Search Results

Search found 62740 results on 2510 pages for 'php function'.

Page 180/2510 | < Previous Page | 176 177 178 179 180 181 182 183 184 185 186 187  | Next Page >

  • PHP Function needed for GENERIC sorting of a recordset array

    - by donbriggs
    Somebody must have come up with a solution for this by now. I wrote a PHP class to display a recordset as an HTML table/datagrid, and I wish to expand it so that we can sort the datagrid by whichever column the user selects. In the below example data, we may need to sort the recordset array by Name, Shirt, Assign, or Age fields. I will take care of the display part, I just need help with sorting the data array. As usual, I query a database to get a result, iterate throught he result, and put the records into an assciateiave array. So, we end up with an array of arrays. (See below.) I need to be able to sort by any column in the dataset. However, I will not know the column names at design time, nor will I know if the colums will be string or numeric values. I have seen a ton of solutions to this, but I have not seen a GOOD and GENERIC solution Can somebody please suggest a way that I can sort the recordset array that is GENERIC, and will work on any recordset? Again, I will not know the fields names or datatypes at design time. The array presented below is ONLY an example. Array ( [0] = Array ( [name] = Kirk [shrit] = Gold [assign] = Bridge ) [1] => Array ( [name] => Spock [shrit] => Blue [assign] => Bridge ) [2] => Array ( [name] => Uhura [shrit] => Red [assign] => Bridge ) [3] => Array ( [name] => Scotty [shrit] => Red [assign] => Engineering ) [4] => Array ( [name] => McCoy [shrit] => Blue [assign] => Sick Bay ) )

    Read the article

  • Execute javascript in PHP

    - by Andreas Bonini
    I'm generating your typical Web 2.0 HTML page with PHP: it contains a lot of <script> tags and javascript code that will substantially change the DOM after the load event. Is there a way to get the final HTML code directly from PHP, without opening the page with any browser? For example, let's say the HTML for the page is (it's just an example): <html> <head> <script>...the jquery library code...</script> <script>$(document).ready(function() { $("body").append("<p>Hi!</p>");</script> </head> <body> </body> </html> This HTML is saved in the $html PHP variable. Now, I want to pass that variable to some function that will return $result = <html>....<body><p>Hi!</p></body></html>. Is this possible?

    Read the article

  • Automatic facebook login

    - by Marc Böhmer
    I manage 4 Facebook fan pages. Whenever users make a news on his website, I post it on Facebook. Now I have made it that I only have to press a button and the news were posted. I would like to automate this. How can I do this? My problem is that it is associated with my Facebook account and then it is not running in cron job or nothing is posted. Can I use my login data to a file which the Cronjob can always log in?

    Read the article

  • Why is my PHP upload script not working?

    - by Turner
    Hello all, I am doing some simple work with uploading a file. I am ignoring error checking and exceptions at this point just to get my uploads working. I have this HTML form: <form action='addResult.php' method='post' enctype='multipart/form-data' target='results_iFrame' onsubmit='startUpload();'> Entry: <input type='text' id='entry' /> Stop: <input type='text' id='stop' /> Final: <input type='text' id='final' /> Chart: <input type='file' id='chart' /> <input type='submit' value='Add' /></form> As you can see, it calls 'addResult.php' within the iFrame 'results_iFrame'. The Javascript is just for animation purposes and to tell me when things are finished. addResult.php has this code in it (along with processing the other inputs): $upload_dir = "../img/"; $chart_loc = $upload_dir.basename($_FILES['chart']['name']); move_uploaded_file($_FILES['chart']['tmp_name'], $chart_loc); print_r($_FILES); It uses the 'chart' input from the form and tries to upload it. I have the print_r() function to display some information on $_FILES, but the array is empty, thus making this fail. What could I be doing wrong?

    Read the article

  • ORDER BY column_name help (via link in HTML table view) (PHP MySQL

    - by Derek
    My output for my table in HTML has several columns such as userid, name, age, dob. The table heading is simply the title of the column name, I want this to be a link, and when clicked, the selected column is sorted in order, ASC, and then DESC (on next click). I thought this was pretty straight forward but I'm having some difficulty. So far, I have produced this, and no output is taken, apart from the URL works by displaying 'users.php?orderby=userid' <?php if(isset($_GET['orderby'])){ $orderby = $_GET['orderby']; $query_sv = "SELECT * FROM users BY ".mysql_real_escape_string($orderby)." ASC"; } //default query else{ $query_sv = "SELECT * FROM users BY user_id DESC"; } ?> <tr> <th><a href="<?php echo $_SERVER['php_SELF']."?orderby=userid";?>">User ID</a></th> Hoefully if I get this working, I can sort the users by D.O.B. next also using the same principles. Does anyone have any ideas?

