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  • Cannot install xdebug with WAMP SERVER 2.1

    - by Jimmy Nguyen
    Hi all, I use WAMP SERVER 2.1 and select PHP 5.3.3 for my system, so I select xDebug with php_xdebug-2.1.0-5.3-vc6.dll and changed name becoming php_xdebug.dll for easy way to use. Following the instructions: php.ini (in Apache folder) extension=php_xdebug.dll ... zend_extension = "C:/wamp/bin/php/php5.3.3/ext/php_xdebug.dll" xdebug.remote_enable=on xdebug.remote_handler=dbgp xdebug.remote_host=localhost xdebug.remote_port=9000 xdebug.idekey="netbeans-xdebug" However, nothing happens, there are no information related to xdebug from phpinfo. Also xdebug announce that xdebug have not installed yet (http://xdebug.org/find-binary.php). I am so worried causing too much time for configuration. I got crazy and totally gave up. Anyone have ideas to solve it, I am so appreciated what you help me. Thanks

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  • How to access my local host server from internet

    - by rahul.p.33
    How to access my local host server from internet have installed WAMP server on my Windows XP, And i had created a index.php file in my root folder, and i assigned a virtual name to my localhost. eg: earlier i accessed my index.php via: Code: http:// localhost/ index.php but now i can access like : Code: http: //www. mysite. com/index.php but the problem is that i can access this from my computer only.. how can i use it from internet.. please help me.... Thanks.

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  • htaccess Redirect 301 problem .. all redirects with one string fail to redirect and 404

    - by Marty
    So I have moved a website and am trying to 301 redirect everything, which I do quite often so this is a weird problem but probably something stupid I'm not seeing. ALL of my redirects are working fine, except any redirect that the first string starts with "/Dining" or "/dining" are failing. For example, this redirect works fine- Redirect 301 /healthfitness/teeth.cfm /healthcare/pretty-teeth ...as well as 100s of others. But all of these are failing (many more than I'm showing)- Redirect 301 /Dining/diningreviews/vawines.cfm /shopping/wines-2004 Redirect 301 /Dining/diningathome/carrotcake.cfm /home-garden/carrot-cake-2003 Redirect 301 /Dining/diningathome/oldvarolls.cfm /home-garden/virginia-rolls-2003 Redirect 301 /Dining/diningathome/pumpkincake.cfm /home-garden/pumpkin-cake-2003 The top of my .htaccess file looks like this- RewriteEngine On RewriteBase / #uploaded files RewriteRule ^(.*/)?files/$ index.php [L] RewriteCond %{REQUEST_URI} !.*wp-content/plugins.* RewriteRule ^(.*/)?files/(.*) wp-content/blogs.php?file=$2 [L] # add a trailing slash to /wp-admin RewriteCond %{REQUEST_URI} ^.*/wp-admin$ RewriteRule ^(.+)$ $1/ [R=301,L] RewriteCond %{REQUEST_FILENAME} -f [OR] RewriteCond %{REQUEST_FILENAME} -d RewriteRule . - [L] RewriteRule ^([_0-9a-zA-Z-]+/)?(wp-.*) $2 [L] RewriteRule ^([_0-9a-zA-Z-]+/)?(.*\.php)$ $2 [L] RewriteRule . index.php [L] <IfModule mod_security.c> <Files async-upload.php> SecFilterEngine Off SecFilterScanPOST Off </Files> </IfModule> #Everything below here are Redirect 301s

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  • Redirecting a HTTP rqueust and response code/headers

    - by Bill Zimmerman
    Hi, I have a loosely coupled web app (one part uses PHP, the other uses WGSI). The WSGI/python framework shares the authentication with the PHP app, meaning that generally, the user should Log in via the PHP interface Now the user can access any of the WSGI pages [this part works if the user has logged in] What I want to do though, is if a user tries to access a WSGI page while not logged in (maybe from a previous bookmark), I would like to redirect him to the login page, and after logging in redirect him back to the orignal URL. I'm not very experienced with server-side programming, so here are my questions. How should I redirect the user back to the PHP login page? What should the HTTP status code be? Do I need to set any extra header information? What is a good way/best practice method to pass the original URL to the login page, and then after logging have it redirect the user back. Thank you!

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  • 404 ErrorDocument Problem

    - by Wayne
    I have .htaccess already configured: ErrorDocument 404 error.php But whenever I go to an invalid page, it should display the 404 page, but it just shows: "error.php" Just blank white, just the text of the php file only... The file and .htaccess does exist.

