Search Results

Search found 32007 results on 1281 pages for 'php openid'.

Page 531/1281 | < Previous Page | 527 528 529 530 531 532 533 534 535 536 537 538  | Next Page >

  • how close a connection from server side that created by ajax from client side

    - by saeid
    I have this sample php code: <?php function response($a) { @ob_end_clean(); header("Connection: close"); ob_start(); echo $a; $length = ob_get_length(); header("Content-Length: $length"); ob_end_flush(); flush(); } response("test1"); sleep(5); echo "test2"; ?> now I make a connection to server from client by ajax. now I want to server only send "test1" to me and then close the connection. but connection not closed and wait 5 seconds and send me "test1 test2" !! this is response headers from server to me : X-Powered-By: PHP/5.3.27 Connection: close Vary: Accept-Encoding,User-Agent Content-Encoding: gzip Cache-Control: private, no-cache, no-store, ------revalidate, no-transform Transfer-Encoding: chunked Content-Type: text/html in response headers I do not see Content-Length header. is problem for this ?? or if not what I must do now for solving problem ??

    Read the article

  • XAMPP, MAMP, MySQL, PDO - A deadly combination?

    - by Rich
    Hey folks, Previously I've worked on a Symfony project (MySQL PDO based) with XAMPP, with no problems. Since then, I've moved to MAMP - I prefer this - but have hit a snag with my database connection. I've created a test.php like this: <?php try { $dbh = new PDO('mysql:host=localhost;dbname=xxx;port=8889', 'xxx', 'xxx'); foreach($dbh->query('SELECT * from FOO') as $row) { print_r($row); } $dbh = null; } catch (PDOException $e) { print "Error!: " . $e->getMessage() . "<br/>"; die(); } ?> Obviously the *xxx*s are real db connection details. Which when served by MAMP seems to work fine. From terminal however I keep getting the following error when running the file: Error!: SQLSTATE[28000] [1045] Access denied for user 'xxx'@'localhost' (using password: YES) Not sure if the terminal is aiming at a different MySQL socket or something along those lines; but I've tried pointing it to the MAMP socket with a local php.ini file. Any help would be greatly appreciated.

    Read the article

  • Why does PDO print my password when the connection fails?

    - by Joe Hopfgartner
    I have a simple website where I establish a connection to a Mysql server using PDO. $dbh = new PDO('mysql:host=localhost;dbname=DB;port=3306', 'USER', 'SECRET',array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8")); I had some traffic on my site and the servers connection limit was reached, and the website throw this error, with my PLAIN password in it! Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[08004] [1040] Too many connections' in /home/premiumize-me/html/index.php:64 Stack trace: #0 /home/premiumize-me/html/index.php(64): PDO-__construct('mysql:host=loca...', 'USER', 'SECRET', Array) #1 {main} thrown in /home/premiumize-me/html/index.php on line 64 Ironically I switched to PDO for security reasons, this really shocked me. Because this exact error is something you can provoke very easily on most sites using simple http flooding. I now wrapped my conenction into a try/catch clause, but still. I think this is catastrophic! So I am new to PDO and my questino is: What do I have to consider to be safe! How to I establish a connection in a secure way? Are there other known security holes like this one that I have to be aware of?

    Read the article

  • Parse a CSV file extracting some of the values but not all.