    Read the article

  • Create facebook object each and every time?

    - by oshirowanen
    I have a login page which will log a user into my webapp based on their facebook login details. I then create a session to remember who they are. What I want to know is, should I be creating and/or checking the facebook credential on every single page of my webapp, or should I simply use the session I create at the beginning to login? For example, once they have logged in, I would like to allow them to post a message onto their own facebook wall from my app. Should I check the login credentials before they can post by recreating the facebook object, or should I simply use the stored login details already in my session and use that to post to their facebook wall?

    Read the article

  • Comparing two ISO8601 dates strings in PHP

    - by oompahloompah
    I need to compare (actually rank/sort) dates in a PHP script. The dates are ISO-8601 Date format i.e. YYYY-MM-DD I wrote a comparison function which splits the dates and compares by year/month/day. However, it seems this may be overkill and I could just as easily done a simple string comparison like: if ($date1 < $date2) // do something elseif( $date1 > $date2) //do something else else //do yet another thing Is my assumption about (ISO-8601) Date string comparison correct - i.e. can I get rid of my function (to save a few clock cycles on the server), or is it safer to explicity do the comparison in a custom function?

    Read the article

  • Javascript + PHP $_POST array empty

    - by Peterim
    While trying to send a POST request via xmlhttp.open("POST", "url", true) (javascript) to the server I get an empty $_POST array. Firebug shows that the data is being sent. Here is the data string from Firebug: a=1&q=151a45a150.... But $_POST['q'] returns nothing. The interesting thing is that file_get_contents('php://input') does have my data (the string above), but PHP somehow doesn't recognize it. Tried both $_POST and $_REQUEST, nothing works. Headers being sent: POST /test.php HTTP/1.1 Host: website.com User-Agent: Mozilla/5.0 (Windows; U; Windows NT 6.1; en-US; rv:1.9.2.3) Gecko/20100401 Firefox/3.6.3 Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8 Accept-Language: en-us;q=0.7,en;q=0.3 Accept-Encoding: gzip,deflate Accept-Charset: utf-8;q=0.7,*;q=0.7 Keep-Alive: 115 Connection: keep-alive Referer: http://website.com/ Content-Length: 156 Content-Type: text/plain; charset=UTF-8 Pragma: no-cache Cache-Control: no-cache Thank you for any suggestions.

    Read the article

  • PHP Class Question

    - by Jerry
    Hi all I am trying to build a PHP class to check the username and password in MySql. I am getting "mysql_query(): supplied argument is not a valid MySQL-Link resource in D:\2010Portfolio\football\main.php on line 38" database has errors:" message. When I move my userQuery code out of my Class, it works fine. Only if it is inside my Class. I am not sure if the problem is that I don't build the connection to mysql inside my Class or not. Thanks for any helps!! Here is my code <?php $userName=$_POST['userName']; $userPw=$_POST['password']; class checkUsers { public $userName; public $userPw; function checkUsers($userName='', $userPw=''){ $this->userName=$userName; $this->userPw=$userPw; } function check1(){ if (empty($this->userName) || empty($this->userPw)){ $message="<strong>Please Enter Username and Password</strong>"; }else{ //line 38 is next line $userQuery=mysql_query("SELECT userName, userPw FROM user WHERE userName='$this->userName' and userPw='$this->userPw'", $connection); if (!$userQuery){ die("database has errors: ". mysql_error()); } if(mysql_num_rows($userQuery)==0){ $message="Please enter valid username and password"; } }//end empty check return $message; }//end check1 method }//end Class $checkUser=new checkUsers($userName, $userPw); echo $checkUser->check1(); ?> I appreciate any helps!!!!

