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  • Converting a certain SQL query into relational algebra

    - by Fumler
    Just doing an assignment for my database course and I just want to double check that I've correctly wrapped my head around relational algebra. The SQL query: SELECT dato, SUM(pris*antall) AS total FROM produkt, ordre WHERE ordre.varenr = produkt.varenr GROUP BY dato HAVING total >= 10000 The relational algebra: stotal >= 10000( ?R(dato, total)( sordre.varenr = produkt.varenr( datoISUM(pris*antall(produkt x ordre)))) Is this the correct way of doing it?

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  • DEADLOCK_WRAP error when using Berkeley Db in python (bsddb)

    - by JiminyCricket
    I am using a berkdb to store a huge list of key-value pairs but for some reason when i try to access some of the data later i get this error: try: key = 'scrape011201-590652' contenttext = contentdict[key] except: print the error <type 'exceptions.KeyError'> 'scrape011201-590652' in contenttext = contentdict[key]\n', ' File "/usr/lib64/python2.5/bsddb/__init__.py", line 223, in __getitem__\n return _DeadlockWrap(lambda: self.db[key]) # self.db[key]\n', 'File "/usr/lib64/python2.5/bsddb/dbutils.py", line 62, in DeadlockWrap\n return function(*_args, **_kwargs)\n', ' File "/usr/lib64/python2.5/bsddb/__init__.py", line 223, in <lambda>\n return _DeadlockWrap(lambda: self.db[key]) # self.db[key]\n'] I am not sure what DeadlockWrap is but there isnt any other program or process accessing the berkdb or writing to it (as far as i know,) so not sure how we could get a deadlock, if its referring to that. Is it possible that I am trying to access the data to rapidly? I have this function call in a loop, so something like for i in hugelist: #try to get a value from the berkdb #do something with it I am running this with multiple datasets and this error only occurs with one of them, the largest one, not the others.

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  • MYSQ request | GROUP BY DAY

    - by user889349
    I have a table: orders, and need to make a request and get other table. My DB table: id close 1 2012-05-29 03:11:15 2 2012-05-30 03:11:40 3 2012-05-31 03:12:10 4 2012-05-31 03:14:13 5 2012-05-31 03:16:50 6 2012-05-31 03:40:07 7 2012-05-31 05:22:18 8 2012-05-31 05:22:22 9 2012-05-31 05:22:50 ... I need to make a request and get this table (GROUP BY DAY(close)): 1 2012-05-29 03:11:15 2 2012-05-30 03:11:40 9 2012-05-31 05:22:50 /*This is a last record on this day (05-31)*/ Thanks!

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  • Why is str_replace not replacing this string?