    - by Yallaa
    Good day, I have a local csv file with values that change daily called DailyValues.csv I need to extract the value field of category2 and category4. Then combine, sort and remove duplicates (if any) from the extracted values. Then save it to a new local file NewValues.txt. Here is an example of the DailyValues.csv file: category,date,value category1,2010-05-18,value01 category1,2010-05-18,value02 category1,2010-05-18,value03 category1,2010-05-18,value04 category1,2010-05-18,value05 category1,2010-05-18,value06 category1,2010-05-18,value07 category2,2010-05-18,value08 category2,2010-05-18,value09 category2,2010-05-18,value10 category2,2010-05-18,value11 category2,2010-05-18,value12 category2,2010-05-18,value13 category2,2010-05-18,value14 category2,2010-05-18,value30 category3,2010-05-18,value16 category3,2010-05-18,value17 category3,2010-05-18,value18 category3,2010-05-18,value19 category3,2010-05-18,value20 category3,2010-05-18,value21 category3,2010-05-18,value22 category3,2010-05-18,value23 category3,2010-05-18,value24 category4,2010-05-18,value25 category4,2010-05-18,value26 category4,2010-05-18,value10 category4,2010-05-18,value28 category4,2010-05-18,value11 category4,2010-05-18,value30 category2,2010-05-18,value31 category2,2010-05-18,value32 category2,2010-05-18,value33 category2,2010-05-18,value34 category2,2010-05-18,value35 category2,2010-05-18,value07 I've found some helpful parsing examples at http://www.php.net/manual/en/function.fgetcsv.php and managed to extract all the values of the value column but don't know how to restrict it to only extract the values of category2/4 then sort and clean duplicate. The solution needs to be in php, perl or shell script. Any help would be much appreciated. Thank you in advance.

    Read the article

  • JQuery datepicker icon trigger broken after inserting data to database

    - by crisgine callano
    at first i got to load my icon for the JQuery datepicker but after i insert the value to my database the icon image is broken. im using Codeigniter framework. here's my html code: <!DOCTYPE HTML> <html lang="en-US"> <head> <meta charset="UTF-8"> <title>CRUD</title> <?php echo link_tag(base_url() . 'css/ui-lightness/jquery-ui-1.8.23.custom.css'); ?> </head> <body> <div> <?php echo form_label('BDay:', 'bday'); $data = array( 'name' => 'bday', 'id' => 'datepicker' ); echo form_input($data); ?> </div> </body> </html> and here's my jscript: <script> $(function() { $( "#datepicker" ).datepicker({ changeMonth: true, changeYear: true, showOn: "button", buttonImage: "<?php echo base_url() . '/images/icons/calendar_24.png' ?>", buttonImageOnly: true, onSelect: function(dateText, inst){ $("input[name='bday']").val(dateText); } }); }); am i missing something? thanks!

    Read the article

  • How to insert form data into MySQL database table

    - by Richard
    So I have this registration script: The HTML: <form action="register.php" method="POST"> <label>Username:</label> <input type="text" name="username" /><br /> <label>Password:</label> <input type="text" name="password" /><br /> <label>Gender:</label> <select name="gender"> <optgroup label="genderset"> <option value="Male">Male</option> <option value="Female">Female</option> <option value="Hermaphrodite">Hermaphrodite</option> <option value="Not Sure!!!">Not Sure!!!</option> </optgroup> </select><br /> <input type="submit" value="Register" /> </form> The PHP/SQL: <?php $username = $_POST['username']; $password = $_POST['password']; $gender = $_POST['gender']; mysql_query("INSERT INTO registration_info (username, password, gender) VALUES ('$username', '$password', '$gender') ") ?> The problem is, the username and password gets inserted into the "registration_info" table just fine. But the Gender input from the select drop down menu doesn't. Can some one tell me how to fix this, thanks.

    Read the article

  • Only show link if conditional equals true

    - by Dave Morin
    I want the link to appear only when $data['block'] equals to 1, 3 or 4. Not if it equals to 2 or 5. <td style="font-size:18px;color:#f0cb01;"> <a href="kickcodes.php?id='.$data["block"].'">Reason Codes</a> </td> EDIT while ($data = mysql_fetch_array($query)) { echo ' <tr style="background-color:#576c11;"> <td style="font-size:18px; color:#f0cb01;">'.$data["keyword"].'</td> <td style="font-size:18px;color:#f0cb01;">'.$data["block"].'</td> <td style="font-size:18px;color:#f0cb01;">'.$data["phone"].'</td> <td style="font-size:18px;color:#f0cb01;">'.$data["Reason"].'</td> <td style="font-size:18px;color:#f0cb01;"><a href="kickcodes.php?id='.$data ["block"].'">Kickcodes</a></td>' echo '<td style="font-size:18px;color:#f0cb01;">'; if( $data['block'] == 1 || $data['block'] == 3 || $data['block'] == 4) { echo '<a href="kickcodes.php?id='.$data["block"].'">Reason Codes</a>'; } else { echo '<span>Reason Codes</span>'; // Or echo nothing } echo '</td>';

    Read the article

  • Prevent empty form input array from being posted?

