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  • HOW TO FIX "The requested URL /phpMyAdmin was not found on this server"

    - by user1392840
    I have install apache,php and mysql on Mac 10.8.1. After this, in my web brower i type this, it give the error message Not Found The requested URL /News-2012-Academy-Awards-53.html was not found on this server. Apache/2.2.22 (Unix) mod_fastcgi/2.4.6 mod_ssl/2.2.22 OpenSSL/0.9.8r DAV/2 PHP/5.3.13 with Suhosin-Patch mod_wsgi/3.3 Python/2.7.2 Server at clontarf.girlsacademy.com.au Port 80" Please help me to solve it.

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  • How to connect two files and use the radio button?

    - by Stupefy101
    I have here a set of form from the index.php to upload a zip file, select an option then perform a converter process. <form action="" method="post" accept-charset="utf-8"> <p class="buttons"><input type="file" value="" name="zip_file"/></p> </form> <form action="index.php" method="post" accept-charset="utf-8" name="form1"> <h3><input type="radio" name="option" value="option1"/> Option1 </h3> <h3><input type="radio" name="option" value="option2"/> Option2 </h3> <h3><input type="radio" name="option" value="option3"/> Option3 </h3> <p class="buttons"><input type="submit" value="Convert"/></p> </form> In the other hand, this is my code for the upload.php that will extract the Zip file. <?php if($_FILES["zip_file"]["name"]) { $filename = $_FILES["zip_file"]["name"]; $source = $_FILES["zip_file"]["tmp_name"]; $type = $_FILES["zip_file"]["type"]; $name = explode(".", $filename); $accepted_types = array('application/zip', 'application/x-zip-compressed', 'multipart/x-zip', 'application/x-compressed'); foreach($accepted_types as $mime_type) { if($mime_type == $type) { $okay = true; break; } } $continue = strtolower($name[1]) == 'zip' ? true : false; if(!$continue) { $message = "The file you are trying to upload is not a .zip file. Please try again."; } $target_path = "C:xampp/htdocs/themer/".$filename; // change this to the correct site path if(move_uploaded_file($source, $target_path)) { $zip = new ZipArchive(); $x = $zip->open($target_path); if ($x === true) { $zip->extractTo("C:xampp/htdocs/themer/"); // change this to the correct site path $zip->close(); unlink($target_path); } $message = "Your .zip file was uploaded and unpacked."; } else { $message = "There was a problem with the upload. Please try again."; } } ?> How can i connect both files that will perform the extracting process? And how to include the codes for radio button after submission? Please Help.

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  • problem with setting a cookieless domain

    - by Pablo
    Here is the header from firebug that shows the scope of the PHP Session cookie: Set-Cookie PHPSESSID=f0e2dfe56cc78be718c8154ac80d1ae2; path=/; domain=pix-all.com But still the PHP Session cookie is been sent for any requests to static.pix-all.com Cookie PHPSESSID=f0e2dfe56cc78be718c8154ac80d1ae2; What could be the problem?

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  • Good article about File- and Folder Permissions on production server?

    - by Camran
    I have a classifieds website, and users may post classifieds, add images, remove classifieds etc etc... I have no idea what to set the permissions to on folders. For instance, a php script which I have uploads a file to a directory. What would you have set the directory permissions to? Nobody need access to the directory, only the php script... Just wonder if anybody has a good (brief) article about setting the "right" permissions? Thanks

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  • Image edit and mysql

    - by Felicita
    I have a simple table for reference page: id name description image In reference.php, A form upload image to a folder and save image's name in image section. In reference.php?action=edit page I want to edit the image. What is correct way to edit? Uploading another image and update the table? Thanks

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  • exim redirect mail to a script

    - by DevilWAH
    I am looking to set up a mail relay so that any mail sent to for example @example.com gets parsed be a PHP script. I know to do this for indervidual address I could do some thing like (from the following web site http://evolt.org/incoming_mail_and_php) Con figure an Aliases such as script: |/our/script.php then any emails sent to [email protected] will be passed to the script given.. but how can I make it that every email to @example.com is passed to the script? thank you

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  • A way to share the return value of fsockopen() between multiple pages?

    - by Chilln
    Hey, im using a connection to a server in my php script, opened with fsockopen() and i want it to share between different pages so i serialized it and saved it in a session variable but it seems that that ia a bad idea because when i do this nothing happens... Not even an error. The problem is that this connection requires a handshake so i cant reconnect everytime Another question, whats the timeout of fsockopen or does the connection stay alive if the. original php script which called it is closed?

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  • what are best practices on asking user to add the facebook app to their page?

    - by simple
    Hello I am looking for a best way to ask/forward user so he/she adds my app to their page one way is to make them follow the link http://www.facebook.com/add.php?api_key=[your application api key]&pages (http://www.facebook.com/add.php?api_key=1fc2946c634702dfc75cce79c97c8cec&pages -real life example) wrapping up the question: as facebook has made a lot of changes maybe the above method is the outdated one(though it is supported), and is there are any more ways to get same result?