    - by Niall
    I have the following PHP code which should load the data from a CSS file into a variable, search for the old body background colour, replace it with the colour from a submitted form, resave the CSS file and finally update the colour in the database. The problem is, str_replace does not appear to be replacing anything. Here is my PHP code (stored in "processors/save_program_settings.php"): <?php require("../security.php"); $institution_name = mysql_real_escape_string($_POST['institution_name']); $staff_role_title = mysql_real_escape_string($_POST['staff_role_title']); $program_location = mysql_real_escape_string($_POST['program_location']); $background_colour = mysql_real_escape_string($_POST['background_colour']); $bar_border_colour = mysql_real_escape_string($_POST['bar_border_colour']); $title_colour = mysql_real_escape_string($_POST['title_colour']); $url = $global_variables['program_location']; $data_background = mysql_query("SELECT * FROM sents_global_variables WHERE name='background_colour'") or die(mysql_error()); $background_output = mysql_fetch_array($data_background); $css = file_get_contents($url.'/default.css'); $str = "body { background-color: #".$background_output['data']."; }"; $str2 = "body { background-color: #".$background_colour."; }"; $css2 = str_replace($str, $str2, $css); unlink('../default.css'); file_put_contents('../default.css', $css2); mysql_query("UPDATE sents_global_variables SET data='{$institution_name}' WHERE name='institution_name'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$staff_role_title}' WHERE name='role_title'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$program_location}' WHERE name='program_location'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$background_colour}' WHERE name='background_colour'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$bar_border_colour}' WHERE name='bar_border_colour'") or die(mysql_error()); mysql_query("UPDATE sents_global_variables SET data='{$title_colour}' WHERE name='title_colour'") or die(mysql_error()); header('Location: '.$url.'/pages/start.php?message=program_settings_saved'); ?> Here is my CSS (stored in "default.css"): @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; } I've run some checks using the following code in the PHP file: echo $css . "<br><br>" . $str . "<br><br>" . $str2 . "<br><br>" . $css2; exit; And it outputs (as you can see it's not changing anything in the CSS): @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; } body { background-color: #CCCCFF; } body { background-color: #FF5719; } @charset "utf-8"; /* CSS Document */ body,td,th { font-family: Arial, Helvetica, sans-serif; font-size: 14px; color: #000; } body { background-color: #CCCCFF; } .main_table th { background:#003399; font-size:24px; color:#FFFFFF; } .main_table { background:#FFF; border:#003399 solid 1px; } .subtitle { font-size:20px; } input#login_username, input#login_password { height:30px; width:300px; font-size:24px; } input#login_submit { height:30px; width:150px; font-size:16px; } .timetable_cell_lesson { width:100px; font-size:10px; } .timetable_cell_tutorial_a, .timetable_cell_tutorial_b, .timetable_cell_break, .timetable_cell_lunch { width:100px; background:#999; font-size:10px; }

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  • PHP: table structure

    - by A3efan
    I'm developing a website that has some audio courses, each course can have multiple lessons. I want to display each course in its own table with its different lessons. This is my sql statement: Table: courses id, title Table: lessons id, cid (course id), title, date, file $sql = "SELECT lessons.*, courses.title AS course FROM lessons INNER JOIN courses ON courses.id = lessons.cid GROUP BY lessons.id ORDER BY lessons.id" ; Can someone help me with the PHP code? This is the I code I have written: mysql_select_db($database_config, $config); mysql_query("set names utf8"); $sql = "SELECT lessons.*, courses.title AS course FROM lessons INNER JOIN courses ON courses.id = lessons.cid GROUP BY lessons.id ORDER BY lessons.id" ; $result = mysql_query($sql) or die(mysql_error()); while ($row = mysql_fetch_assoc($result)) { echo "<p><span class='heading1'>" . $row['course'] . "</span> </p> "; echo "<p class='datum'>Posted onder <a href='*'>*</a>, latest update on " . strftime("%A %d %B %Y %H:%M", strtotime($row['date'])); } echo "</p>"; echo "<class id='text'>"; echo "<p>...</p>"; echo "<table border: none cellpadding='1' cellspacing='1'>"; echo "<tr>"; echo "<th>Nr.</th>"; echo "<th width='450'>Lesso</th>"; echo "<th>Date</th>"; echo "<th>Download</th>"; echo "</tr>"; echo "<tr>"; echo "<td>" . $row['nr'] . "</td>"; echo "<td>" . $row['title'] . "</td>"; echo "<td>" . strftime("%d/%m/%Y", strtotime($row['date'])) . "</td>"; echo "<td><a href='audio/" . rawurlencode($row['file']) . "'>MP3</a></td>"; echo "</tr>"; echo "</table>"; echo "<br>"; } ?>

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  • Eclipse jvm.dll error when loading

    - by Dan James Palmer
    Trying to open Eclipse after a couple of months and get this error: So I checked that folder to see if it existed, and it did: I checked my PATH Was correct and it was also correct: When this error first occurred I had 3 Java installations. JRE 7 Update 10, JDK 7 Update 7 32bit and 64 bit. I uninstall ALL and restarted my machine. Eclipse then stated, as expected that I needed a JRE or a JDK. So I downloaded and installed the latest JDK and now I get this error, despite everything appearing to be correct. Any ideas?