    - by user355295
    Sorry if this has been answered somewhere; I'm not quite sure how to phrase the problem to even look for help. Anyway, I have a form with three text input boxes, where the user will input three song titles. I have simple PHP set up to treat those input boxes as an array (because I may want, say, 100 song titles in the future) and write the song titles to another document. <form method="post"> <input type="text" name="songs[]" value="" /> <input type="text" name="songs[]" value="" /> <input type="text" name="songs[]" value="" /> <button type="submit" name="submit" value="submit">Submit</button> </form> <?php if (isset($_POST['submit'])) { $open = fopen("test.html", "w"); if(empty($_POST['songs'])) { } else { $songs = $_POST['songs']; foreach($songs as $song) { fwrite($open, $song."<br />"); }; }; }; ?> This correctly writes the song titles to an external file. However, even when the input boxes are empty, the external file will still be written to (just with the <br />'s). I'd assumed that the if statement would ensure nothing would happen if the boxes were blank, but that's obviously not the case. I guess the array's not really empty like I thought it was, but I'm not really sure what implications that comes with. Any idea what I'm doing wrong? (And again, I am clueless when it comes to PHP, so forgive me if this has been answered a million times before, if I described it horribly, etc.)

    Read the article

  • javascript unable to locate a form using the ID tag

    - by ihake
    Here's my problem: I'm trying to set up a simple mobile contact form with a captcha built in. The page I'm working on can be found here: http://m.lancasterpainting.com/contact.php I'm using the following php contact form: http://www.html-form-guide.com/contact-form/php-email-contact-form.html I want to first say that I'm not the only one to run into this problem. After googling the issue, I've found multiple people struggling with this, but no-one seems to have an answer. Now for the problem... As you can see if you visit the page, each time the page is accessed, an error appears that says "Error: couldnot get Form object contact_form". I cannot--for the life of me--figure out why the javascript can't find the form I pass it. I call the function that generates this error at the top of the page: var frmvalidator = new Validator("contact_form"); The form I'm referencing is as follows in the HTML code: <div data-role="page" data-theme="e" id="contact_form" name="contact_form" data-position="inline"> ... And the function that is called that generates the error can be found in an external .js file here: http://m.lancasterpainting.com/scripts/gen_validatorv31.js Is there something that I am simply not seeing? Why can't the javascript locate the form? Thanks so much to anyone that helps with this.

    Read the article

  • How can i get my delete messages function just appear for the user's own messages left on their friends page?

    - by Hannah_B
    I had been working on this trying the delete message button to work on my own profile page of my site. When I delete a message left by a friend it not only deletes it from the screen but deletes it from the database. The messages in the database have 4 fields: message_id, from, to and message. Here is my profile view that shows how Im deleting messages from my friends: if(!empty($messages)){ foreach($messages as $message): $delete = $message['message_id']; //var_dump($message); ?> <li><?=$message['from']?> says...: "<?=$message['message']?>"(<?=anchor("home/deleteMsg/$delete", 'delete')?>)</li> //this is where the delete button appears beside messages left <?php endforeach?> <?php }else{ ?> <?php echo 'No messages left yet !!!'; }?> Here is my controller showing the deleteMsg function called: function deleteMsg($messageid) { $this->messages->deleteMsg($messageid); redirect('home'); } Here is the messages model showing the deleteMsg model itself: function deleteMsg($message_id) { $this->db->where(array('message_id' => $message_id)); $this->db->delete('messages'); } Here is my friendprofile view where I want to implement the delete message command just so the button appears for messages Ive left and I can delete them. The delete button will not appear beside other friends comments on this page: <li><?=$message['from']?> says...: "<?=$message['message']?>"</li> Now I've tried creating a new delete Message function to no success so far, am I better off doing this than calling the same function? As this didnt work either.