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  • Send multiple fields with the same name via jQuery

    - by Swell
    Hi, how to send multiple fields with the same name via jQuery like this: <input type="file" name="file[]" /> <input type="file" name="file[]" /> <input type="file" name="file[]" /> jQuery: function upload() { $.post('upload.php', { file[]: uplaodForm.file[].value }, function(output) { $('#result').html(output).show(); }); } upload.php: $file[$i] = $_FILES['file']['name'][$i]; thank you,

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  • How would you show a file with cURL?

    - by Kyle
    There is an epic lack of PHP cURL love on the Internet for beginners like me. I was wondering how to use cURL to download & display an ICS file (They're plain text to me...) in my PHP code. Unless fopen() is 1,000 times easier, I'd like to stick with cURL for this one.

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  • MySQL only returning one result. Im Baffled.

    - by Tapha
    Here is te code: <?php //Starting session session_start(); //Includes mass includes containing all the files needed to execute the full script //Also shows homepage elements without customs require_once ('includes/mass.php'); $username = $_SESSION['username']; if (isset($username)) { //Query database for the users networths $sq_l = "SELECT * FROM user"; $sql_query_worth = mysql_query($sq_l); while ($row = mysql_fetch_assoc($sql_query_worth)) { $dbusername = $row['username']; } echo $dbusername; } ?>

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  • form validation without reset

    - by Paulo Bueno
    Hi guys Is there a way to check the data sent by a form to a PHP page return to the form page WITHOUT resetting the data sent and show a error? The form has 20 fields and I need to check one of them on a bd. If it fails the user may be redirected to the form page with the form populated and displaying a error message on the field which is 'wrong'. I would like any advice of a technique instead of populating each field using PHP.

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  • directory_map() not working outside of application

    - by iamdadude
    Is there a reason as to why $map = directory_map('./', TRUE); works perfectly and, $map = directory_map('../', TRUE); doesn’t work at all? I tried creating a directory with the basic mkdir(); PHP function (http://php.net/mkdir) and it didn’t work either. Can anyone set me in the right direction as to what I’m doing wrong and how I can fix it? Thanks!

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  • Storing nested arrays in a cookie

    - by morpheous
    I am trying to store nested arrays in a cookie. I decided to store the array as a JSON string. However, I am getting this warning: PHP Warning: Cookie values can not contain any of the following ',; \t\r\n\013\014' in foobar.php Is there a recommended way of storing nested arrays in a cookie?

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  • Autopopulate from Select box from database

    - by Chris Spalton
    hope you can help, please forgive any poor coding or anytihng, I'm new to this and just hacking my way through to get things to work. That said, on one of my projects I have this code, which successfully populates the dropdown from a database when the page is loaded: <select name="Region" id="Region"> <option value="">-- Select Region --</option> <?php $region=$POST['Region']; if ($region); { $regionquery = "SELECT DISTINCT REGION FROM Sales_Execs "; $regionresult = mysql_query($regionquery); while($row = mysql_fetch_array($regionresult)) { echo "<option value=\"".$row['REGION']."\">".$row['REGION']."</option>\n "; } } ?> <script type="text/javascript"> document.getElementById('Region').value = <?php echo json_encode(trim($_POST['Region']));?>; </script> </select> On my next project that I'm working on now, I need to do the same thing, so I copied the above code amended, and placed in my new project: <select name="Sales_Exec" id="Sales_Exec"> <option value="">-- Select SE --</option> <?php $salesexec=$POST['Sales_Exec']; if ($salesexec); { $salesexecquery = "SELECT DISTINCT Assigned FROM Data "; $salesexecresult = mysql_query($salesexecquery); while($row = mysql_fetch_array($salesexecresult)) { echo "<option value=\"".$row['ASSIGNED']."\">".$row['ASSIGNED']."</option>\n "; } } ?> <script type="text/javascript"> document.getElementById('Sales_Exec').value = <?php echo json_encode(trim($_POST['Sales_Exec']));?>; </script> </select> This second chunk of code doesn't work... and I can't work out why as it seems I've copied it all and amended all the neccersary parts, can anyone spot what is wrong? Thankyou!

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  • How would I add an if statement into MSQLI query?