    Read the article

  • change last key name from array in php

    - by robertdd
    i want to be able to change the last key from array i try with this function i made: function getlastimage($newkey){ $arr = $_SESSION['files']; $oldkey = array_pop(array_keys($arr)); $arr[$newkey] = $arr[$oldkey]; unset($arr[$oldkey]); $results = end($arr); //echo json_encode($results); print_r($arr); } if i call the function getlastimage('newkey') it change the key!but after if i print the $_SESSION the key is not changed? why this?

    Read the article

  • Setting up a LAMP VM server for Development and Testing?

    - by TdotThomas
    Info: I would like to set up a VM server on my local computer which will serve pages in the exact same way as my current hosting (but only to me on my local computer). I currently pay a big web hosting company to host my website & web store and they are doing a great job, but I would like to be able to work on my Web site and its corresponding MySQL DB, HTML, and PHP code without being at risk of messing something completely up on the live servers. My current plan of action: Set up a VM webserver with Debian, MySQL, PHP, Apache. Copy web store (PHP/HTML) code to VM server. Copy my current MySQL databases from my hosting provider and install on VM server. Modify and test new features on VM server. Upload MySQL DB and HTML/PHP code back to web host's server where it should work as before but with new modifications. Questions: Now I'm pretty sure I have steps one and two down correctly but I can't for the life of me figure out how to proceed next, so here are my questions. I have my /etc/host file set up so www.MySite.test redirects to the IP address of the local VM webserver. Once I import my PHP/HTML files and MySQL file whats the best way to navigate around the fact that all of my files and DBs will reference www.MySite.com. I can export my MySQL dbs but do I also have to export my MySQL users and passwords to access those db or are those coded into my html/php code?

    Read the article

  • Filter sub-categories like in layered navigation

    - by russjman
    I created a new template file catalog/category/list.phtml. This is to display all sub categories of the current category. I have layered navigation which is displaying sub-categories as one of the filters, but I want this new template to work with these filters as well. Right now when i click the subcategory filter, it filters all products on the page, but still displays all categories of the parent category. $_filters is how i am trying to access these filters, but i get nothing. Is there something i am not initializing correctly to have access to these filters from the layered navigation. <?php $_helper = $this->helper('catalog/output'); $_filters = $this->getActiveFilters(); echo $_filters; if (!Mage::registry('current_category')) return ?> <?php $_categories=$this->getCurrentChildCategories() ?> <?php $_count = is_array($_categories)?count($_categories):$_categories->count(); ?> <?php if($_count): ?> <?php foreach ($_categories as $_category): ?> <?php if($_category->getIsActive()): ?> <?php $cur_category=Mage::getModel('catalog/category')->load($_category->getId()); $layer = Mage::getSingleton('catalog/layer'); $layer->setCurrentCategory($cur_category); $_imgHtml = ''; if ($_imgUrl = $this->getCurrentCategory()->getImageUrl()) { $_imgHtml = '<img src="'.$_imgUrl.'" alt="'.$this->htmlEscape($_category->getName()).'" title="'.$this->htmlEscape($_category->getName()).'" class="category-image" />'; $_imgHtml = $_helper->categoryAttribute($_category, $_imgHtml, 'image'); } echo $_category->getImageUrl(); ?> <div class="category-image-box"> <div class="category-description clearfix" > <div class="category-description-textbox" > <h2><span><?php echo $this->htmlEscape($_category->getName()) ?></span></h2> <p><?php echo $this->getCurrentCategory()->getDescription() ?></p> </div> <a href="<?php echo $this->getCategoryUrl($_category) ?>" class="collection-link<?php if ($this->isCategoryActive($_category)): ?> active<?php endif ?>" >See Entire Collection</a> <a href="<?php echo $this->getCategoryUrl($_category) ?>"><?php if($_imgUrl): ?><?php echo $_imgHtml ?><?php else: ?><img src="/store/skin/frontend/default/patio_theme/images/category-photo.jpg" class="category-image" alt="collection" /><?php endif; ?></a> </div> <?php echo '<pre>'.print_r($_category->getData()).'</pre>';?> </div> <?php endif; ?> <?php endforeach ?> <?php endif; ?>

