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  • Wordpress - Total User Count who only have posts

    - by knightrider
    I want to display total number of user who only have posts at Wordpress. I can get all users by this query <?php $user_count = $wpdb->get_var("SELECT COUNT(*) FROM $wpdb->users;"); echo $user_count ?> But for the user count only with posts, i think i might need to join another table, does anyone have snippets ? Thanks.

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  • openssl error wehn compiling Ruby from source

    - by Florian Salihovic
    Prelude: I don't want to use rvm. I installed ruby 2 with the following configuration on Mac OS X 10.8.5 ./configure --prefix=/usr/local \ --enable-pthread \ --with-readline-dir=/usr/local \ --enable-shared It is installed, version is printed correctly ... Now, when invoking gem install jekyll I get the following error: ERROR: Loading command: install (LoadError) cannot load such file -- openssl ERROR: While executing gem ... (NoMethodError) undefined method `invoke_with_build_args' for nil:NilClass I installed openssl into /usr/local but i'm really banging my head against the wall on how installing gems. It can't be that big of a deal right?

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  • ASP.net error message when using REST starter kit

    - by jonhobbs
    Hi all, I've written some code using the REST starter kit and it works fine on my development machine. However, when I upload it to our server the page gives me the following error message... CS1684: Warning as Error: Reference to type 'System.Runtime.Serialization.Json.DataContractJsonSerializer' claims it is defined in 'c:\WINNT\assembly\GAC_MSIL\System.ServiceModel.Web\3.5.0.0__31bf3856ad364e35\System.ServiceModel.Web.dll', but it could not be found I've removed code line by line and it appears that the following line of code is triggering the error... HttpContent newOrganizationContent = HttpContentExtensions.CreateXmlSerializable(newOrganizationXml); Really haven't got a clue how to fix it. I assumed it might be because it needs a newer version of the framework to run, but looking in IIS it says it's running version 2.0.50727 which I think is the lates version because it says that even when we're using framework 3.5 Very confused, any ideas? Jon

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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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  • sql query question / count

    - by scheibenkleister
    Hi, I have houses that belongs to streets. A user can buy several houses. How do I find out, if the user owns an entire street? street table with columns (id/name) house table with columns (id/street_id [foreign key] owner table with columns (id/house_id/user_id) [join table with foreign keys] So far, I'm using count which returns the result: select count(*), street_id from owner left join house on owner.house_id = house.id group by street_id where user_id = 1 count(*) | street_id 3 | 1 2 | 2 A more general count: select count(*) from house group by street_id returns: count(*) | street_id 3 | 1 3 | 2 How can I find out, that user 1 owns the entire street 1 but not street 2? Thanks.

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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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  • INSERT INTO ... SELECT FROM ... ON DUPLICATE KEY UPDATE

    - by dnagirl
    I'm doing an insert query where most of many columns would need to be updated to the new values if a unique key already existed. It goes something like this: INSERT INTO lee(exp_id, created_by, location, animal, starttime, endtime, entct, inact, inadur, inadist, smlct, smldur, smldist, larct, lardur, lardist, emptyct, emptydur) SELECT id, uid, t.location, t.animal, t.starttime, t.endtime, t.entct, t.inact, t.inadur, t.inadist, t.smlct, t.smldur, t.smldist, t.larct, t.lardur, t.lardist, t.emptyct, t.emptydur FROM tmp t WHERE uid=x ON DUPLICATE KEY UPDATE ...; //update all fields to values from SELECT, except for exp_id, created_by, location, animal, starttime, endtime I'm not sure what the syntax for the UPDATE clause should be. How do I refer to the current row from the SELECT clause?

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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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  • Is it possible to LIMIT results from a JOIN query?

    - by Arms
    I've got a query that currently queries a Post table while LEFT JOINing a Comment table. It fetches all Posts and their respective Comments. However, I want to limit the number of Comments returned. I tried adding a sub-select, but ran into errors if I didn't LIMIT the results to 1. I'm really not sure how to go about this while still using only one query. Is this possible?

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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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  • 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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  • Procedure in converting int to decimal data type?

    - by Fedor
    I have an int(11) column which is used to store money. I read some of the answers on SO and it seems I just need to update it to be a decimal (19,4) data type. Are there any gotchas I should know about before I actually do the converting? My application is in PHP/Zend and I'm not using an ORM so I doubt I would need to update any sort of class to consistently identify the data type.

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  • Minimizing SQL queries using join with one-to-many relationship

    - by Brian
    So let me preface this by saying that I'm not an SQL wizard by any means. What I want to do is simple as a concept, but has presented me with a small challenge when trying to minimize the amount of database queries I'm performing. Let's say I have a table of departments. Within each department is a list of employees. What is the most efficient way of listing all the departments and which employees are in each department. So for example if I have a department table with: id name 1 sales 2 marketing And a people table with: id department_id name 1 1 Tom 2 1 Bill 3 2 Jessica 4 1 Rachel 5 2 John What is the best way list all departments and all employees for each department like so: Sales Tom Bill Rachel Marketing Jessica John Pretend both tables are actually massive. (I want to avoid getting a list of departments, and then looping through the result and doing an individual query for each department). Think similarly of selecting the statuses/comments in a Facebook-like system, when statuses and comments are stored in separate tables.