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  • Sql script, create a database

    - by Blanca
    Hi! I have the next file: create_mysql.sql DROP DATABASE IF EXISTS playence_media; CREATE DATABASE playence_media; USE playence_media; GRANT ALL PRIVILEGES ON . TO 'media'@'localhost' IDENTIFIED BY 'media' WITH GRANT OPTION; But I don't know how to create this database. I would like to do it with my terminal, no other graphics interfaces. Thanks

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  • Ordering Wordpress posts by most recent comment

    - by James
    I'm wanting to order Wordpress posts by the most recent comment. To the best of my knowledge this isn't possible using the WP_Query object, and would require a custom $wpdb query, which I can easily write. However, I then don't know how to setup the loop to run off this object. Can anyone help?

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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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  • Error with default argument in Source.getLines (Scala 2.8.0 RC1)

    - by Derek
    assuming I running Scala 2.8.0 RC1, the following scala code should print out the content of the file "c:/hello.txt" for ( line<-Source.fromPath( "c:/hello.txt" ).getLines ) println( line ) However, when I run it, I get the following error <console>:10: error: missing arguments for method getLines in class Source; follow this method with `_' if you want to treat it as a partially applied function Error occured in an application involving default arguments. val it = Source.fromPath("c:/hello.scala").getLines From what I understand, Scala should use the default argument "compat.Platform.EOL" for "getLines". I am wondering if I did wrong or is it a bug in scala 2.8 Thanks

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  • social networking website database management

    - by Anup Prakash
    This could be very basic type of question for you! But for me it is very important. 1) How these(orkut, facebook or other) website store the images in server? Options: a) Keeping all the images in database by converting into bytecode/binary. b) Making a new folder for each user and saving photographs according to their library name. c) Or something else which i(Anup) didn't guess yet. Please reply me. Sayiing thanx to see my question and any many many thanx for answering my question.

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  • Preg_match differences?

    - by sky
    Hi, i want to ask, what is the meaning or difference between these two line? if( preg_match_all('/\#([?-?À-ÿ?-??-?a-z0-9\-_]{1,50})/iu', $message, $matches, PREG_PATTERN_ORDER) ) { if( preg_match_all('/\#([?-?a-z0-9\-_\x{4e00}-\x{9fa5}]{1,50})/iu', $message, $matches, PREG_PATTERN_ORDER) ) { and what does the number 3 mean in this line? (Arrow pointing) if( preg_match_all('/\@([a-zA-Z0-9\-_\x{4e00}-\x{9fa5}]{->3,30})/u', $message, $matches, PREG_PATTERN_ORDER) ) { Thanks!

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  • Use Zip to Pre-Populate City/State Form with jQuery AJAX

    - by Paul
    I'm running into a problem that I can solve fine by just submitting a form and calling a db to retrieve/echo the information, but AJAX seems to be a bit different for doing this (and is what I need). Earlier in a form process I ask for the zip code like so: <input type="text" maxlength="5" size="5" id="zip" /> Then I have a button to continue, but this button just runs a javascript function that shows the rest of the form. When the rest of the form shows, I want to pre-populate the City input with their city, and pre-populate the State dropdown with their state. I figured I would have to find a way to set city/state to variables, and echo the variables into the form. But I can't figure out how to get/set those variables with AJAX as opposed to a form submit. Here's how I did it without ajax: $zip = mysql_real_escape_string($_POST['zip']); $q = " SELECT city FROM citystatezip WHERE zip = $zip"; $r = mysql_query($q); $row = mysql_fetch_assoc($r); $city = $row['city']; Can anybody help me out with using AJAX to set these variables? Thanks!

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  • Managing Foreign Keys

    - by jwzk
    So I have a database with a few tables. The first table contains the user ID, first name and last name. The second table contains the user ID, interest ID, and interest rating. There is another table that has all of the interest ID's. For every interest ID (even when new ones are added), I need to make sure that each user has an entry for that interest ID (even if its blank, or has defaults). Will foreign keys help with this scenario? or will I need to use PHP to update each and every record when I add a new key?

