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  • current_date casting

    - by Armen Mkrtchyan
    Hi. string selectSql = "update " + table + " set state_" + mode + "_id=1 WHERE stoping_" + mode + " < current_date;"; when i call current_date, it return yyyy-MM-dd format, but i want to return dd.MM.yyyy format, how can i do that. please help. my program works fine when i am trying string selectSql = "update " + table + " set state_" + mode + "_id=1 WHERE stoping_" + mode + " < '16.04.2010';";

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  • adding other parameter to function

    - by Ronnie Chester Lynwood
    hello. i got a function that listing downloads in a table with foreach. it's also lists searched term, search type. public function fetchDownloads($displaySite=true) { $downloads = array(); $sqlWhere = ""; if(isset($this->q)) { if(strlen($this->q) <= $this->recents_length && !empty($this->q)) { $insertRecent = $this->processDataHook("insertRecent",$this->q); if($insertRecent) { if(!@mysql_query("INSERT INTO wcddl_recents (query) VALUES ('".$this->qSQL."')")) { @mysql_query("UPDATE wcddl_recents SET searches = searches+1 WHERE query = '".$this->qSQL."'"); } } } if($this->search_type == "narrow") { $sqlWhere = " WHERE title LIKE '%".mysql_real_escape_string(str_replace(" ","%",$this->q))."%'"; } elseif($this->search_type == "wide") { $qExp = explode(" ",$this->q); $sqlWhere = array(); foreach($qExp as $fq) $sqlWhere[] = "title LIKE '%".mysql_real_escape_string($fq)."%'"; $sqlWhere = implode(" OR ",$sqlWhere); $sqlWhere = " WHERE (".$sqlWhere.")"; } } if(isset($this->type)) { if(!empty($sqlWhere)) { $sqlWhere .= " AND type = '".$this->typeSQL."'"; } else { $sqlWhere = " WHERE type = '".$this->typeSQL."'"; } } $sqlWhere = $this->processDataHook("fetchDownloadsSQLWhere",$sqlWhere); $this->maxPages = mysql_query("SELECT COUNT(*) FROM wcddl_downloads".$sqlWhere.""); $this->maxPages = mysql_result($this->maxPages,0); $this->numRows = $this->maxPages; $this->maxPages = ceil($this->maxPages/$this->limit); $sqlMain = "SELECT id,sid,title,type,url,dat,views,rating FROM wcddl_downloads".$sqlWhere." ORDER BY ".(isset($this->sqlOrder) ? mysql_real_escape_string($this->sqlOrder) : "id DESC")." LIMIT ".$this->pg.",".$this->limit.""; $sqlMain = $this->processDataHook("whileFetchDownloadsSQL",$sqlMain); $sqlMain = mysql_query($sqlMain); $this->processHook("whileFetchDownloads"); while($row = mysql_fetch_assoc($sqlMain)) { if($displaySite) { $downloadSite = mysql_query("SELECT name as sname, url as surl, rating as srating FROM wcddl_sites WHERE id = '".$row['sid']."'"); $downloadSite = mysql_fetch_assoc($downloadSite); $row = array_merge($row,$downloadSite); } $downloads_current = $this->mapit($row,array("stripslashes","strip_tags")); $downloads_current = $this->processDataHook("fetchDownloadsRow",$downloads_current); $downloads[] = $downloads_current; } $this->pageList = $this->getPages($this->page,$this->maxPages); $this->processHook("endFetchDownloads"); return $downloads; } I want to add if $_REQUEST['site'] is set, order downloads by sname that catching from wcddl_sites.

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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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  • 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 to get rank based on SUM's?

