Search Results

Search found 31657 results on 1267 pages for 'php 5 2'.

Page 521/1267 | < Previous Page | 517 518 519 520 521 522 523 524 525 526 527 528  | Next Page >

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • 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

  • Looping Through Database Query

    - by DrakeNET
    I am creating a very simple script. The purpose of the script is to pull a question from the database and display all answers associated with that particular question. We are dealing with two tables here and there is a foreign key from the question database to the answer database so answers are associated with questions. Hope that is enough explanation. Here is my code. I was wondering if this is the most efficient way to complete this or is there an easier way? <html> <head> <title>Advise Me</title> <head> <body> <h1>Today's Question</h1> <?php //Establish connection to database require_once('config.php'); require_once('open_connection.php'); //Pull the "active" question from the database $todays_question = mysql_query("SELECT name, question FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Variable to hold $todays_question aQID $questionID = mysql_query("SELECT commentID FROM approvedQuestions WHERE status = active") or die(mysql_error()); //Print today's question echo $todays_question; //Print comments associated with today's question $sql = "SELECT commentID FROM approvedQuestions WHERE status = active"; $result_set = mysql_query($sql); $result_num = mysql_numrows($result_set); for ($a = 0; $a < $result_num; $a++) { echo $sql; } ?> </body> </html>

    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

  • Parameter error with Mysqli

    - by Morgan Green
    When I run this Query I recieve Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /home/morgan58/public_html/wow/includes/index/index_admin.php on line 188 SELECT * FROM characters WHERE id=5 Warning: mysqli_free_result() expects parameter 1 to be mysqli_result, boolean given in /home/morgan58/public_html/wow/includes/index/index_admin.php on line 194 The Query is running and it is strying to select the correct information, but for on the actual output it's giving me a fetch_array error; if anyone can see where the error lies it'd be much appreciated. Thank you. <?php $adminid= $admin->get_id(); $characterdb= 'characters'; $link = mysqli_connect("$server", "$user", "$pass", "$characterdb"); if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } $query = "SELECT * FROM characters WHERE id=$adminid"; $result = mysqli_query($link, $query); while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { echo $query; echo $row['name']; } mysqli_free_result($result); mysqli_close($link); ?>

    Read the article

  • destructor being called by subclass

    - by zero
    I'm currently learning more about php objects and constructors/destructors, but i've noticed in my code that the parent class's destructor is being called twice, I thought it was because i was extending the first class to my second class and that the second class was calling it, but this is what the php docs say about that: Like constructors, parent destructors will not be called implicitly by the engine. In order to run a parent destructor, one would have to explicitly call parent::__destruct() in the destructor body. so if it is not being called by the subclass then is it because by extended the first class that i've made a reference to the parent class making it call itself twice or I'm I way off base here? the code: <?php class test{ public $test1 = "this is a test of a pulic property"; private $test2 = "this is a test of a private property"; protected $test3 = "this is a test of a protected property"; const hello = 900000; function __construct($h){ //echo 'this is the constructor test '.$h; } function x($x2){ echo ' this is fn x'.$x2; } function y(){ print "this is fn y"; } } $obj = new test("this is an \"arg\" sent to instance of test"); class hey extends test{ function hey(){ $this->x('<br>from the host with the most'); echo ' <br>from hey class'.$this->test3; } } $obj2 = new hey(); echo $obj2::hello; ?>

    Read the article

  • Sort the $_POST variables

    - by Jerry
    Hello guys Might be an easy for you guys. I am trying to sort the $_POST variables that were sent by a form and update the sorted result in mysql. I am not sure how to do it and appreciate it anyone can help me about it. My main.php //I have a loop here. (omitted) //$k will be increased by 1 every time the loop starts, so I will know the total times of the loops //the form will be submitted to update.php echo "<input type='hidden' name='pickTotal' value='".$k."' />"; echo "<input type='hidden' id='point' name='earnedPoint".$k."' value='".$point."' />"; echo "<input type='hidden' id='users' name='userName".$k."' value='".$userPick['user']."' />"; //loop ends My update.php if(isset($_POST['submit'])){ $pickTotal=$_POST['pickTotal']; //get the total loop for ($p=0;$p<=$pickTotal;$p++){ $userToBeUpdated=$_POST['userName'.$p]; $userPoint=$_POST['earnedPoint'.$p]; //sort the $userPoint here. //I need to find out who got the most points //and list user's place. 1st, 2nd, 3rd...etc. //update my mysql } Thanks for any helps.

    Read the article

  • problem when uploading file

    - by Syom
    i have the form, and i want to upload two files. here is the script <form action="form.php" method="post" enctype="multipart/form-data" /> <input type="file" name="video" /> <input type="file" name="picture" > <input type="submit" class="input" value="?????" /> <input type="hidden" name="MAX_FILE_SIZE" value="100000000" /> </form> form.php: <? print_r($_FILES); $video_name = $_FILES["video"]["name"]; $image_name = $_FILES["picture"]["name"]; echo "video",$video_name; echo "image",$image_name; //returns Array ( ) videoimage ?> when i try to upload the file greater than 10MB, it doesn't happen. i try in many browsers. maybe i must change some field in php.ini? but i haven't permission to change them on the server. so what can i do? thanks

