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  • Jquery Ajaxform with input type=file and multifile

    - by crunchingnumbers
    Im using ajax form with multifile. However ajaxform just seems to quietly die and does not do anything if using multifile. Multifile is just a jquery plugin that builds a list of input type=file so that you can upload multiple files at once, before which you can add/delete your file upload list. I've looked at multifile and made one change to ensure that it generated a unique name as well as id which it didn't do before and i've checked the form elements multifile is building which all appear to be correct. From looking at ajaxform, it seems that it shoulld just generate the iframe as normal and post the multiple input type=file but it does nothing. Has anyone else experienced problems with multiple input type=file uploads

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  • Does retrieving an object from Doctrine2 cause __construct() of the model class to run?

    - by jiewmeng
    When I retrieve an object say by $em->find('Application\Models\User', 1); or other methods like DQL, findBy*() cause the __construct() of the model class to run? I am having a problem where I set variables there like reference to EntityManager and I find that its not set. I tried putting a die() in __construct() and it doesn't halt the application. Can I say that if I want to set other properties/fields like EntityManager $em I have to do it some other way? Perhaps something like protected function getEm() { if (!isset($this->em)) { $this->em = \Zend_Registry::get('em'); } return $this->em; }

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  • mysql_connect()

    - by Jacksta
    I am trying to connect to mysql and am getting an error. I put my servers ip address in and used port 3306 whihch post should be used? <?php $connection = mysql_connect("serer.ip:port", "user", "pass") or die(mysql_error()); if ($connection) {$msg = "success";} ?> <html> <head> </head> <body> <? echo "$msg"; ?> </body> </html> Here is the error its producing Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'admin'@'server1.myserver.com' (using password: YES) in /home/admin/domains/mydomain.com.au/public_html/db_connect.php on line 3 Access denied for user 'admin'@'server1.myserver.com' (using password: YES)

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  • unknown column in where clause

    - by ranzy
    $result = mysql_query("SELECT * FROM Volunteers WHERE Volunteers.eventID = " . $var); $sql = mysql_query("SELECT * FROM Members WHERE Members.pid = " . $temp); I am also doing or die(mysql_error()) at the end of both statements if that matter. My problem is that the first statement executes perfectly but in that table I store an attribute called pid. So the second statement is supposed to take that and return the row where it equals that pid so I can get the name. I get an error that says unknown column in 'a2' in 'where clause' where a2 the pid attribute returned from the first statement. Thanks for any help!

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  • thumbs.db messing up my upload routine

    - by Scott B
    I'm getting the following error while uploading a zip archive. Warning: ZipArchive::extractTo(C:\xampplite\htdocs\testsite/wp-content/themes/mytheme//styles\mytheme/Thumbs.db) [ziparchive.extractto]: failed to open stream: Permission denied in C:\xampplite\htdocs\testsite\wp-content\themes\mythem\uploader.php on line 17 The thing I can't quite figure is that I don't see a thumbs.db file in either the zip archive or the destination folder that was created (the upload still processes, I just get these errors). The function is below, line 17 is commented... function openZip($file_to_open) { global $target; $zip = new ZipArchive(); $x = $zip->open($file_to_open); if($x === true) { $zip->extractTo($target); //this is line 17 $zip->close(); unlink($file_to_open); } else { die("There was a problem. Please try again!"); } }

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  • Form Security (discussion)

    - by Eray Alakese
    I'm asking for brain storming and sharing experience. Which method you are using for form submiting security ? For example , for block automatically sended POST or GET datas, i'm using this method : // Generating random string <?php $hidden = substr(md5(microtime()) ,"-5"); ?> <form action="post.php" .... // assing this random string to a hidden input <input type="hidden" value="<?php echo $hidden;" name="secCode> // and then put this random string to a session variable $_SESSION["secCode"] = $hidden; **post.php** if ($_POST["secCode"] != $_SESSION["secCode"]) { die("You have to send this form, on our web site"); }

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  • Which is more efficient/faster when calling a cached image?

    - by andufo
    Hi, i made an image resizer in php. When an image is resized, it caches a new jpg file with the new dimensions. Next time you call the exact img.php?file=hello.jpg&size=400 it checks if the new jpg has already been created. If it has NOT been created yet, it creates the file and then prints the output (cool). If it ALREADY exists, no new file needs to be generated and instead, it just calls the already cached file. My question is regarding the second scenario. Which of these is faster? redirecting: header('Location: cache/hello_400.jpg');die(); grabbing data and printing the cached file: $data = file_get_contents('cache/hello_400.jpg'); header('Content-type: '.$mime); header('Content-Length: '.strlen($data)); echo $data; Any other ways to improve this?