    Read the article

  • Is it possible to send back a json array along with seperate variables

    - by Scarface
    Hey guys I have an ajax jquery function that receives data from a php script. I want to return an array with all the online users which is retrieved from a mysql statement, and I want to send other separate variables I need for other purposes along with it. If anyone has any ideas, I would greatly appreciate it. NOTE: the example below is to illustrate what I want to do, I understand that json-encoding the array with other variables is dysfunctional. JQUERY $.ajax({ type: "POST", data: "parameters", url: "retrieval.php", dataType: 'json', success: function(json) { $('#div1').html(json.array); $('#div2').html(json.variable1); $('#div3').html(json.variable2); } }) PHP $qryuserscount1="SELECT * FROM active_users"; $userscount1=mysql_query($qryuserscount1); while ($row = mysql_fetch_array($userscount1)) { $onlineuser= $row['username']; $id=$row['id']; $data[]=$onlineuser.$id; //for example there are 3 users, should send 3 entries back } $data['variable1']='something'; $data['variable2']='something else'; $out = json_encode($data); print $out;

    Read the article

  • allowing bb code but not java script

    - by user1405062
    Hello im trying to get the hang of using bb codes onto my normal php site ( not forum or anything just a normal site ) I have seen a few posts like this one http://www.pixel2life.com/forums/index.php?/topic/10659-php-bbcode-parser/ which says i need a bb parser. I was just wondering can anyone show me how i would use one ? I have a status box were users can update there status but i would like to allow bb codes but not java script. So when im doing my inserting into the db i strip the status like so.. $status= mysql_real_escape_string($_POST['status']); $status2= strip_tags($status); And that stops the java script and tags from getting though but i need the bb code tags to come though but carry on blocking the java script code is there anyway to do this ? Also then i just echo it out echo $status2 ; But just plain text shows so was just wondering if anyone knows how to let though bb code and stop java script and could some one show me how to use the bb parasher ? also need to know how to echo out the bb coding...

    Read the article

  • How to evaluate json member using variable ?

    - by Miftah
    Hi i've got a problem evaluating json. My goal is to insert json member value to a function variable, take a look at this function func_load_session(svar){ var id = ''; $.getJSON('data/session.php?load='+svar, function(json){ eval('id = json.'+svar); }); return id; } this code i load session from php file that i've store beforehand. i store that session variable using dynamic var. <?php /* * format ?var=[nama_var]&val=[nilai_nama_var] */ $var = $_GET['var']; $val = $_GET['val']; $load = $_GET['load']; session_start(); if($var){ $_SESSION["$var"] = $val; echo "Store SESSION[\"$var\"] = '".$_SESSION["$var"]."'"; }else if($load){ echo $_SESSION["$load"]; } ?> using firebug, i get expected response but i also received error uncaught exception: Syntax error, unrecognized expression: ) pointing at this eval('id = json.'+svar); i wonder how to solve this

    Read the article

  • problem with accessing a php page

    - by EquinoX
    So I have a info.php page which is located on the folder /var/www/nginx-default, however when I go to my ip address/info.php, it always redirects me to this site: http://www.iana.org/domains/example/ is this because I have a virtual host that I called example? Here is my config for the example website: server { listen 80; server_name www.example.com; rewrite ^/(.*) http://example.com/$1 permanent; } server { listen 80; server_name example.com; access_log /var/www/example.com/logs/access.log; error_log /var/www/example.com/logs/error.log; location / { root /var/www/example.com/public/; index index.html; } } The way I access this site is by changing my /var/hosts in my macbook so that example.com is mapped to my server IP address... however now when I do xxx.xxx.xxx.xxx/info.php.. it redirects me to that site I posted above

