sometimes it does) So are they really so bad. Is there a way to use nested queries without temp tables and filesort

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  • Query broke down and left me stranded in the woods

    - by user1290323
    I am trying to execute a query that deletes all files from the images table that do not exist in the filters tables. I am skipping 3,500 of the latest files in the database as to sort of "Trim" the table back to 3,500 + "X" amount of records in the filters table. The filters table holds markers for the file, as well as the file id used in the images table. The code will run on a cron job. My Code: $sql = mysql_query("SELECT * FROM `images` ORDER BY `id` DESC") or die(mysql_error()); while($row = mysql_fetch_array($sql)){ $id = $row['id']; $file = $row['url']; $getId = mysql_query("SELECT `id` FROM `filter` WHERE `img_id` = '".$id."'") or die(mysql_error()); if(mysql_num_rows($getId) == 0){ $IdQue[] = $id; $FileQue[] = $file; } } for($i=3500; $i<$x; $i++){ mysql_query("DELETE FROM `images` WHERE id='".$IdQue[$i]."' LIMIT 1") or die("line 18".mysql_error()); unlink($FileQue[$i]) or die("file Not deleted"); } echo ($i-3500)." files deleted."; Output: 0 files deleted. Database contents: images table: 10,000 rows filters table: 63 rows Amount of rows in filters table that contain an images table id: 63 Execution time of php script: 4 seconds +/- 0.5 second Relevant DB structure TABLE: images id url etc... TABLE: filter id img_id (CONTAINS ID FROM images table) etc...

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  • how to set a status

    - by ejah85
    hello guys..here i've a problem where i want to set the status whether it is approved or reject.. the condition are if admin select the registration number and driver name, that means the status is approve otherwise, if admin fill up the reason, that means the request is reject.. here is the code to set status if ($reason =='null'){ $query2 = "UPDATE usage SET status ='APPROVED' WHERE '$bookingno'=bookingno"; $result2 = @mysql_query($query2); } elseif (($regno =='null')&&($d_name =='null')) { $query3 = "UPDATE usage SET status ='REJECT' WHERE '$bookingno'=bookingno"; $result3 = @mysql_query($query3); } when i save the data, the status field are not updates..

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  • Can a binary tree or tree be always represented in a Database as 1 table and self-referencing?

    - by Jian Lin
    I didn't feel this rule before, but it seems that a binary tree or any tree (each node can have many children but children cannot point back to any parent), then this data structure can be represented as 1 table in a database, with each row having an ID for itself and a parentID that points back to the parent node. That is in fact the classical Employee - Manager diagram: one boss can have many people under him... and each person can have n people under him, etc. This is a tree structure and is represented in database books as a common example as a single table Employee.

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  • How to print contents from a session variable by looping in a foreach statement

    - by itsover9000
    im trying to write a code where can print and loop through the contents of my session variable by using a foreach statement here is my code <form class="form form-inline" method = "post" action="reportmaker.php"> <select name="rfield"> <option value="">--Select Field--</option> <?php $sc2=mysql_query("SELECT * from searchcolumn s left join report_fields r on s.scol_id=r.field_id where s.category != 'wh'"); foreach($sc2 as $sc){ ?> <option value="<?php echo $sc[advsearch_col]; ?>"><?php echo $sc[advsearch_name]; ?></option> <?php } ?> </select> <button type="submit" value = "submit" id="add" name="add" class="btn pull-right">Add More</button> </form> <?php if(isset($_POST['add'])) { $_SESSION['temp'][]=$_POST['rfield']; } if($_SESSION[temp][]!=""){ foreach($_SESSION[temp][] as $temp) { echo $temp; } } ?> the error that appears with this code is Fatal error: Cannot use [] for reading the line where the error is is this if($_SESSION[temp][]!=""){ i need to print the contents of the session array and this is the only way i know how is there a way to fix this? thanks =========EDIT thanks for the answers guys i finally got it

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  • Rails: Generated tokens missing occasionally

    - by Vincent Chan
    We generate an unique token for each user and store it on database. Everything is working fine in the local environment. However, after we upload the codes to the production server on Engine Yard, things become weird. We tried to register an account right after the deploy. It is working fine and we can see the token in the db. But after that, when we register new accounts, we cannot see any tokens. We only have NULL in the db. Not sure what caused this problem because we can't re-produce this in the local machine. Thanks for your help.

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  • PHP coding a price comparaison tool

    - by Tristan
    Hello, it's the first time I developp such tool you all know (the possibility to compare articles according to price and/or options) Since I never did that i want to tell me what do you think of the way i see that : On the database we would have : offer / price / option 1 / option 2 / option 3 / IDseller / IDoffer best buy / 15$ / full FTP / web hosting / php.ini / 10 / 1 .../..../.... And the request made by the client : "SELECT * FROM offers WHERE price <= 20 AND option1 = fullFTP"; I don't know if it seems OK to you. Plus i was wondering, how to avoid multiples entries for the same seller. Imagine you have multiple offers with a price <= 20 with the option FullFTP for the same seller, i don't want him to be shown 5 times on the comparator. If you have any advices ;) Thanks

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  • How to stop looking in a database after X rows are found?

    - by morningface
    I have a query to a database that returns a number X of results. I am looking to return a maximum of 10 results. Is there a way to do this without using LIMIT 0,9? I'll use LIMIT if I have to, but I'd rather use something else that will literally stop the searching, rather than look at all rows and then only return the top 10.

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  • Get number of posts in a topic PHP

    - by Wayne
    How do I get to display the number of posts on a topic like a forum. I used this... (how very noobish): function numberofposts($n) { $sql = "SELECT * FROM posts WHERE topic_id = '" . $n . "'"; $result = mysql_query($sql) or die(mysql_error()); $count = mysql_num_rows($result); echo number_format($count); } The while loop of listing topics: <div class="topics"> <div class="topic-name"> <p><?php echo $row['topic_title']; ?></p> </div> <div class="topic-posts"> <p><?php echo numberofposts($row['topic_id']); ?></p> </div> </div> Although it is a bad method of doing this... All I need is to know what would be the best method, don't just point me out to a website, do it here, because I'm trying to learn much. Okay? :D Thanks.

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