    Read the article

  • Finding cause of memory leaks in large PHP stacks

    - by Mike B
    I have CLI script that runs over several thousand iterations between runs and it appears to have a memory leak. I'm using a tweaked version of Zend Framework with Smarty for view templating and each iteration uses several MB worth of code. The first run immediately uses nearly 8MB of memory (which is fine) but every following run adds about 80kb. My main loop looks like this (very simplified) $users = UsersModel::getUsers(); foreach($users as $user) { $obj = new doSomethingAwesome(); $obj->run($user); $obj = null; unset($obj); } The point is that everything in scope should be unset and the memory freed. My understanding is that PHP runs through its garbage collection process at it's own desire but it does so at the end of functions/methods/scripts. So something must be leaking memory inside doSomethingAwesome() but as I said it is a huge stack of code. Ideally, I would love to find some sort of tool that displayed all my variables no matter the scope at some point during execution. Some sort of symbol-table viewer for php. Does anything like that or any other tools that could help nail down memory leaks in php exist?

    Read the article

  • Android Camera takePicture function does not call Callback function

    - by Tomáš 'Guns Blazing' Frcek
    I am working on a custom Camera activity for my application. I was following the instruction from the Android Developers site here: http://developer.android.com/guide/topics/media/camera.html Everything seems to works fine, except the Callback function is not called and the picture is not saved. Here is my code: public class CameraActivity extends Activity { private Camera mCamera; private CameraPreview mPreview; private static final String TAG = "CameraActivity"; /** Called when the activity is first created. */ @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.camera); // Create an instance of Camera mCamera = getCameraInstance(); // Create our Preview view and set it as the content of our activity. mPreview = new CameraPreview(this, mCamera); FrameLayout preview = (FrameLayout) findViewById(R.id.camera_preview); preview.addView(mPreview); Button captureButton = (Button) findViewById(R.id.button_capture); captureButton.setOnClickListener(new OnClickListener() { @Override public void onClick(View v) { Log.v(TAG, "will now take picture"); mCamera.takePicture(null, null, mPicture); Log.v(TAG, "will now release camera"); mCamera.release(); Log.v(TAG, "will now call finish()"); finish(); } }); } private PictureCallback mPicture = new PictureCallback() { @Override public void onPictureTaken(byte[] data, Camera camera) { Log.v(TAG, "Getting output media file"); File pictureFile = getOutputMediaFile(); if (pictureFile == null) { Log.v(TAG, "Error creating output file"); return; } try { FileOutputStream fos = new FileOutputStream(pictureFile); fos.write(data); fos.close(); } catch (FileNotFoundException e) { Log.v(TAG, e.getMessage()); } catch (IOException e) { Log.v(TAG, e.getMessage()); } } }; private static File getOutputMediaFile() { String state = Environment.getExternalStorageState(); if (!state.equals(Environment.MEDIA_MOUNTED)) { return null; } else { File folder_gui = new File(Environment.getExternalStorageDirectory() + File.separator + "GUI"); if (!folder_gui.exists()) { Log.v(TAG, "Creating folder: " + folder_gui.getAbsolutePath()); folder_gui.mkdirs(); } File outFile = new File(folder_gui, "temp.jpg"); Log.v(TAG, "Returnng file: " + outFile.getAbsolutePath()); return outFile; } } After clicking the Button, I get logs: "will now take picture", "will now release camera" and "will now call finish". The activity finishes succesfully, but the Callback function was not called during the mCamera.takePicture(null, null, mPicture); function (There were no logs from the mPicture callback or getMediaOutputFile functions) and there is no file in the location that was specified. Any ideas? :) Much thanks!

    Read the article

  • PHP - upload.class image and $_FILES

    - by Ockonal
    Hello, I'm using class.upload.php to upload pictures onto the server. Here is my form: <form action="<?="http://".$_SERVER['SERVER_NAME'].$_SERVER['REQUEST_URI'];?>" method="post" enctype="multipart/form-data"> <table style="width: 100%; padding-top: 20px;"> <tr> <td>Image file:</td> <td><input type="file" name="image_file" /></td> </tr> <tr> <td>&nbsp;</td> <td align="right"><input type="submit" name="add_new" value="Add image" /><td></td> </tr> </table> </form> In php code I do: if( array_key_exists('add_new', $_POST) ) { echo 'add new is in array'; echo '<pre>'; print_r($_POST); print_r($_FILES); echo '</pre>'; $handle = new Upload($_FILES['image_file']); ... } Here is an output of print_r: Array ( [image_file] => somefile.png [add_new] => Add image ) Array ( ) As you can see second array is empty ($_FILES), so image doesn't upload. Why? operating system : Linux PHP version : 5.2.12 GD version : 2.0.34 supported image types : png jpg gif bmp open_basedir : /home/httpd/vhosts/kz-gbi.ru/httpdocs:/tmp