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  • CakePHP Test Fixtures Drop My Tables Permanently After Running A Test Case

    - by Frank
    I'm not sure what I've done wrong in my CakePHP unit test configuration. Every time I run a test case, the model tables associated with my fixtures are missing form my test database. After running an individual test case I have to re-import my database tables using phpMyAdmin. Here are the relevant files: This is the class I'm trying to test comment.php. This table is dropped after the test. App::import('Sanitize'); class Comment extends AppModel{ public $name = 'Comment'; public $actsAs = array('Tree'); public $belongsTo = array('User' => array('fields'=>array('id', 'username'))); public $validate = array( 'text' = array( 'rule' =array('between', 1, 4000), 'required' ='true', 'allowEmpty'='false', 'message' = "You can't leave your comment text empty!") ); database.php class DATABASE_CONFIG { var $default = array( 'driver' = 'mysql', 'persistent' = false, 'host' = 'project.db', 'login' = 'projectman', 'password' = 'projectpassword', 'database' = 'projectdb', 'prefix' = '' ); var $test = array( 'driver' = 'mysql', 'persistent' = false, 'host' = 'project.db', 'login' = 'projectman', 'password' = 'projectpassword', 'database' = 'testprojectdb', 'prefix' = '' ); } My comment.test.php file. This is the table that keeps getting dropped. <?php App::import('Model', 'Comment'); class CommentTestCase extends CakeTestCase { public $fixtures = array('app.comment', 'app.user'); function start(){ $this-Comment =& ClassRegistry::init('Comment'); $this-Comment-useDbConfig = 'test_suite'; } This is my comment_fixture.php class: <?php class CommentFixture extends CakeTestFixture { var $name = "Comment"; var $import = 'Comment'; } And just in case, here is a typical test method in the CommentTestCase class function testMsgNotificationUserComment(){ $user_id = '1'; $submission_id = '1'; $parent_id = $this-Comment-commentOnModel('Submission', $submission_id, '0', $user_id, "Says: A"); $other_user_id = '2'; $msg_id = $this-Comment-commentOnModel('Submission', $submission_id, $parent_id, $other_user_id, "Says: B"); $expected = array(array('Comment'=array('id'=$msg_id, 'text'="Says: B", 'submission_id'=$submission_id, 'topic_id'='0', 'ack'='0'))); $result = $this-Comment-getMessages($user_id); $this-assertEqual($result, $expected); } I've been dealing with this for a day now and I'm starting to be put off by CakePHP's unit testing. In addition to this issue -- Servral times now I've had data inserted into by 'default' database configuration after running tests! What's going on with my configuration?!

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  • Error in Jaclplus component

    - by Aruna
    Hi, I am working on Joomla 1.5 and our site is using the Bluehost server.Till March 9th, 2010 our site was working perfectly.And on March 10th,2010 , the following error was shown in the site as Warning: Unexpected character in input: '' (ASCII=28) state=1 in /home1/tcscoinc/public_html/tcsrnd/administrator/components/com_jaclplus/jac lplus.class.php on line 73 Parse error: syntax error, unexpected T_STRING in /home1/tcscoinc/public_html/tcsrnd/administrator/components/com_jaclplus/jac lplus.class.php on line 73 For time being, i have just renamed the file jaclplus.class.php to make the site to function. As we are using Jaclplus component and we have purchased the Jaclplus component from Byos Tech , i have checked in the membership details.And we found that our Membership has been expired on 03/10/2010 . I have asked them but they had replied that it may not be a problem on membership. Please anyone share what can be the reason for the error.I am not able to open that file(jaclplus.class.php) since its a binary file.

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  • JQuery text display problem?