    Read the article

  • how to fetch data using jquery

    - by user3566029
    I have tried to connect my index.php file in 000webhost.com using jQuery. From the site menu I have connected with 000webhost using FTP details available in 000webhost.com. When I tried to read data from jQuery it doesn’t working. Anyone please tell me what should I need to change? Should I need to connect my Dreamweaver database to webhost? If yes please explain? This is what I have done. index.php $mysql_host = "mysql0.000webhost.com"; $mysql_database = "a000000_mydb"; $mysql_user = "a000000_root"; $mysql_password = "******"; $conn=@mysql_connect ( $mysql_host,$mysql_user,$mysql_password )or die('aa'); mysql_select_db($mysql_database,$conn) or die('eoor on db'); $quer=mysql_query("SELECT * FROM sam"); $res=array(); while($row=mysql_fetch_row($quer)) { $res[]=$row; } print(json_encode($res)); return json_encode($res); mysql_close(); index.html $.get('public_html/index.php', function( data ) { alert( 'Successful' +data); });

    Read the article

  • inserts 'Array' into mysql table

    - by Noah Smith
    i want to insert an array into a mysql table. The array is produced by script scanning all the links, converting into absolute links and then displaying them in an array. i decided to mysql_query the array into the table but now i am stuck. it only posts 'Array', instead of every row from the array into a different row. Any ideas??! <?php require_once('simplehtmldom_1_5/simple_html_dom.php'); require_once('url_to_absolute/url_to_absolute.php'); $connect = mysql_connect("xxxx", "xxxx", "xxx") or die('Couldn\'t connect to MySQL Server: ' . mysql_error()); mysql_select_db("xxxx", $connect ) or die('Couldn\'t Select the database: ' . mysql_error( $connect )); $links = Array(); $URL = 'http://www.theqlick.com'; // change it for urls to grab // grabs the urls from URL $file = file_get_html($URL); foreach ($file->find('a') as $theelement) { $links[] = url_to_absolute($URL, $theelement->href); } print_r($links); mysql_query("INSERT INTO pages (url) VALUES ('$links[]')"); mysql_close($connect);

    Read the article

  • Page does update with details from the database after i hit a button

    - by swathi
    I have a code and the way it should work is,when they click on NEW CUSTOMER,it takes them to test1.php where in they enter the details and they hit submit.it saves all the details in properly in the database and when i go back and hit REFRESH ,it should come up with the customer details which they had entered in previously. But what happens is, when i click on the REFRESH,it refreshes the same old page which is empty.I wanted to find out where am i missing the logic.Thanks in advance. The sample code would be <tr> <td class="tdvisitbig" colspan="5">THIS IS A TEST</td> </tr> <tr> <td class='tdvisitbig' colspan="5"><input type="button" onClick="openVisit('test1.php?id=<?=$key?>&name=<?=$name?>');return false;" value="NEW CUSTOMER" class="submit">&nbsp;<input type="button" value="REFRESH" name="add_xyz" class="submit" onClick="document.add.target='_self';document.add.action='test3.php?redirect=visit&section=test page';document.add.submit();"></td> </tr> <? $q = "SELECT address,customernum,status FROM customer WHERE name='$name' ORDER BY customernum"; $r = mysql_query( $q , $Link ); while( $rw = mysql_fetch_assoc( $r ) ) { extract( $rw ); ?> <tr> <? } ?>

    Read the article

  • Adding an integer at the end of an input's name to get a specific url

    - by Gadgetster
    I am trying to get a url where I can retrieve the selected values from. For example, if I put a check mark on a and b then sumbit, I will get: index.php?category=1&&category=2 I want to get this instead: index.php?category0=1&&category1=2 So that I can later get this specific value with $_GET['category0'] Is there a way to add a counter for the selected checkboxes and add 0,1,2,3.. at the end of the name of its input? <form action="" method="get"> <!-- this will be a php loop instead of hardcored which will retrieve data from the db --> <label><input type="checkbox" name="category" value="1">a</label> <label><input type="checkbox" name="category" value="2">b</label> <label><input type="checkbox" name="category" value="3">c</label> <label><input type="checkbox" name="category" value="4">d</label> <label><input type="checkbox" name="category" value="5">e</label> <input type="submit"> </form>

    Read the article

  • Apache mod_rewrite : How to REWRITE (or whatever) child directories to parent?

    - by ????
    Actually i am trying to make a PHP MVC like application. A basic one. The current milestone i am reaching already includes: Basic RESTful Routing Means, if i type: www.example.com/items/book/8888 .. it properly just stays there as it is and i can already slice out the URL by slashes / and loads the responsible Controllers .... etc from the top single index.php file. I mean, so it is OK for the backend PHP. But the only problem is, it still CAN NOT process the REWRITES properly. For example, the CSS & JS are BROKEN as if i VIEW PAGE SOURCE of the page www.example.com/items/book/8888, the asset files are being called as: www.example.com/items/book/8888/css/main.css www.example.com/items/book/8888/js/jquery.js .. which really are PROBLEMS because in the code is like: <link type="text/css" rel="stylesheet" media="all" href="css/main.css"> <script type="text/javascript" src="js/jquery.js"></script> So the question is: How can i use Apache REWRITE (or whatever approach) to make sure every ASSET FILES to be correctly being called from the DOCROOT. For example, if i am in the URL: www.example.com/items/book/8888 My ASSET FILES should still be called as: www.example.com/css/main.css www.example.com/js/jquery.js Or is there any other methods i need to follow? Please kindly help suggest. Thank you.

    Read the article

< Previous Page | 517 518 519 520 521 522 523 524 525 526 527 528  | Next Page >