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  • LINQ expression precedence with Skip(), Take() and OrderBy()

    - by Robert Koritnik
    I'm using LINQ to Entities and display paged results. But I'm having issues with the combination of Skip(), Take() and OrderBy() calls. Everything works fine, except that OrderBy() is assigned too late. It's executed after result set has been cut down by Skip() and Take(). So each page of results has items in order. But ordering is done on a page handful of data instead of ordering of the whole set and then limiting those records with Skip() and Take(). How do I set precedence with these statements? My example (simplified) var query = ctx.EntitySet.Where(/* filter */).OrderBy(/* expression */); int total = query.Count(); var result = query.Skip(n).Take(x).ToList();

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  • PHP / Zend Framework: Force prepend table name to column name in result array?

    - by Brian Lacy
    I am using Zend_Db_Select currently to retrieve hierarchical data from several joined tables. I need to be able to convert this easily into an array. Short of using a switch statement and listing out all the columns individually in order to sort the data, my thought was that if I could get the table names auto-prepended to the keys in the result array, that would solve my problem. So considering the following (assembled) SQL: SELECT user.*, contact.* FROM user INNER JOIN contact ON contact.user_id = user.user_id I would normally get a result array like this: [username] => 'bob', [contact_id] => 5, [user_id] => 2, [firstname] => 'bob', [lastname] => 'larsen' But instead I want this: [user.user_id] => 2, [user.username] => 'bob', [contact.contact_id] => 5, [contact.firstname] => 'bob', [contact.lastname] => 'larsen' Does anyone have an idea how to achieve this? Thanks!

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  • tiny mce sql error when adding links

    - by Anders Kitson
    I am using tiny mce with a script I built for uploading some content to a blog like system. Whenever I add a link via tiny mce I get this error. The field type in mysql for $content which is the one carrying the link is longblob if that helps. here is the link error first and then my code You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'google test" href="http://www.google.ca" target="_blank"google est laborum' at line 4 /* GRAB FORM DATA */ $title = $_POST['title']; $date = $_POST['date']; $content = $_POST['content']; $imageName1 = $_FILES["file"]["name"]; $date = date("Y-m-d"); $sql = "INSERT INTO blog (title,date,content,image)VALUES( \"$title\", \"$date\", \"$content\", \"$imageName1\" )"; $results = mysql_query($sql)or die(mysql_error());

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  • How to select DISTINCT rows without having the ORDER BY field selected

    - by JannieT
    So I have two tables students (PK sID) and mentors (PK pID). This query SELECT s.pID FROM students s JOIN mentors m ON s.pID = m.pID WHERE m.tags LIKE '%a%' ORDER BY s.sID DESC; delivers this result pID ------------- 9 9 3 9 3 9 9 9 10 9 3 10 etc... I am trying to get a list of distinct mentor ID's with this ordering so I am looking for the SQL to produce pID ------------- 9 3 10 If I simply insert a DISTINCT in the SELECT clause I get an unexpected result of 10, 9, 3 (wrong order). Any help much appreciated.

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  • Unique Alpha numeric generator

    - by AAA
    Hi, I want to give our users in the database a unique alpha-numeric id. I am using the code below, will this always generate a unique id? Below is the updated version of the code: old php: // Generate Guid function NewGuid() { $s = strtoupper(md5(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(); echo $Guid; echo "<br><br><br>"; New PHP: // Generate Guid function NewGuid() { $s = strtoupper(uniqid("something",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(); echo $Guid; echo "<br><br><br>"; Will the second (new php) code guarantee 100% uniqueness. Final 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(); echo $Guid; $alphabet = '123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ'; function base_encode($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 $decoded; } echo base_encode($Guid, $alphabet); } So for more stronger uniqueness, i am using the $Guid as the key generator. That should be ok right?

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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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