    - by SLAPme
    Kind of new to JQuery and I was wondering how can I state that the users submitted info was saved when they click the submit button by displaying the message Changes saved at the top of the form and then have it disappear when the user leaves the web page and return back to it? Right now my code only displays that changes were saved at the bottom of the form outside of the lists and will not disappear when the users leave the web page and return back to it. Here is the JQuery code. $(function() { $(".save-button").click(function() { $.post($("#contact-form").attr("action"), $("#contact-form").serialize(), function(html) { $("div.contact-info-form").html(html); $('#contact-form').append('<li>Changes saved!</li>'); }); return false; // prevent normal submit }); }); Here is the html code. <div id="contact-info-form" class="form-content"> <h2>Contact Information</h2> <form method="post" action="index.php" id="contact-form"> <fieldset> <ul> <li><label for="address">Address 1: </label><input type="text" name="address" id="address" size="25" class="input-size" value="<?php if (isset($_POST['address'])) { echo $_POST['address']; } else if(!empty($address)) { echo $address; } ?>" /></li> <li><label for="address_two">Address 2: </label><input type="text" name="address_two" id="address_two" size="25" class="input-size" value="<?php if (isset($_POST['address_two'])) { echo $_POST['address_two']; } else if(!empty($address_two)) { echo $address_two; } ?>" /></li> <li><label for="city_town">City/Town: </label><input type="text" name="city_town" id="city_town" size="25" class="input-size" value="<?php if (isset($_POST['city_town'])) { echo $_POST['city_town']; } else if(!empty($city_town)) { echo $city_town; } ?>" /></li> <li><label for="state_province">State/Province: </label> <?php echo '<select name="state_province" id="state_province">' . "\n"; foreach($state_options as $option) { if ($option == $state_province) { echo '<option value="' . $option . '" selected="selected">' . $option . '</option>' . "\n"; } else { echo '<option value="'. $option . '">' . $option . '</option>'."\n"; } } echo '</select>'; ?> </li> <li><label for="zipcode">Zip/Post Code: </label><input type="text" name="zipcode" id="zipcode" size="5" class="input-size" value="<?php if (isset($_POST['zipcode'])) { echo $_POST['zipcode']; } else if(!empty($zipcode)) { echo $zipcode; } ?>" /></li> <li><label for="country">Country: </label> <?php echo '<select name="country" id="country">' . "\n"; foreach($countries as $option) { if ($option == $country) { echo '<option value="' . $option . '" selected="selected">' . $option . '</option>' . "\n"; } else if($option == "-------------") { echo '<option value="' . $option . '" disabled="disabled">' . $option . '</option>'; } else { echo '<option value="'. $option . '">' . $option . '</option>'."\n"; } } echo '</select>'; ?> </li> <li><label for="email">Email Address: </label><input type="text" name="email" id="email" size="25" class="input-size" value="<?php if (isset($_POST['email'])) { echo $_POST['email']; } else if(!empty($email)) { echo $email; } ?>" /><br /><span>We don't spam or share your email with third parties. We respect your privacy.</span></li> <li><input type="submit" name="submit" value="Save Changes" class="save-button" /> <input type="hidden" name="contact_info_submitted" value="true" /> <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li> </ul> </fieldset> </form> </div>

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  • Python 2.5.2: trying to open files recursively

    - by user248959
    Hi, the script below should open all the files inside the folder 'pruebaba' recursively but i get this error: Traceback (most recent call last): File "/home/tirengarfio/Desktop/prueba.py", line 8, in f = open(file,'r') IOError: [Errno 21] Is a directory This is the hierarchy: pruebaba folder1 folder11 test1.php folder12 test1.php test2.php folder2 test1.php The script: import re,fileinput,os path="/home/tirengarfio/Desktop/pruebaba" os.chdir(path) for file in os.listdir("."): f = open(file,'r') data = f.read() data = re.sub(r'(\s*function\s+.*\s*{\s*)', r'\1echo "The function starts here."', data) f.close() f = open(file, 'w') f.write(data) f.close() Any idea? Regards Javi

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  • joomla .htaccess file rewrite rule explanation required

    - by Vivek Chandraprakash
    Hi, I'm trying to understand the following lines in joomla's .htaccess file. Can someone explain this please #RewriteCond %{REQUEST_FILENAME} !-f #RewriteCond %{REQUEST_FILENAME} !-d #RewriteCond %{REQUEST_URI} !^/index.php #RewriteCond %{REQUEST_URI} (/|\.php|\.html|\.htm|\.feed|\.pdf|\.raw|/[^.]*)$ [NC] #RewriteRule (.*) index.php #RewriteRule .* - [E=HTTP_AUTHORIZATION:%{HTTP:Authorization},L] I want to do some custom redirects for example if a url is like this example.com/subdirectory1 i want to redirect to some article. tried adding this line in the .htaccess file RewriteRule ^somepath index.php?option=com_content&view=article&id=1&Itemid=12 but for some reason the article shows the title even though it's not supposed to show. when i access using the long url string the title doesn't appear if i rewrite it appears. Please help. -Vivek

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  • cakephp VS codeigniter VS zend framework

    - by i need help
    Very possibly very related: What PHP framework would you choose for a new application and why? Zend or CakePHP? Which one is better? Some people say CakePHP is better for php 4, what do you think? In my case, I would like the following: Lesser code to write, have really strong library and plugin base. Always have new library etc added in from contributor, eg: google map and etc... Ability to use together with the templating system like smarty. Have ACL that can control all the permission level issue. Load class when needed, unload when not needed. Load class once and use globally. Can run in windows environment (I am using xampp to run my php in windows.) After the site done, I will upload all codes into windows 2008 server (using php 5)

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  • The post thumbnail size issues

    - by user195257
    Hey Im using this in my wordpress functions.php add_theme_support( 'post-thumbnails' ); set_post_thumbnail_size( 150, 150 ); Im then using it like this <a href="<?php the_permalink(); ?>"><?php the_post_thumbnail(); ?></a> However, the image is displayed <img width="150" height="109" ... The images themselves are 300px x 300px, anyone know whats going on here?