    Read the article

  • Executing system command in php, differs in using broswer and in using command line

    - by Amit
    Hi, I have to execute a Linux "more" command in php from a particular offset, format the result and display the result in Browser. My Code for the above is : <html> <head> <META HTTP-EQUIV=REFRESH CONTENT=10> <META HTTP-EQUIV=PRAGMA CONTENT=NO-CACHE> <title>Runtime Access log</title> </head> <body> <?php $moreCommand = "more +3693 /var/log/apache2/access_log | grep -v -e '.jpg' -e '.jpeg' -e '.css' -e '.js' -e '.bmp' -e '.ico'| wc -l"; exec($moreCommand, $accessDisplay); echo "<br/>No of lines are : $accessDisplay[0] <br/>"; ?> The output at the browser is :: No of lines are : 3428 (This is wrong) While executing the same command using command line gives a different output. My code snippet for the same is : <?php $moreCommand = "more +3693 /var/log/apache2/access_log | grep -v -e '.jpg' -e '.jpeg' -e '.css' -e '.js' -e '.bmp' -e '.ico'| wc -l"; exec($moreCommand, $accessDisplay); echo "No of lines are : $accessDisplay[0] \n"; ? The output at the command line is :: No of lines are : 279 (This is correct) While executing the same command directly in command line, gives me output as 279. I am unable to understand why the output of the same command is wrong in the browser. Its actually giving the word count of lines, ignoring the offset parameter. Please help !! Thanks, Amit

    Read the article

  • Create an array from mysql with column names and values [on hold]

    - by ScaZ
    i'm trying to create an array with PHP and MySQL, but i always get errors. The code i'm using function db_listar_usuarios(){ $link=db_connect(); $query = "select * from usuarios" or die("Problemas en el select: " . mysqli_error($link)); $result = $link->query($query); while($row = mysqli_fetch_assoc($result)) { $row['nombre'] . array(; foreach ($row as $col => $val) { $col => $val; } } } And what I want to create with this code is: array( 'john' => array('address' => 'st 123', 'age' => '25', 'surname' => 'doe'), 'ane' => array('address' => 'av 456', 'age'=> '32', 'surname' => 'smith'), ); To use then like something like this: private $contacts = db_listar_usuarios(); I use 2 files: functions.php and server.php server.php is a downloaded file example to do a REST API. Here are both of them. server.php - pastebin.com/5j54m1Mz functions.php - pastebin.com/N7jMhSBa Thank you in advance!

    Read the article

  • Returning a local object from a function

    - by pocoa
    Is this the right way to return an object from a function? Car getCar(string model, int year) { Car c(model, year); return c; } void displayCar(Car &car) { cout << car.getModel() << ", " << car.getYear() << endl; } displayCar(getCar("Honda", 1999)); I'm getting an error, "taking address of temporary". Should I use this way: Car &getCar(string model, int year) { Car c(model, year); return c; }

    Read the article

  • Use of date function in PHP to output a user-friendly date

    - by Jamie
    I have a MySQL database column named DateAdded. I'd like to echo this as a readable date/time. Here is a simplified version of the code I currently have: $result = mysql_query(" SELECT ListItem, DateAdded FROM lists WHERE UserID = '" . $currentid . "' "); while($row = mysql_fetch_array($result)) { // Make the date look nicer $dateadded = date('d-m-Y',$row['DateAdded']); echo $row['ListItem'] . ","; echo $dateadded; echo "<br />"; } Is the use of the date function the best way to output a user-friendly date? Thanks for taking a look,