    Read the article

  • retrieve only sub-pages (wordpress)

    - by Radek
    I want to list all sub-pages only one level though of one particular page. I was reading Function Reference/get pages and thought that $pages = get_pages( array( 'child_of' => $post->ID, 'parent' => $post->ID)) ; will do the trick but it is not working. It lists all pages on the same level like the page I call that code from. If I omit parent option I will get all pages even with sub-pages that I want. But I want only sub-pages. The whole function is like function about_menu(){ if (is_page('about')){ $pages = get_pages( array( 'child_of' => $post->ID, 'parent' => $post->ID)) ; foreach($pages as $page) { ?> <h2><a href="<?php echo get_page_link($page->ID) ?>"><?php echo $page->post_title ?></a></h2> <?php } } } and mine is second one

    Read the article

  • How to document an existing small web site (web application), inside and out?

    - by Ricket
    We have a "web application" which has been developed over the past 7 months. The problem is, it was not really documented. The requirements consisted of a small bulleted list from the initial meeting 7 months ago (it's more of a "goals" statement than software requirements). It has collected a number of features which stemmed from small verbal or chat discussions. The developer is leaving very soon. He wrote the entire thing himself and he knows all of the quirks and underlying rules to each page, but nobody else really knows much more than the user interface side of it; which of course is the easy part, as it's made to be intuitive to the user. But if someone needs to repair or add a feature to it, the entire thing is a black box. The code has some minimal comments, and of course the good thing about web applications is that the address bar points you in the right direction towards fixing a problem or upgrading a page. But how should the developer go about documenting this web application? He is a bit lost as far as where to begin. As developers, how do you completely document your web applications for other developers, maintainers, and administrative-level users? What approach do you use, where do you start, do you have a template? An idea of magnitude: it uses PHP, MySQL and jQuery. It has about 20-30 main (frontend) files, along with about 15 included files and a couple folders of some assets. So overall it's a pretty small application. It interfaces with 7 MySQL tables, each one well-named, so I think the database end is pretty self-explanatory. There is a config.inc.php file with definitions of consts like the MySQL user details, some from/to emails, and URLs which PHP uses to insert into emails and pages (relative and absolute paths, basiecally). There is some AJAX via jQuery. Please comment if there is any other information that would help you help me and I will be glad to edit it in.

    Read the article

  • How can I make a new CTP (CakePHP) file in NetBeans?

    - by John Isaacks
    I have found a lot of info about adding CTP file support for NetBeans, but this is usually talking about code highlighting and treating a ctp file like a php file. This can be done at: Tools - Options - Miscellaneous - Files I have done this. However, when I try to create a NEW ctp file. I do not have the option. I tried going to Tools -> Templates to add a ctp template. There is no "new" button just an "add" button that looks for a file. I created a file on my desktop called cake_template.ctp on my desktop. I added it to the PHP templates in the template manager. I called the template "PHP Cake Template". Still when I go to create a new file, the option is not there. I restarted NetBeans, still the same. I just want to create a new .ctp file, it shouldn't be this difficult. Does anyone know how? I am using version 6.9.1

    Read the article

  • Pass Element value to $ajax->link in cakephp

    - by TwoThumbs
    I need to pass the value of an element to an $ajax-link without using a form/submit structure. (because I have a dynamically set number of clickable links through which I am triggering the action) I used to do this in Ruby using the Prototype javascript function $F like this: <%= link_to_remote "#{item.to_s}", :url => { :action => :add_mpc }, :with => "'category=' + $F('mpc_category')" -%> But this does not seem to work in Cakephp: <?php echo $ajax->link(substr($vehicle['vehicles']['year'], -2), array('action' => 'add_mpc', 'category' => '$F("mpc_category")'), array('update' => 'results', 'position' => 'top')); ?> PHP sees $F as a variable instead of a call to javascript. I'm not too familiar with Javascript, but is there another way to pass the value of the 'mpc_category' input element to the controller through this link? I have been looking for a couple days and can't find anyone dealing with this specific issue. Thanks for any assistance. Edit: fixed syntax in php statement.