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  • jeditable not updating browser display - leaves "click to edit..." after succesful edit

    - by Enoch
    I am using jeditable fairly simply and it all works fine, updates the database, etc. The only problem I have is after the user types the new value data and hits enter it doesn't update the field in the browser to show the new value - instead it puts "Click to edit..." in place of it. Am I missing something like a return value from my php file? the php fucntion just takes the args and updates the database - and it works fine. Enoch the jquery\jeditable code $('.edit').editable('update.php',{ id: 'field', name: 'val', indicator: 'Saving...', tooltip: 'Click to edit...', select : true, submitdata : { db : "pers", kn : "key", rec : "?php echo $rec; ?" } }); the div <div class="edit" id="svc_ad1"><?php echo $row-svc_ad1; ?>< /div> i also have a css class for pEdit edit{ float:left; width:200px; height:15px; margin-bottom:5px; border-bottom:1px solid #aaaaaa; }

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  • How do I get apache RewriteRule working correctly for a subdomain?

    - by mike
    I just setup a subdomain with the following RewriteCond: RewriteCond $1 !^search.php$ RewriteCond %{REQUEST_FILENAME} !-f RewriteRule ^/?([^/]+)$ search.php?q=$1 [L,NS] I'm using the same rewrite condition on my main domain and it works perfectly. However, when I set it up on the subdomain, it simply outputs "index.php" when going to http://sub.domain.com Every page on the subdomain outputs the page name in the body instead of processing the code, except for the search page, which appears to be working correctly. What can I do to correct this issue?

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  • Using phpFlickr, how would one display the primary photo from each photoset?

    - by Michael
    Referring to this question: http://stackoverflow.com/questions/2561475/flickr-phpflickr-api how would you display a primary photo from a photoset rather than all photos and photosets? this is the code I have so far: photosets_getList($user); ? <?php $photoset_id = $ph_set['id']; $photos = $f->photosets_getPhotos($photoset_id); foreach ($photos['photoset']['photo'] as $photo): ?> <?php if($parentID == $ph_set['parent']): ?> <a rel="lightbox[album<?=$count;?>]" href="<?= $f->buildPhotoURL($photo, 'medium') ?>" title="<?= $photo['title'] ?>"> <?php endif;?> <img src="<?= $f->buildPhotoURL($photo, 'square') ?>" alt="<?= $photo['title'] ?>" width="75" height="75" title="<?= $photo['title'] ?>" /> <h3><?=$ph_set['title']?></h3> <?php if($parentID == $ph_set['parent']): ?> </a> </div> <?php endif;?>

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  • Perl - MySQL connection problem in windows

    - by dexter
    I have two folders php and perl they contain index.php and index.pl respectivly in index.pl my perl code looks like: #!/usr/bin/perl use Mysql; print "Content-type: text/html\n\n"; print "<h2>PERL-mySQL Connect</h2>"; print "page info"; $host = "localhost"; $database = "cdcol"; $user = "root"; $password = ""; $db = Mysql->connect($host, $database, $user, $password); $db->selectdb($database); when i run above code (ie type: http://localhost:88/perl/ in browser) following error comes Error message: Can't locate Mysql.pm in @INC (@INC contains: C:/xampp/perl/site/lib/ C:/xampp/perl/lib C:/xampp/perl/site/lib C:/xampp/apache) at C:/xampp/htdocs/perl/index.pl line 2. BEGIN failed--compilation aborted at C:/xampp/htdocs/perl/index.pl line 2. while this works: in browser http://localhost:88/php/ where index.php has: <?php $con = mysql_connect("localhost","root",""); if($con) { if(mysql_select_db("cdcol", $con)) { $sql="SELECT Id From products"; if(mysql_query($sql)) { $result = mysql_query($sql); if ($result)............