    Read the article

  • PHP Outputting File Attachments with Headers

    - by OneNerd
    After reading a few posts here I formulated this function which is sort of a mishmash of a bunch of others: function outputFile( $filePath, $fileName, $mimeType = '' ) { // Setup $mimeTypes = array( 'pdf' => 'application/pdf', 'txt' => 'text/plain', 'html' => 'text/html', 'exe' => 'application/octet-stream', 'zip' => 'application/zip', 'doc' => 'application/msword', 'xls' => 'application/vnd.ms-excel', 'ppt' => 'application/vnd.ms-powerpoint', 'gif' => 'image/gif', 'png' => 'image/png', 'jpeg' => 'image/jpg', 'jpg' => 'image/jpg', 'php' => 'text/plain' ); // Send Headers //-- next line fixed as per suggestion -- header('Content-Type: ' . $mimeTypes[$mimeType]); header('Content-Disposition: attachment; filename="' . $fileName . '"'); header('Content-Transfer-Encoding: binary'); header('Accept-Ranges: bytes'); header('Cache-Control: private'); header('Pragma: private'); header('Expires: Mon, 26 Jul 1997 05:00:00 GMT'); readfile($filePath); } I have a php page (file.php) which does something like this (lots of other code stripped out): // I run this thru a safe function not shown here $safe_filename = $_GET['filename']; outputFile ( "/the/file/path/{$safe_filename}", $safe_filename, substr($safe_filename, -3) ); Seems like it should work, and it almost does, but I am having the following issues: When its a text file, I am getting a strange symbol as the first letter in the text document When its a word doc, it is corrupt (presumably that same first bit or byte throwing things off). I presume all other file types will be corrupt - have not even tried them Any ideas on what I am doing wrong? Thanks - UPDATE: changed line of code as suggested - still same issue.

    Read the article

  • How Can I optimize this RewriteEngine Code?

    - by Lucki Mile
    I have server overload, server admin said that this issue is caused from htaccess file This is the code: RewriteEngine On RewriteBase /here/ RewriteRule ^top/?$ index.php?mode=top [QSA] RewriteRule ^top/video/?$ index.php?mode=top&cat=vids [QSA] RewriteRule ^top/picture/?$ /index.php?mode=top&cat=pics [QSA] RewriteRule ^random$ index.php?mode=random [QSA] RewriteRule ^random/video/?$ index.php?mode=random&cat=vids [QSA] RewriteRule ^random/picture/?$ index.php?mode=random&cat=pics [QSA] RewriteRule ^new/?$ index.php [QSA] RewriteRule ^new/video/?$ index.php?mode=&cat=vids [QSA] RewriteRule ^new/picture/?$ index.php?mode=&cat=pics [QSA] RewriteRule ^video/([0-9]+)_(.*)$ item.php?cat=vids&id=$1 [QSA] RewriteRule ^picture/([0-9]+)_(.*)$ item.php?cat=pics&id=$1 [QSA] ErrorDocument 404 /item.php

    Read the article

  • php form submit and the resend infromation screen

    - by Para
    Hello, I want to ask a best practice question. Suppose I have a form in php with 3 fields say name, email and comment. I submit the form via POST. In PHP I try and insert the date into the database. Suppose the insertion fails. I should now show the user an error and display the form filled in with the data he previously inserted so he can correct his error. Showing the form in it's initial state won't do. So I display the form and the 3 fields are now filled in from PHP with echo or such. Now if I click refresh I get a message saying "Are you sure you want to resend information?". OK. Suppose after I insert the data I don't carry on but I redirect to the same page but with the necessary parameters in the query string. This makes the message go away but I have to carry 3 parameters in the query string. So my question is: How is it better to do this? I want to not carry around lots of parameters in the query string but also not get that error. How can this be done? Should I use cookies to store the form information.

    Read the article

  • CURL issue in PHP while getting location list

    - by Ajay
    I am retrieving the nearest locations available from a given address (Longitude/Latitude) from geolocation website. It works fine, but for some places it gives junk characters in the name. Moreover, in browser I am getting different characters compared to my PHP CURL functionality. Here is the URL http://www.geoplugin.net/extras/nearby.gp?lat=17.7374669&long=83.3214858&limit=5&radius=50&format=php One of the location is "Sitampeta" in original location name, but in browser I am getting "Sitammapeta" where as in CURL function I am getting "Sītammapeta". Please tell me why this difference. I wrote a function to convert browser output to original which works fine. function convert ($old) { $n=""; for ($i=0; $i<strlen($old); $i++) { $n .= chr(ord(substr($old,$i,1))); } return $n; } But I dont understand how I convert the CURL output to original name.

    Read the article

< Previous Page | 176 177 178 179 180 181 182 183 184 185 186 187  | Next Page >