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  • Export XML with only one MySQL request ?

    - by mere-teresa
    I want to export in XML format some data from 7 tables (MySQL database), and then I want to import in another database. And I have a update or insert rule for data. I already have a SQL query retrieving all data, with JOINs on my 7 tables. But...when I try to put data in XML format, I reach a limit. My PHP loop can catch each row, but I would like to benefit from hierachical structure of the XML, and all I have are rows with the same data repeated. It is better to query once and to construct the XML tree in PHP or to query each time I want access to a lower level ?

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  • Need help with regular expressions - URL redirection

    - by paperless
    Hello everyone. I'm trying to redirect an easy to remember url to a php file but I'm having some trouble with the regex. Here's what I have at the moment: RewriteRule ^tcb/([a-zA-Z0-9]{1,})/([a-zA-Z0-9]{1,})/([a-zA-Z0-9]{1,}) /tcb/lerbd.php?autocarro=$1&tipo=$2&dsd=$3 It is working but only if I supply all 3 arguments. I want the last two arguments to be optional so it either works with only the first or all three. I'm hoping you can help me with this. Thank you very much.

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  • Simple question about javascript history.go

    - by Camran
    I have a classifieds website. In every classified, there is a back link which simply takes the browser back one step. This is because when users search classifieds, and click on one to view it, they can easily go back with a link also (instead of only the browser back button). Here is the problem, if the classified is entered directly into the adress bar of a browser, or if somebody bookmarked a classified, then this back-link would take them someplace else... Is there any way of making sure that the previous page is a certain page (index.php in my case)? This way I would only display the back link if the previous page was index.php... Thanks

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  • Create a picture with GD containing other images

    - by Jensen
    Hi, I would like to create a picture in PHP with GD composed by different other pictures. For example I have 6 pictures (or more) and I would like to create ONE picture who contain these different pictures. The Difficulty is that my final picture must have a fixed width and height (304x179), so if the different pictures are too big they must be cut. This is an example from IconFinder : This picture is composed by 6 images, but the 3rd bird (green) is cutted, and the 4, 5 and 6 are cutted in the bottom. This is what I want, can you give me some help to write this code in PHP ? Thanks

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  • Placing .htaccess file in var/www folder messes my website up...

    - by Camran
    I am playing with mod_rewrite now, and have successfully enabled it. However, I need to put a htaccess file inside var/www/ in order to achieve what I want, which is to rename Urls simply... When I place it my website becomes very strange and nothing basically works... Is there any code I need to put into the htaccess file in order for things to act normally? Here is the htaccess file I have so far: Options +FollowSymLinks Options +Indexes RewriteEngine On RewriteCond %{REQUEST_URI} !^/ad\.php RewriteRule ^(.*)$ ad.php?ad_id=$1 [L] My DocumentRoot is also set to var/www/ and my entire website root is there... (index.html etc etc)... What am I missing about the htaccess? If you need more input let me know...

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  • Warning: session_start(): Cannot send session cache limiter - headers already sent

    - by shyam
    I am receiving this warning in a page while I try starting the session. <?php session_start(); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" > <head> <title>Video</title> </head> -this is the part of the code. There are no characters (not even space) before the first line. This page is reached after logging in from another page, through redirection. Any help? (PHP version is 5.2)

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  • Won't connect to the database

    - by user1657958
    I'm confused...I'm using the same code in a different document and in there it's not a problem to get a connection to the database. But in the new document it's just not working...(password, username, database name...all is checked and correct) :-/ <?php define ("DB_HOST", "db1234567.db.hello.com"); // set database host define ("DB_USER", "db1234567"); // set database user define ("DB_PASS","password123"); // set database password define ("DB_NAME","db1234567"); // set database name $link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Couldn't make connection."); $db = mysql_select_db(DB_NAME, $link) or die("Couldn't select database"); ?> In the browser I get this: "Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'db1234567'@'123.123.12.12 (using password: YES) in /homepages/12/1234567/test/test.php on line 8 Couldn't make connection." Would be cool if someone could help me :) I'm not seeing any error... Thx!

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  • How to access constant defined in child class from parent class functions?

    - by kavoir.com
    I saw this example from php.net: <?php class MyClass { const MY_CONST = "yonder"; public function __construct() { $c = get_class( $this ); echo $c::MY_CONST; } } class ChildClass extends MyClass { const MY_CONST = "bar"; } $x = new ChildClass(); // prints 'bar' $y = new MyClass(); // prints 'yonder' ?> But $c::MY_CONST is only recognized in version 5.3.0 or later. The class I'm writing may be distributed a lot. Basically, I have defined a constant in ChildClass and one of the functions in MyClass (father class) needs to use the constant. Any idea?

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  • Add a custom class name to Wordpress body tag?