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  • SIMPLE reverse geocoding using Nominatim

    - by tony gil
    i am developing an online mapping application using OpenLayers + OpenStreetMaps. i need help implementing a simple reverse geocoding function in javascript (or php) that receives Latitude and Longitude and returns an Address. i would like to work with Nominatim, if possible. i do NOT want to use Google, Bing or CloudMade or other proprietary solutions. this link returns a reasonable response and i used simple_html_dom.php to break down the result but it is sort of an ugly solution. <?php include('simple_html_dom.php'); $url = "http://nominatim.openstreetmap.org/reverse?format=xml&lat=-23.56320001&lon=-46.66140002&zoom=27&addressdetails=1"; $html = file_get_html($url); foreach ($html->find('road') as $element ) { echo $element; } ?> any suggestions of a more elegant solution?

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  • JQuery form submit Question?

    - by SLAPme
    I'm kind of new to JQuery and was wondering how can I have the following text appear when the user submits there changes <p>Your changes have been saved.</p>? How do I fix my code so it displays this message? JQuery script. $(function() { $(".save-button").click(function() { $.post($("#contact-form").attr("action"), $("#contact-form").serialize(), function(html) { $("div.contact-info-form").html(html); }); return false; // prevent normal submit }); }); Here is the html. <div id="contact-info-form" class="form-content"> <h2>Contact Information</h2> <form method="post" action="index.php" id="contact-form"> <fieldset> <ul> <li><label for="address">Address 1: </label><input type="text" name="address" id="address" size="25" class="input-size" value="<?php if (isset($_POST['address'])) { echo $_POST['address']; } else if(!empty($address)) { echo $address; } ?>" /></li> <li><label for="address_two">Address 2: </label><input type="text" name="address_two" id="address_two" size="25" class="input-size" value="<?php if (isset($_POST['address_two'])) { echo $_POST['address_two']; } else if(!empty($address_two)) { echo $address_two; } ?>" /></li> <li><label for="city_town">City/Town: </label><input type="text" name="city_town" id="city_town" size="25" class="input-size" value="<?php if (isset($_POST['city_town'])) { echo $_POST['city_town']; } else if(!empty($city_town)) { echo $city_town; } ?>" /></li> <li><label for="state_province">State/Province: </label> <?php echo '<select name="state_province" id="state_province">' . "\n"; foreach($state_options as $option) { if ($option == $state_province) { echo '<option value="' . $option . '" selected="selected">' . $option . '</option>' . "\n"; } else { echo '<option value="'. $option . '">' . $option . '</option>'."\n"; } } echo '</select>'; ?> </li> <li><label for="zipcode">Zip/Post Code: </label><input type="text" name="zipcode" id="zipcode" size="5" class="input-size" value="<?php if (isset($_POST['zipcode'])) { echo $_POST['zipcode']; } else if(!empty($zipcode)) { echo $zipcode; } ?>" /></li> <li><label for="country">Country: </label> <?php echo '<select name="country" id="country">' . "\n"; foreach($countries as $option) { if ($option == $country) { echo '<option value="' . $option . '" selected="selected">' . $option . '</option>' . "\n"; } else if($option == "-------------") { echo '<option value="' . $option . '" disabled="disabled">' . $option . '</option>'; } else { echo '<option value="'. $option . '">' . $option . '</option>'."\n"; } } echo '</select>'; ?> </li> <li><label for="email">Email Address: </label><input type="text" name="email" id="email" size="25" class="input-size" value="<?php if (isset($_POST['email'])) { echo $_POST['email']; } else if(!empty($email)) { echo $email; } ?>" /><br /><span>We don't spam or share your email with third parties. We respect your privacy.</span></li> <li><p>Changes have been saved</p><input type="submit" name="submit" value="Save Changes" class="save-button" /> <input type="hidden" name="contact_info_submitted" value="true" /> <input type="submit" name="submit" value="Preview Changes" class="preview-changes-button" /></li> </ul> </fieldset> </form> </div>

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  • Simple Custom rule for Jquery validator

    - by thatweblook
    Hi, I read your reply regarding the jQuery validator where you outline a method to check a username against a value in a database. Ive tried implementing this method but no matter what is returned from the PHP file I always get the message that the username is already taken. Here is ths custom method... $.validator.addMethod("uniqueUserName", function(value, element) { $.ajax({ type: "POST", url: "php/get_save_status.php", data: "checkUsername="+value, dataType:"html", success: function(msg) { // if the user exists, it returns a string "true" if(msg == "true") return false; // already exists return true; // username is free to use } })}, "Username is Already Taken"); And here is the validate code... username: { required: true, uniqueUserName: true }, Is there a specific way i am supposed to return the message from php. Thanks A

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