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  • Why does php show error for my SQL query

    - by ZincX
    UPDATE: My mistake - I made a typo. Nevermind this question. I'm using php to update a mysql database. The resultant query I'm using when i print it out on my webpage before executing is as follows: INSERT INTO perch2_content_items (itemOrder, regionID, pageID, itemRev, itemID, itemJSON, itemSearch ) SELECT MAX(itemOrder)+1, 105, 81, 11, 118, 'json', 'search' FROM perch2_content_items WHERE regionID=105 When I copy and paste this query directly into the phpmyadmin SQL interface, it works fine. The table gets updated. However, when I try to execute it using my php code as follows, it throws an error. $insertToPerch = "INSERT INTO perch2_content_items (itemOrder, regionID, pageID, itemRev, itemID, itemJSON, itemSearch ) SELECT MAX(itemOrder)+1, $regionID, $pageID, $regionRev, $newItemID, 'json', 'search' FROM perch2_content_items WHERE regionID=$regionID"; mysql_query(insertToPerch) or die(mysql_error()); The error I'm getting is: 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 'insertToPerch' at line 1 Can anybody help me figure out why it is failing.

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  • WPF custom user widgets. Will UI components be standardized?

    - by Andrew Florko
    There are lots of articles and video lessons that describe how to create your unique user widget (graphical control) with WPF. There are tons of technical details what is behind the scene and I feel people enthusiasm with ability to customize widgets as never before. I remember those days when VCL library (Delphi) appeared and there was the same enthusiasm in VCL widgets area. Ability to create VCL controls was nearly the must when you was applying for a job as Delphi developer. This situation continued for several years till professional sophisticated 3'd party UI libraries appeared. Hardly you'll have to create your own VCL widget nowadays. Will WPF widgets enthusiasm die as VCL one?

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  • Select distinct... in fulltext search

    - by lam3r4370
    <?php session_start(); $user =$_GET['user']; $conn = mysql_connect("localhost","...","..."); mysql_select_db("..."); $sql= "SELECT filter FROM userfilter WHERE user='$user'"; $mksql = mysql_query($sql); while($row =mysql_fetch_assoc($mksql)) { $filter=$row['filter']; $sql2 = "SELECT DISTINCT * FROM rss WHERE MATCH(content,title) AGAINST ('$filter')"; $mksql2 = mysql_query($sql2) or die(mysql_error()); while($rows=mysql_fetch_assoc($mksql2)) { echo ..... } ?> If I have two rows content that contains the $filter ,it outputs me that content but it's repeating. For example: title|content asd |This is a sample content ,number one das |This is a sample content ,number two .... And if my keywords are "sample" and "number" ,it outputs me twice the title and the content.How to prevent that?

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  • receive xml file via post in php

    - by thegunner
    Hi, I'm looking for a PHP script that can accept an XML file via a POST, then send a response.... Does anyone have any code that could do this? So far the only code I have is this but not sure about the response or if indeed I am even going in the right direction as XML characters are not saved correctly. Any ideas? <?php if ( $_SERVER['REQUEST_METHOD'] === 'POST' ){ $postText = file_get_contents('php://input'); } $datetime=date('ymdHis'); $xmlfile = "myfile" . $datetime . ".xml"; $FileHandle = fopen($xmlfile, 'w') or die("can't open file"); fwrite($FileHandle, $postText); fclose($FileHandle); ?> Thanks,

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  • What's the best way to tell if a file exists in a directory?

    - by Nano HE
    I'm trying to move a file but I want to ensure that it exists before I do so. What's the simplest way to do this in Perl? My code is like this. I looked up the open command, but I am not sure it is the simplest way or not. if #Parser.exe exist in directory of Debug { move ("bin/Debug/Parser.exe","Parser.exe"); } elsif #Parser.exe exist in directory of Release { move ("bin/Release/Parser.exe","Parser.exe"); } else { die "Can't find the Parser.exe."; } Thank you.