    Read the article

  • Calling function and running it

    - by devs
    I am quite new to objects and OOP. I really don't know how to explain it well but I'll try. So I am trying to read though JSON with JS, the JSON is passed from PHP. This would be easy if all of the information was on the same html page, but I' am trying something that I am new too. So let me show my code... First is the JS which is in app.js var Donors = function(){ var api = this.list; $(document).ready(function(){ $.getJSON(api, function(){ var donorObj = api.payload; $.each(donorObj, function(i, donor){ //console.log(donor.ign); }); }); }); } What I want this part to do is read from the JSON I'm giving it and console.log each name (or donor.ign) when the document is ready. On the html page, or header.php <script> $(function(){ var list = <?php cbProxy(); ?>; var Dons = new Donors(); Dons.list = list; }); </script> the data that's in list is the below JSON. You already know what the rest does, it just passes the JSON to the Donors() function. JSON example: { "code": 0, "payload": [ { "time": 1349661897, "packages": [ "49381" ], "ign": "Notch", "price": "15.99", "currency": "USD" } I'm use to just making functions and calling it on the same page or file and this is my first doing this kind of function. How can I get the function to run with the data I sent it so it console.log() each name.

    Read the article

  • Zip multiple database PDF blob files

    - by Michael
    I have a database table that contains numerous PDF blob files. I am attempting to combine all of the files into a single ZIP file that I can download and then print. Please help! <?php include '../config.php'; include '../connect.php'; $session = $_GET['session']; $query = "SELECT $tbl_uploads.username, $tbl_uploads.description, $tbl_uploads.type, $tbl_uploads.size, $tbl_uploads.content, $tbl_members.session FROM $tbl_uploads LEFT JOIN $tbl_members ON $tbl_uploads.username = $tbl_members.username WHERE $tbl_members.session = '$session'"; $result = mysql_query($query) or die('Error, query failed'); $files = array(); while(list($username, $description, $type, $size, $content) = mysql_fetch_array($result)) { $files[] = "$username-$description.pdf"; } $zip = new ZipArchive; $zip->open('file.zip', ZipArchive::CREATE); foreach ($files as $file) { $zip->addFile($file); } $zip->close(); header('Content-Type: application/zip'); header('Content-disposition: attachment; filename=filename.zip'); header('Content-Length: ' . filesize($zipfilename)); readfile($zipname); exit(); ?>

    Read the article

  • JQuery, AJAX: Trying AJAX for the first time but cant get this to work...

    - by fwaokda
    Trying to get the code below to work but the success doesn't execute - the error does. How can I get more detailed information on what is exactly going wrong? I'll include the code for next.php in a pastebin link also. Thanks. [next.php: http://pastebin.com/Gnu2AfU8 ] $("a#next").click(function() { $.ajax({ type : 'POST', url : 'next.php', dataType : 'json', data : { nextID : $("a#next").attr("rel") }, success : function ( data ) { $("img#spotlight").attr("src",data.spotlightimage); $("div#showcase h1").text(data.title); $("div#showcase h2").text(data.subtitle); for(var i=0; i < data.size; i++) { $("ul#features").append("<li>").text(data.feature+i).append("</li>"); } $("div#showcase p").text(data.description); for(i=1; j < data.picsize; i++) { $("div.thumbnails ul").append("<li>").text(data.image+i).append("</li>"); } $("a#next").attr("rel", $a("a#next").attr("rel") + 1); }, error : function ( XMLHttpRequest, textStatus, errorThrown) { $("div#showcase h1").text("An error has occured."); } }); });

    Read the article

  • How to reslove mysql_fetch_assoc(): problems!