    - by Kenan
    I have comments table where everything is stored, and i have to SUM everything and add BEST ANSWER*10. I need rank for whole list, and how to show rank for specified user/ID. Here is the SQL: SELECT m.member_id AS member_id, (SUM(c.vote_value) + SUM(c.best)*10) AS total FROM comments c LEFT JOIN members m ON c.author_id = m.member_id GROUP BY c.author_id ORDER BY total DESC LIMIT {$sql_start}, 20

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  • Need to map classes to different databases at runtime in Hibernate

    - by serg555
    I have MainDB database and unknown number (at compile time) of UserDB_1, ..., UserDB_N databases. MainDB contains names of those UserDB databases in some table (new UserDB can be created at runtime). All UserDB have exactly the same table names and fields. How to handle such situation in Hibernate? (database structure cannot be changed). Currently I am planning to create generic User classes not mapped to anything and just use native SQL for all queries: session.createSQLQuery("select * from " + db + ".user where id=1") .setResultTransformer(Transformers.aliasToBean(User.class)); Is there anything better I can do? Ideally I would want to have mappings for UserDB tables and relations and use HQL on required database.

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  • Database design: Using hundred of fields for little values

    - by user964260
    I'm planning to develop a PHP Web App, it will mainly be used by registered users(sessions) While thinking about the DB design, I was contemplating that in order to give the best user experience possible there would be lots of options for the user to activate, deactivate, specify, etc. For example: - Options for each layout elements, dialog boxes, dashboard, grid, etc. - color, size, stay visible, invisible, don't ask again, show everytime, advanced mode, simple mode, etc. This would get like 100s of fields ranging from simple Yes/No or 1 to N values..., for each user. So, is it having a field for each of these options the way to go? or how do those CRMs or CMS or other Web Apps do it to store lots of 1-2 char long values? Do they group them on Text fields separated by a special char and then "explode" them as an array for runtime usage? thank you

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  • Set primary key with two integers

    - by user299196
    I have a table with primary key (ColumnA, ColumnB). I want to make a function or procedure that when passed two integers will insert a row into the table but make sure the largest integer always goes into ColumnA and the smaller one into ColumnB. So if we have SetKeysWithTheseNumbers(17, 19) would return |-----------------| |ColumnA | ColumnB| |-----------------| |19 | 17 | |-----------------| SetKeysWithTheseNumbers(19, 17) would return the same thing |-----------------| |ColumnA | ColumnB| |-----------------| |19 | 17 | |-----------------|

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  • Php INNER JOING jqGrid help

    - by yanike
    I'm trying to get INNER JOIN to work with JQGRID, but I can't get it working. I want the code to get the first_name and last_name from members using the "efrom" from messages that matches the "id" from members. $col = array(); $col["title"] = "From"; $col["name"] = "messages.efrom"; $col["width"] = "70"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "First Name"; $col["name"] = "members.first_name"; $col["width"] = "80"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Last Name"; $col["name"] = "members.last_name"; $col["width"] = "80"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Subject"; $col["name"] = "messages.esubject"; $col["width"] = "300"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $col = array(); $col["title"] = "Date"; $col["name"] = "messages.edatetime"; $col["width"] = "150"; $col["hidden"] = false; $col["editable"] = false; $col["sortable"] = true; $col["search"] = true; $cols[] = $col; $g = new jqgrid(); $grid["sortname"] = 'messages.edatetime'; $g->select_command = "SELECT messages.efrom, messages.esubject, messages.edatetime, members.first_name, members.last_name FROM messages INNER JOIN members ON messages.efrom = members.id";

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  • Weirdest PHP error

    - by yes123
    I found in my error log this [12-Nov-2011 19:22:26] PHP Warning: Attempt to assign property of non-object in on line 1768776801 [12-Nov-2011 19:22:31] PHP Warning: Attempt to assign property of non-object in on line 1768776801 [12-Nov-2011 19:22:39] PHP Warning: Attempt to assign property of non-object in on line 1768776801 [12-Nov-2011 19:22:40] PHP Warning: Attempt to assign property of non-object in on line 1768776801 [12-Nov-2011 19:22:46] PHP Warning: Attempt to assign property of non-object in on line 1768776801 [12-Nov-2011 19:22:51] PHP Warning: Attempt to assign property of non-object in on line 1768776801 Of course I don't have any script with 1 bn lines. PHP Version 5.3.3-7 Apache 2 The other weird thing is that I have a set_error_handler( 'myHandler' ); To write in the error log other inforamtion too, but with this error it seems PHP just ignores my error_handler =/ I don't have any code that can generate this errore before my call to set_error_handler