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  • PHP file copy to another server; Access filesystem on other server

    - by dclowd9901
    I'm trying to write a PHP script to copy the files from your local machine to a server: $destination_directory = 'I:\path\to\file\\' . $theme_number; if(!@opendir($desination_directory)) { echo 'Sorry, the destination directory could not be found.'; die(); } I check the access to the destination folder with that process, and I keep getting the error return. Anyone know what I'm doing wrong? I pretty much have everything else in place. I just don't know how to access this other server.

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  • why this condition not work i put the full code link plz help

    - by migo
    I've noticed that the first condition does not work if (empty($ss)) { echo "please write your search words"; } but the second does else if ($num < 1) { echo "not found any like "; full code <?php require_once "conf.php"; $ss= $_POST["ss"]; $sql2=("SELECT * FROM student WHERE snum = $ss"); $rs2 = mysql_query($sql2) or die(mysql_error()); $num = mysql_num_rows($rs2); if (empty($ss)) { echo "please write your search words"; } else if ($num < 1 ) { echo "not found any like "; }else { $sql=("SELECT * FROM student WHERE snum = $ss "); $rs = mysql_query($sql) or die(mysql_error()); while($data=mysql_fetch_array($rs)) { ?> <div id="name"> <table align="center" border="3" bgcolor="#FF6666"> <tr> <td><? echo $data ["sname"]." "."????? ??????"; ?></td> </tr> </table> </div> <div id="ahmed"> <table width="50%" height="50" align="center" border="2px" bgcolor="#BCD5F8"> <tr> <td width="18%"><strong>???????</strong></td> <td width="13%"><strong>?????</strong></td> <td width="13%"><strong>?????</strong></td> <td width="14%"><strong>????</strong></td> <td width="12%"><strong>????</strong></td> <td width="30%"><strong>??????</strong></td> </tr> <tr> <td>100</td> <td>100</td> <td>100</td> <td>100</td> <td>100</td> <td><strong>?????? ????????</strong></td> </tr> <td><? echo $data['geo']; ?></td> <td><? echo $data['snum']; ?></td> <td><? echo $data['math']; ?></td> <td><? echo $data['arab']; ?></td> <td><? echo $data['history']; ?></td> <td><strong>????? ??????</strong></td> </tr> <tr> <td colspan="5" align="center" valign="middle"> <? $sum= $data['geo'] + $data['snum'] + $data['math'] + $data['arab'] + $data['history']; echo $sum ; ?> </td> <td><strong>????? ???????</strong></td> </tr> <tr> <td colspan="5" align="center"> <? $all=500 ; $sum= $data['geo'] + $data['snum'] + $data['math'] + $data['arab'] + $data['history']; $av=$sum/$all*100 ; echo $av."%" ; ?> </td> <td><strong> ?????? ??????? </strong></td> </tr> </table> </tr> </div> <? } }; the full code link is http://www.mediafire.com/?2d4yzdjiym0

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  • How could I catch an "Unicode non-character"-warning?

    - by sid_com
    How could I catch the "Unicode non-character 0xffff is illegal for interchange"-warning? #!/usr/bin/env perl use warnings; use 5.012; use Try::Tiny; use warnings FATAL => qw(all); my $character; try { $character = "\x{ffff}"; } catch { die "---------- caught error ----------\n"; }; say "something"; Output: # Unicode non-character 0xffff is illegal for interchange at ./perl1.pl line 11.

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  • mysql_query where statment help

    - by Anders Kitson
    I am retrieving values from the url with the GET method and then using a if statement to determine of they are there then query them against the database to only show those items that match them, i get an unknown error with your request. here is my code $province = $_GET['province']; $city = $_GET['city']; if(isset($province) && isset($city) ) { $results3 = mysql_query("SELECT * FROM generalinfo WHERE province = $province AND city = $city ") or die( "An unknown error occurred with your request"); } else { $results3 = mysql_query("SELECT * FROM generalinfo"); } /*if statement ends*/

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  • mysql join default value

    - by andy
    I've been trying to use the IsNull() function to ensure that there is a value for a field. $result = mysql_query(" SELECT crawled.id,IsNull(sranking.score,0) as Score,crawled.url,crawled.title,crawled.blurb FROM crawled LEFT JOIN sranking ON crawled.id = sranking.sid WHERE crawled.body LIKE '%".$term."%' ORDER BY Score DESC LIMIT " . $start . "," . $c . " ") or die(mysql_error()); But I get the error message:Incorrect parameter count in the call to native function 'IsNull' Anybody have any ideas? I'm pretty new to mySQL.