    - by Scott B
    I'd like to place a directive in my theme's functions.php file which appends a classname to the wordpress body tag. Is there a built-in API method for this? For example, my body tag code is... <body <?php if(function_exists("body_class") && !is_404()){body_class();} else echo 'class="page default"'?>> And it results in the following being written to the body tag (depending on the context in which the page is presented (page, post, logged-in, etc) <body class="home blog logged-in"> Depending on the child theme I'm using at the time, I want it to be... <body class="home blog logged-in mychildthemename">

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  • How do stop form posting to mysql if database contains a specific ID?

    - by user342391
    I have a form that I am using to post data to mysql. Before submitting the form I want to check the database and see if there are any fields in the column 'customerid' that equal 'userid' and if so not to post the form. Basically, I am trying to limit my users from posting more than once. Users will be able to login to my system and make ONE post. They will be able to delete and modify their post but are only limited to one post. How would I do this??? Code so far: <?php include '../login/dbc.php'; page_protect(); $userid = $_SESSION['user_id']; $sql="INSERT INTO content (customerid, weburl, title, description) VALUES ('$_POST[userid]','$_POST[webaddress]','$_POST[pagetitle]','$_POST[pagedescription]')"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } echo "1 record added"; ?>

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  • How to know if the website being scraped has changed?

    - by Lost_in_code
    I'm using PHP to scrape a website and collect some data. It's all done without using regex. I'm using php's explode() method to find particular HTML tags instead. It is possible that if the structure of the website changes (CSS, HTML), then wrong data may be collected by the scraper. So the question is - how do I know if the HTML structure has changed? How to identify this before storing any data to my database to avoid wrong data being stored.

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  • Uploading files wont work on production server

    - by Camran
    I have a php script which uploads images to a temporary folder on the server. This works on my local computer with the local (virtual) server. (wampserver). However, on the production server (linux) I cant get it working. Everything is loading as it should, except for that the file doesn't show up on the server. My Q is, is there anything I should think about when moving to a production server with uploading images or files? I have set permissions on the folder where the images are supposed to go to 777 and also the php-code which uploads them to 777. Thanks

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  • CakePHP accessing other controllers

    - by James
    CakePHP Newbie :) I am having trouble accessing another controller and passing that data to a view in one of my controllers: In controllers/landings_controller.php: var $uses = 'User'; function home() { $userdata = $this->User->read(); $this->set(compact('userdata')); } In views/landings/home.ctp: <?php echo $this->userdata; ?> When accessing /landings/home I get the following error: Notice (8): Undefined property: View::$userdata [APP/views/landings/home.ctp, line 38] I don't know what I am doing wrong. Any help? Thanks!

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  • Include a repetitive chunk of HTML

    - by user146780
    I basically have a div on my site that always has the same stuff. However, this div is not present on all pages which is why I won't use the dynamic web template. I was wondering if it was possible for PHP to get the code from a document on the server and put in into the div? Ex: <div id="section... then my text file contains <p>hello</p> Basically I want PHP to put it into the div when the user sees it. If theres a smarter way of acheiving this I'd be open to it as well. Thanks

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  • Is there any way to rename or hide only one HTML tag?

    - by Jason
    Preface: I cannot rename the source tags or edit their IDs. Any changes to the tags must happen after they have been fetched. What I'm doing: using file_get_contents in PHP, I am requesting data from a remote site. This data is just two <p> tags. I need to hide or rename the second of the two <p> tags. Is this possible with PHP or jQuery? What I'm working with: <p>Hello my name is test</p><p>I like studying geology.</p>

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  • Apache crashes after installing mysqli

    - by Marco P.
    System: Apache 2.2 running on Windows 2008 Server with PHP 5.2.17 VC6 Thread-Safe as a Module and MySQL 5.5.17 - all working fine. After installing mysqli using the php package, Apache won't start anymore. There is no error message in the log. What I have tried: Make sure Windows PATH points to libmysql.dll: Yes, done. Make sure extension_dir points to the right point: Yes. Other extensions load fine. Try without mysqli: Yes, Apache loads fine then. Try without mysql: Yes, does not help. Test mysql itself: Restarts server! Overwrite libmysql: Yes, does not help. It seems to me that there is some general problem with MySql, but the DB server seems to be running fine. I'm really out of ideas of things I could try, so I'm desperate for any hints or tricks.

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  • how to list out the submited data in same page where form submitted?