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  • Compare structures of two databases?

    - by streetparade
    Hello, I wanted to ask whether it is possible to compare the complete database structure of two huge databases. We have two databases, the one is a development database, the other a production database. I've sometimes forgotten to make changes in to the production database, before we released some parts of our code, which results that the production database doesn't have the same structure, so if we release something we got some errors. Is there a way to compare the two, or synchronize?

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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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  • Preventing spam bots on site?

    - by Mike
    We're having an issue on one of our fairly large websites with spam bots. It appears the bots are creating user accounts and then posting journal entries which lead to various spam links. It appears they are bypassing our captcha somehow -- either it's been cracked or they're using another method to create accounts. We're looking to do email activation for the accounts, but we're about a week away from implementing such changes (due to busy schedules). However, I don't feel like this will be enough if they're using an SQL exploit somewhere on the site and doing the whole cross site scripting thing. So my question to you: If they are using some kind of XSS exploit, how can I find it? I'm securing statements where I can but, again, its a fairly large site and it'd take me awhile to actively clean up SQL statements to prevent XSS. Can you recommend anything to help our situation?

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  • WordPress on other parts of my site

    - by SHiNKiROU
    I have a WordPress installation on my site, and I want to display WP posts on other parts of my site (that is outside the WP installation). How do I do that with PHP? I tried to search this type of question on Stack Overflow, Google and WP official site but I didn't find anything.

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  • How can I prevent a double submit with jQuery or Javascript?

    - by kielie
    Hi guys, I keep getting duplicate entries in my database because of impatient users clicking the submit button multiple times. I googled and googled and found a few scripts, but none of them seem to be sufficient. How can I prevent these duplicate entries from occurring using javascript or preferably jQuery? Thanx in advance!

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  • to take values of checkbox in table attributes

    - by mwj
    i have a database patient with 3-4 tables n each table has about 8 attributes.... i have a table medical history which has attribute additional info ... under which i have 5 checkboxes.... all the values entered are taken up except the chekbox values..... plz help

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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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  • 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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  • User Getting Logged Out After Making First Comment

    - by John
    Hello, I am using a login system that works well. I am also using a comment system. The comment function does not show up unless the user is logged in (as shown in commentformonoff.php below). When a user makes a comment, the info is passed from the function "show_commentbox" to the file comments2a.php. Then, the info is passed to the file comments2.php. When the site is first pulled up on a browser, after logging in and making a comment, the user is logged out. After logging in a second time during the same browser session, the user is no longer logged out after making a comment. How can I keep the user logged in after making the first comment? Thanks in advance, John Commentformonoff.php: <?php if (!isLoggedIn()) { if (isset($_POST['cmdlogin'])) { if (checkLogin($_POST['username'], $_POST['password'])) { show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl); } else { echo "<div class='logintocomment'>Login to comment</div>"; } } else { echo "<div class='logintocomment'>Login to comment</div>"; } } else { show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl); } ?> Function "show_commentbox": function show_commentbox($submissionid, $submission, $url, $submittor, $submissiondate, $countcomments, $dispurl) { echo '<form action="http://www...com/.../comments/comments2a.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <input type="hidden" value="'.$_SESSION['username'].'" name="u"> <input type="hidden" value="'.$submissionid.'" name="submissionid"> <input type="hidden" value="'.stripslashes($submission).'" name="submission"> <input type="hidden" value="'.$url.'" name="url"> <input type="hidden" value="'.$submittor.'" name="submittor"> <input type="hidden" value="'.$submissiondate.'" name="submissiondate"> <input type="hidden" value="'.$countcomments.'" name="countcomments"> <input type="hidden" value="'.$dispurl.'" name="dispurl"> <label class="addacomment" for="title">Add a comment:</label> <textarea class="checkMax" name="comment" type="comment" id="comment" maxlength="1000"></textarea> <div class="commentsubbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; } Included in comments2a.php: $uid = mysql_real_escape_string($_POST['uid']); $u = mysql_real_escape_string($_POST['u']); $query = sprintf("INSERT INTO comment VALUES (NULL, %d, %d, '%s', NULL)", $uid, $subid, $comment); mysql_query($query) or die(mysql_error()); $lastcommentid = mysql_insert_id(); header("Location: comments2.php?submission=".$submission."&submissionid=".$submissionid."&url=".$url."&submissiondate=".$submissiondate."&comment=".$comment."&subid=".$subid."&uid=".$uid."&u=".$u."&submittor=".$submittor."&countcomments=".$countcomments."&dispurl=".$dispurl."#comment-$lastcommentid"); exit(); Included in comments2.php: if($_SERVER['REQUEST_METHOD'] == "POST"){header('Location: http://www...com/.../comments/comments2.php?submission='.$submission.'&submissionid='.$submissionid.'&url='.$url.'&submissiondate='.$submissiondate.'&submittor='.$submittor.'&countcomments='.$countcomments.'&dispurl='.$dispurl.'');} $uid = mysql_real_escape_string($_GET['uid']); $u = mysql_real_escape_string($_GET['u']);

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