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  • Json / Jsonp not connecting to php (Phonegap + jquerymobile)

    - by Madhulika Mukherjee
    I am trying to make - an android WEB application with phonegap layout with JqueryMobile What Im doing - An html form that takes ID, name, and address as input 'Serialize's this data using ajax makes a json object out of it Should send it to a file called 'connection.php' Where, this data is put into a database (MySql) Other details - My server is localhost, Im using xampp I have already created a database and table using phpmyadmin The problem - My html file, where my json object is created, does not connect to the php file which is hosted by my localhost Here is my COMPLETE html file: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd"> <html> <head> <!-- Change this if you want to allow scaling --> <meta name="viewport" content="width=default-width; user-scalable=no" /> <meta http-equiv="Content-type" content="text/html;charset=utf-8"> <title>Trial app</title> <link rel="stylesheet" href="thestylesheet.css" type="text/css"> <script type="text/javascript" charset="utf-8" src="javascript1.js"></script> <script type="text/javascript" charset="utf-8" src="javascript2.js"></script> <script type="text/javascript" charset="utf-8" src="cordova-1.8.0.js"></script> <script> $(document).ready(function () { $("#btn").click( function() { alert('hello hello'); $.ajax({ url: "connection.php", type: "POST", data: { id: $('#id').val(), name: $('#name').val(), Address: $('#Address').val() }, datatype: "json", success: function (status) { if (status.success == false) { alert("Failure!"); } else { alert("Success!"); } } }); }); }); </script> </head> <body> <div data-role="header"> <h1>Heading of the app</h1> </div><!-- /header --> <div data-role="content"> <form id="target" method="post"> <label for="id"> <input type="text" id="id" placeholder="ID"> </label> <label for="name"> <input type="text" id="name" placeholder="Name"> </label> <label for="Address"> <input type="text" id="Address" placeholder="Address"> </label> <div id="btn" data-role="button" data-icon="star" data-theme="e">Add record</div> <!--<input type="submit" value="Add record" data-icon="star" data-theme="e"> --> </form> </div> </body> </html> And here is my 'connection.php' hosted by my localhost <?php header('Content-type: application/json'); $server = "localhost"; $username = "root"; $password = ""; $database = "jqueryex"; $con = mysql_connect($server, $username, $password); if($con) { echo "Connected to database!"; } else { echo "Could not connect!"; } //or die ("Could not connect: " . mysql_error()); mysql_select_db($database, $con); /* CREATE TABLE `sample` ( `id` int(11) unsigned NOT NULL AUTO_INCREMENT, `name` varchar(45) DEFAULT NULL, `Address` varchar(45) DEFAULT NULL, PRIMARY KEY (`id`) ) */ $id= json_decode($_POST['id']); $name = json_decode($_POST['name']); $Address = json_decode($_POST['Address']); $sql = "INSERT INTO sample (id, name, Address) "; $sql .= "VALUES ($id, '$name', '$Address')"; if (!mysql_query($sql, $con)) { die('Error: ' . mysql_error()); } else { echo "Comment added"; } mysql_close($con); ?> My doubts: No entry is made in my table 'sample' when i view it in phpmyadmin So obviously, i see no success messages either I dont get any errors, not from ajax and neither from the php file. Stuff Im suspecting: Should i be using jsonp instead of json? Im new to this. Is there a problem with my php file? Perhaps I need to include some more javascript files in my html file? I assume this is a very simple problem so please help me out! I think there is just some conceptual error, as i have only just started with jquery, ajax, and json. Thank you.

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  • PHP How to access constant defined outside class?

    - by Ashley Ward
    I have defined some constants eg: define('DB_HOSTNAME', 'localhost', true); define('DB_USERNAME', 'root', true); define('DB_PASSWORD', 'root', true); define('DB_DATABASE', 'authtest', true); now when I try to do this: class Auth{ function AuthClass() { $this->db_link = mysql_connect(DB_HOSTNAME, DB_USERNAME, DB_PASSWORD) or die(mysql_error()); } } I get an error. Why is this and what do I need to do? See, I've tried using (for example) global DB_HOSTNAME but this fails with an error.