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  • Jetty 6: Unknown Error 99

    - by Silvio Donnini
    The system I'm developing is comprised of a jetty server (v6.1.2rc4) and a php frontend that sends http requests to jetty via curl_exec. The server and the client are on the same machine. The requests I send can be both POSTs and GETs, I get the same error for either which is: Failed to connect to 127.0.0.1: Unknown error 99 This is rather cryptic. It seems that after the first problematic request, some of the following (unrelated) requests also get corrupted. It looks like jetty is simply refusing the connection, but I can't read more than that into the error message. I thought it was a problem with the server's configuration, so I tried changing jetty's maxIdleTimeMs, but without success. Any idea about what to do is welcome thanks, Silvio

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  • Oracle XE + ODP.NET ANNOYING VIEW ERROR

    - by Alex
    Sup guys, heres my view: CREATE OR REPLACE VIEW SISTEMA.VWTELA AS SELECT TEL_DLTELA AS Tela, TEL_DLDESCRICAO As Descricao, TEL_DLTABELA As Tabela, CASE WHEN to_char(TEL_STATIVO) = to_char(1) THEN to_char('Yes') ELSE to_char('No') END as Ativo, TEL_IDTELA AS IDTEL FROM SISTEMA.TEL_TELA; When i do a SELECT * FROM SISTEMA.VWTELA it works fine from PL/SQL Developer but when i launch the query from my VB.NET application it throws me a super annoying error ORA-01722. Any ideas? The Application code works perfecty with any query so its not application code error but prolly some "super cool feature" from ODP.NET. Already tried to_number, to_whatever and same error always happens.

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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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  • 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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  • Table not Echoing out if another Table has a Zero value

    - by John
    Hello, The table below with mysql_query($sqlStr3) (the one with the word "Joined" in its row) does not echo if the result associated with mysql_query($sqlStr1) has a value of zero. This happens even if mysql_query($sqlStr3) returns a result. In other words, if a given loginid has an entry in the table "login", but not one in the table "submission", then the table associated with mysql_query($sqlStr3) does not echo. I don't understand why the "submission" table would have any effect on mysql_query($sqlStr3), since the $sqlStr3 only deals with another table, called "login", as seen below. Any ideas why this is happening? Thanks in advance, John W. <?php echo '<div class="profilename">User Profile for </div>'; echo '<div class="profilename2">'.$profile.'</div>'; $tzFrom = new DateTimeZone('America/New_York'); $tzTo = new DateTimeZone('America/Phoenix'); $profile = mysql_real_escape_string($_GET['profile']); $sqlStr = "SELECT l.username, l.loginid, s.loginid, s.submissionid, s.title, s.url, s.datesubmitted, s.displayurl FROM submission AS s INNER JOIN login AS l ON s.loginid = l.loginid WHERE l.username = '$profile' ORDER BY s.datesubmitted DESC"; $result = mysql_query($sqlStr); $arr = array(); echo "<table class=\"samplesrec1\">"; while ($row = mysql_fetch_array($result)) { $dt = new DateTime($row["datesubmitted"], $tzFrom); $dt->setTimezone($tzTo); echo '<tr>'; echo '<td class="sitename3">'.$dt->format('F j, Y &\nb\sp &\nb\sp g:i a').'</a></td>'; echo '<td class="sitename1"><a href="http://www.'.$row["url"].'">'.$row["title"].'</a></td>'; echo '</tr>'; } echo "</table>"; $sqlStr1 = "SELECT l.username, l.loginid, s.loginid, s.submissionid, s.title, s.url, s.datesubmitted, s.displayurl, l.created, count(s.submissionid) countSubmissions FROM submission AS s INNER JOIN login AS l ON s.loginid = l.loginid WHERE l.username = '$profile'"; $result1 = mysql_query($sqlStr1); $arr1 = array(); echo "<table class=\"samplesrec2\">"; while ($row1 = mysql_fetch_array($result1)) { echo '<tr>'; echo '<td class="sitename5">Submissions: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$row1["countSubmissions"].'</td>'; echo '</tr>'; } echo "</table>"; $sqlStr2 = "SELECT l.username, l.loginid, c.loginid, c.commentid, c.submissionid, c.comment, c.datecommented, l.created, count(c.commentid) countComments FROM comment AS c INNER JOIN login AS l ON c.loginid = l.loginid WHERE l.username = '$profile'"; $result2 = mysql_query($sqlStr2); $arr2 = array(); echo "<table class=\"samplesrec3\">"; while ($row2 = mysql_fetch_array($result2)) { echo '<tr>'; echo '<td class="sitename5">Comments: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$row2["countComments"].'</td>'; echo '</tr>'; } echo "</table>"; $tzFrom3 = new DateTimeZone('America/New_York'); $tzTo3 = new DateTimeZone('America/Phoenix'); $sqlStr3 = "SELECT created, username FROM login WHERE username = '$profile'"; $result3 = mysql_query($sqlStr3); $arr3 = array(); echo "<table class=\"samplesrec4\">"; while ($row3 = mysql_fetch_array($result3)) { $dt3 = new DateTime($row3["created"], $tzFrom3); $dt3->setTimezone($tzTo3); echo '<tr>'; echo '<td class="sitename5">Joined: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;'.$dt->format('F j, Y').'</td>'; echo '</tr>'; } echo "</table>"; ?> </body> </html>