    - by sky
    When i use the code below, im getting this error: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource when returning the data, anyone can fix it? Thanks! <?php $mysql_server_name="localhost"; $mysql_username=""; $mysql_password=""; $mysql_database=""; $conn=mysql_connect($mysql_server_name, $mysql_username, $mysql_password); ?> <?php $result = mysql_query("SELECT * FROM users"); $arrays = array(); while ($row = mysql_fetch_assoc($result)) { foreach ($row as $key => $val) { if (!array_contains_key($key)) { $arrays[$key] = array(); } $arrays[$key][] = $val; } } ?> <script type="text/javascript"> <?php foreach ($arrays as $key => $val) { print 'var ' . $key . ' = ' . json_encode($val) . ";\r\n"; } ?> </script>

    Read the article

  • Send form with jQuery using keypress()

    - by kexxcream
    What I have so far: I have a form in PHP that contain the following: <input type="submit" id="0" name="0" value="This button"> <input type="submit" id="1" name="1" value="This button"> <input type="hidden" id="response" name="response" value="150"> The problem: I would like to send the response value together with either name=1 or name=0, depending on which button the user submit. However, this should be done using the keyboard letters A and S. If a user press the letter A then the value 0 should be submitted, and if S is pressed then the value 1 should be sent. The jQuery code: $(document).ready(function() { // listens for any navigation keypress activity $(document).keypress(function(e) { switch(e.which) { // user presses the "a" case 97: submitViaKeypress("0"); break; // user presses the "s" key case 115: submitViaKeypress("1"); break; } }); }); // shows a given element and hides all others function submitViaKeypress(element_id) { var response = $('#response').attr('value'); $.ajax({ type: "POST", url: "initiate.php", data: "response=" + response + "&" + element_id + "=" + element_id }); } Goal: That "initiate.php receive two POST variables (response and either 0 or 1).

    Read the article

  • Only first word of two strings gets added to db

    - by dkgeld
    When trying to add words to a database via php, only the first word of both strings gets added. I send the text via this code: public void sendTextToDB() { valcom = editText1.getText().toString(); valnm = editText2.getText().toString(); t = new Thread() { public void run() { try { url = new URL("http://10.0.2.2/HB/hikebuddy.php?function=setcomm&comment="+valcom+"&name="+valnm); h = (HttpURLConnection)url.openConnection(); if( h.getResponseCode() == HttpURLConnection.HTTP_OK){ is = h.getInputStream(); }else{ is = h.getErrorStream(); } h.disconnect(); } catch (Exception e) { // TODO Auto-generated catch block e.printStackTrace(); Log.d("Test", "CONNECTION FAILED 1"); } } }; t.start(); } When tested with spaces and commas etc. in a browser, the php function adds all text. The strings also return the full value when inserted into a dialog. How do I fix this? Thank you.

    Read the article

  • Strange mod_rewrite problem; Website works partially

    - by Camran
    I have Ubuntu 9.10 Server... I need to get mod_rewrite working... the mod_rewrite module IS LOADED. On my server, the httpd.conf is empty, instead everything (almost) is in a file called apache2.conf. Anyways, I have also read I have to change the AllowOverride None to AllowOverride All in some file... My httpd.conf is empty as you know, but I have a folder called sites-enabled which contains a 000-default file. This is where I have set: AllowOverride All Now my goal as I stated in the last Q is to turn this link: http://mydomain.com/ad.php?ad_id=Bmw_nice_M3_497379462 into this: http://mydomain.com/Bmw_nice_M3_497379462 So as I got an answer in the last Q i inserted this into the htaccess file: Options +FollowSymLinks Options +Indexes RewriteEngine On RewriteCond %{REQUEST_URI} !^/ad\.php RewriteRule ^(.*)$ ad.php?ad_id=$1 [L] Now, this works (no fully) when entering the url manually in the adress bar, but my website isn't working now for some reason. It is like the website is locked down or something, and unless I change AllowOverride to None it will act like that. Any ideas why? Also another note, the links inside the rewritten url doesn't work properly (images are not shown, while some are shown)...

    Read the article

< Previous Page | 527 528 529 530 531 532 533 534 535 536 537 538  | Next Page >