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  • Rails: translate ActiveRecord error template headers for a single model

    - by Chris
    Hi, I'm trying to rename the authlogic error messages in a Rails 3 app. The general format I found out working in Rails 3: de: errors: template: header: one: "Konnte {{model}} nicht speichern: ein Fehler." other: "Konnte {{model}} nicht speichern: {{count}} Fehler." body: "Bitte überprüfen Sie die folgenden Felder: But I want to change this for the authlogic user session model (and only for this one) because when the Login fails, the message "Could not save user session" does not make very much sense. How can I do that?

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  • splitting on all kind of spaces and tabs

    - by rockyurock
    hello, i am reading some parameters(from user input) from a .txt file and want to make sure that my script could read it even a space or tab is left before that particular parameter by user. also if i want to add a comment for each parameter followed by # , after the parameter (e.g 7870 #this is default port number) to let the user know about the parameter how can i achieve it in same file ? here is wat i am using (/\|\s/) code:: $data_file="config.txt"; open(RAK, $data_file)|| die("Could not open file!"); @raw_data=; @Ftp_Server =split(/\|\s/,$raw_data[32]); config.txt (user input file) PING_TTL | 1 CLIENT_PORT | 7870 FTP_SERVER | 192.162.522.222 could any body suggest me a robust way to do it? /rocky

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  • having trouble with a mysql query

    - by chuck akers
    this keeps saying Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in directory here the error is near the login_query variable, can someone help me fix it. <?php if (isset($_POST['login_username'], $_POST['login_password'])) { $login_username = trim(mysql_real_escape_string(htmlentities($_POST['login_username']))); $login_password = md5(trim(mysql_real_escape_string(htmlentities($_POST['login_password'])))); if (!empty($login_username) && !empty($login_password)) { $login_query = mysql_query("SELECT user_id FROM username WHERE username='".$login_username."' AND password='".$login_password."'"); if (mysql_num_rows($login_query)==1) { $user_id = mysql_result($login_query, 0, 'user_id'); $_SESSION['user_id'] = $user_id; header('Location: index.php'); die(); } }} ?

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  • Is GOTO really as evil as we are led to believe?

    - by RoboShop
    I'm a young programmer, so all my working life I've been told GOTO is evil, don't use it, if you do, your first born son will die. Recently, I've realized that GOTO actually still exists in .NET and I was wondering, is GOTO really as bad as they say, or is it just because everyone says you shouldn't use it, so that's why you don't. I know GOTO can be used badly, but are there any legit situations where you may possibly use it. The only thing I can think of is maybe to use GOTO to break out of a bunch of nested loops. I reckon that might be better then having to "break" out of each of them but because GOTO is supposedly always bad, I would never use it and it would probably never pass a peer review. What are your views? Is GOTO always bad? Can it sometimes be good? Has anyone here actually been gutsy enough to use GOTO for a real life system?

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  • mysqli prepare statment error?

    - by user310850
    Hi all, $mysqli = new mysqli("localhost", "root", "", "test"); $mysqli->query('PREPARE mid FROM "SELECT name FROM test_user WHERE id = ?"'); //$mysqli->query('PREPARE mid FROM "SELECT name FROM test_user" '); $res = $mysqli->query( 'EXECUTE mid 1;') or die(mysqli_error($mysqli)); while($resu = $res->fetch_object()) { echo '<br>' .$resu->name; } Error: 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 '1' at line 1 my php version is PHP Version 5.3.0 and mysql mysqlnd 5.0.5-dev - 081106 - $Revision: 1.3.2.27 $

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  • connect perl with mssql

    - by Bharanikumar
    i have user id, password,databasename, datasource details, i want to connect perl with mssql server, i just used following snippet, but getting error, #!/usr/bin/perl -w use strict; use DBI; my $data_source = q/dbi:ODBC:192.168.3.137/; my $user = q/bharani/; my $password = q/123456/; # Connect to the data source and get a handle for that connection. my $dbh = DBI->connect($data_source, $user, $password) or die "Can't connect to $data_source: $DBI::errstr"; My Error DBI connect('192.168.3.137','bharani',...) failed: [Microsoft][ODBC Driver Manag er] Data source name not found and no default driver specified (SQL-IM002) at my sqlconnect.pl line 14 Can't connect to dbi:ODBC:192.168.3.137: [Microsoft][ODBC Driver Manager] Data s ource name not found and no default driver specified (SQL-IM002) at mysqlconnect .pl line 14. Information: my sql server present in another system, am just trying to connect with above details, plz tellme, i should crease DSN in my system, or anything i missed in my program

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