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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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  • declaring constraint to consider prog logic

    - by shantanuo
    I can open a trip only once but can close it multiple times. I can not declare the Trip_no + status as primary key since there can be multiple entries while closing the trip. Is there any way that will assure me that a trip number is opened only once? For e.g. there should not be the second row with "Open" status for trip No. 3 since it is already there in the following table. Trip No | Status 1 Open 1 Close 1 Close 2 Open 2 Close 3 Open 3 Close 3 Close 3 Close 3 Close

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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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  • 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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  • Favouriting things in a database - most efficient method of keeping track?

    - by a2h
    I'm working on a forum-like webapp where I'd like to allow users to favourite an item so that they can keep track of it, and also so that others can see how many times an item's been favourited. The problem is, I'm unsure on the best practices for databases, which includes this situation. I have two ideas in my head on how to do this: Add an extra column to the user table and store things like so: "|2|5|73|" Add an extra table with at least two columns, one for referencing an item, the other for referencing a user. I feel uncomfortable about going for the second method as it involves an extra table, and potentially more queries would be required. Perhaps these beliefs aren't an issue, as I have little understanding of databases beyond simply working with table layouts and basic queries.

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  • how to link table to table

    - by Niño Seymour L. Rodriguez
    I am a comsci student and I'm taking up database now. I got a problem in or should I say I dont know how to link table to table. It is not like you'll just use a foreign key and connect it to the primary key. The outcome should be like this: In the table Course there are three fields namely "course_id", "Description" and "subjects". When you click the name field Subject, a table named Subject should appear. Can you help me with this? hope you understnd my grammar, hehe..im not good in english......it will be a big help if you can answer it.........thank you po..............

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  • update myqsl table

    - by Simon
    how can i write the query, to update the table videos, and set the value of field name to 'something' where the average is max(), or UPDATE the table, where average has the second value by size!!! i think the query must look like this!!! UPDATE videos SET name = 'something' WHERE average IN (SELECT `average` FROM `videos` ORDER BY `average` DESC LIMIT 1) but it doesn't work!!!

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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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  • What's wrong with this SQL query?

    - by ThinkingInBits
    I have two tables: photographs, and photograph_tags. Photograph_tags contains a column called photograph_id (id in photographs). You can have many tags for one photograph. I have a photograph row related to three tags: boy, stream, and water. However, running the following query returns 0 rows SELECT p.* FROM photographs p, photograph_tags c WHERE c.photograph_id = p.id AND (c.value IN ('dog', 'water', 'stream')) GROUP BY p.id HAVING COUNT( p.id )=3 Is something wrong with this query?

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