    - by Josh
    Okay so I'm just learning mysqli and I'm having a little trouble putting this code together. I've posted the mysqli query below and then below that is the code I'm trying to combine with the mysqli query and I can't seem to get it to work. Maybe what I'm doing isn't possible, but the third section below is how I had the query written for mysql and it's working fine. Answers in code are appreciated! Thanks! MYSQLI QUERY: <?php require("../config.php"); if ($stmt = $mysqli->prepare("SELECT firstname,lastname,spousefirst,phonecell,email,date,contacttype,status FROM contacts WHERE contacttype IN ('Buyer','Seller','Buyer / Seller','Investor') ORDER BY date DESC")) { $stmt->execute(); $stmt->bind_results($firstname,$lastname,$spousefirst,$phonecell,$email,$date,$contacttype,$status); while ($stmt->fetch()) { echo ''.$firstname.' '." ".' '.$lastname.' '.",".' '.$spousefirst.' '.",".' '.$phonecell.' '.",".' '.$email.' '.",".' '.$date.' '.",".' '.$contacttype.' '.",".' '.$status.'</br>'; } $stmt->close(); } $mysqli->close(); ?> WHAT I'M TRYING TO COMBINE THE ABOVE WITH: if (($_GET['date'] == 'today')) { $sql = "SELECT * FROM contacts WHERE contacttype IN ('Buyer','Seller','Buyer / Seller','Investor') AND date = DATE(NOW()) ORDER BY date DESC"; } WHAT I HAD BEFORE WITH MYSQL THAT WORKS: <?php require("../config.php"); $sql = "SELECT * FROM contacts WHERE contacttype IN ('Buyer','Seller','Buyer / Seller','Investor') AND status = 'New' ORDER BY date DESC"; if (($_GET['date'] == 'today')) { $sql = "SELECT * FROM contacts WHERE contacttype IN ('Buyer','Seller','Buyer / Seller','Investor') AND date = DATE(NOW()) ORDER BY date DESC"; } ?>

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  • An internal error occurred during runtime. Smarty

    - by rag
    WARNING [2] include(somepath/templates_c/%%B0^B01^B019F522%%login.htm.php) [function.include]: failed to open stream: No such file or directory on Line No 1871 in somepath/Smarty.class.php Error!: An internal error occurred during runtime. Any body please tell me why this error is occuring..

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  • Require_once to variable

    - by swamprunner7
    I want to call require_once("test.php") but not display result and save it into variable like this: //pseudocode $test = require_once('test.php'); //some operations like $test = preg_replace(…); echo $test; Is it possible?

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  • getting function's argument names

    - by Gotys
    in PHP Consider this function: function test($name, $age) {} I need to somehow extract the parameter names (for generating custom documentations automatically) so that I could do something like: get_func_argNames('test'); and it would return Array['name','age'] . Is this even possible in PHP ?

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  • login not working when changing from mysql to mysqli

    - by user1438647
    I have a code below where it logs a teacher in by matching it's username and password in the database, if correct, then log in, if incorrect, then display a message. <?php session_start(); $username="xxx"; $password="xxx"; $database="mobile_app"; $link = mysqli_connect('localhost',$username,$password); mysqli_select_db($link, $database) or die( "Unable to select database"); foreach (array('teacherusername','teacherpassword') as $varname) { $$varname = (isset($_POST[$varname])) ? $_POST[$varname] : ''; } ?> <form action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>" method="post" id="teachLoginForm"> <p>Username</p><p><input type="text" name="teacherusername" /></p> <!-- Enter Teacher Username--> <p>Password</p><p><input type="password" name="teacherpassword" /></p> <!-- Enter Teacher Password--> <p><input id="loginSubmit" type="submit" value="Login" name="submit" /></p> </form> <?php if (isset($_POST['submit'])) { $query = " SELECT * FROM Teacher t WHERE (t.TeacherUsername = '".mysqli_real_escape_string($teacherusername)."') AND (t.TeacherPassword = '".mysqli_real_escape_string($teacherpassword)."') "; $result = mysqli_query($link, $query); $num = mysqli_num_rows($result); $loged = false; while($row = mysqli_fetch_array($result)) { if ($_POST['teacherusername'] == ($row['TeacherUsername']) && $_POST['teacherpassword'] == ($row['TeacherPassword'])) { $loged = true; } $_SESSION['teacherforename'] = $row['TeacherForename']; $_SESSION['teachersurname'] = $row['TeacherSurname']; $_SESSION['teacherusername'] = $row['TeacherUsername']; } if ($loged == true){ header( 'Location: menu.php' ) ; }else{ echo "The Username or Password that you Entered is not Valid. Try Entering it Again."; } mysqli_close($link); } ?> Now the problem is that even if the teacher has entered in the correct username and password, it still doesn't let the teacher log in. When the code above was the old mysql() code, it worked fine as teacher was able to login when username and password match, but when trying to change the code into mysqli then it causes login to not work even though username and password match. What am I doing wrong?

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  • Where's the best place to find good senior web developers?

    - by bokani
    We are looking for a senior web developer for a business start up based in London Mayfair? • Demonstrable experience developing Web 2.0 projects • Complete fluency in HTML, Javascript, CSS, php and MySQL • Experience of jQuery, AJAX and php interaction • Ability to develop applications making use of APIs (Google Maps, Facebook, bespoke CRMs and similar) • Good design aesthetic, including familiarity with Photoshop and CSS • Substantial experience hand-coding • Familiarity with server administration including cPanel • Ability to design HTML newsletters • Progressive enhancement • AJAX application state-memory Salary : £30,000 to £40,000

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