    - by OM The Eternity
    I have a form with 3 text values and one image.. I want to save these values such that i can display these records in the list below.. how can i do that... I am using osCommerce For example: <form method="post" id="fm-form" action ="" enctype="multipart/form-data"> <label>Name:</label> <input type="text" id="fm-name" name="fm-name" value="" /> <label>Email:</label> <input type="text" id="fm-email" name="fm-email" value="" /> <label>Birthdate:</label> <input type="text" id="fm-birthdate" name="fm-birthdate" value="" /> <input type="file" id="fm-image" name="fm-image"/> <input type="submit" id="fm-submit" value="Save it"> </form> <table width="100%" border="0" cellspacing="0" cellpadding="0"> <tr > <td align="center" class="productListing-heading">Product(s)</td> <td align="center" class="productListing-heading">Edit</td> <td align="center" class="productListing-heading">Delete</td> </tr> <?php for($i=0;$i<$count_image;$i++){?> <tr> <td align="left" class="productListing-data1"> <?php echo tep_image(DIR_WS_IMAGES . $file_realname, $save_image[$i], '110', '110');?> </td> <td align="center" class="productListing-data1">Edit</td> <td align="center" class="productListing-data1">Delete</td> </tr> <tr><td>&nbsp;</td></tr> <?php }?> </table> In the above format as the form is submitted the image has to be stored in a count_image array variable... and the on its count, the list below the form is displayed.. but i cannot get it worked.. could u pls help in doing this...

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  • PDOException “could not find driver”

    - by Mike
    I have just installed Debian Lenny with Apache, MySQL, and PHP and I am receiving a PDOException could not find driver. This is the specific line of code it is referring to: $dbh = new PDO('mysql:host=' . DB_HOST . ';dbname=' . DB_NAME, DB_USER, DB_PASS) DB_HOST, DB_NAME, DB_USER, and DB_PASS are constants that I have defined. It works fine on the production server (and on my previous Ubuntu Server setup). Is this something to do with my PHP installation? Searching the internet has not helped, all I get is experts-exchange and examples, but no solutions.

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  • Keeping some files private

    - by user490895
    Hi, I was trying to create a directory of private files that could only be accessed when a user logs in. To do this, I used a folder outside the web directory, and then php to access it, if allowed. Here's an example: function display_movie($file){ printf("<video id='movie' width='960' height='416' controls='controls' onerror='fix()'> <source src='movie.php?file=%s' type='video/ogg; codecs=\"theora, vorbis\"'> </video>", rawurlencode($file)); } This works great for images, but breaks the media player. Also, I've only tested this locally on a Linux machine. Any ideas? Thanks.

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  • jQuery.post not working when using data type json

    - by swift
    I have been trying to utilize json in this jQuery.post because I need to return two values from my executed php. The code was working when I was not implementing json. I need to see if a promo code entered is valid for a particular broker. The two variables I need back are the instant message whether or not it's valid (this is displayed to the user) and I need to update a hidden field that will be used later while updating the database. The jQuery.post does not seem to be firing at all, but the code directly above it (the ajax-loader.gif) is working. I did re-write the whole thing at one point using jQuery.ajax, and had issues there too. Granted, I have probably been looking at this too long and have tried to re-write too many times, but any help is greatly appreciated!! Here's the jQuery.post <!-- Below Script is for Checking Promo Code Against Database--> <script type="text/javascript"> jQuery(document).ready(function() { jQuery("#promocode").keyup(function (e) { //removes spaces from PromoCode jQuery(this).val(jQuery(this).val().replace(/\s/g, '')); var promocode = jQuery(this).val(); var brokerdealerid = document.getElementById("BrokerDealerId").value; if(promocode.length > 0 ){ jQuery("#promo-result").html('<img src="../imgs/ajax-loader.gif" />'); jQuery.post( '../check_promocode.php', {promocode:promocode, brokerdealerid:brokerdealerid}, function(data) { $("#promo-result").html(data.promoresult); $("#promo-result-valid").html(data.promovalid); }, "json"); } }); }); </script> <!-- End Script is for Checking Promo Code Against Database--> Here's relevant code from check_promocode.php: //sanitize incoming parameters if (isset($_POST['brokerdealerid'])) $brokerdealerid = sanitizeMySQL($_POST['brokerdealerid']); $promocode = sanitizeMySQL($promocode); //check promocode in db $results = mysql_query("SELECT PromotionCodeIdentifier FROM PromotionCode WHERE PromotionCodeIdentifier='$promocode' AND BrokerDealerId='$brokerdealerid' AND PromotionCodStrtDte <= CURDATE() AND PromotionCodExpDte >= CURDATE()"); //return total count $PromoCode_exist = mysql_num_rows($results); //total records //if value is more than 0, promocode is valid if($PromoCode_exist) { echo json_encode(array("promoresult"=>"Promotion Code Valid", "promovalid"=>"Y")); exit(); }else{ echo json_encode(array("promoresult"=>"Invalid Promotion Code", "promovalid"=>"N")); exit(); }

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  • Stopping users posting more than once

    - by user342391
    Before posting my form I am checking the database to see if there are any previous posts from the user. If there are previous posts then the script will kick back a message saying you have already posted. The problem is that what I am trying to achieve isn't working it all goes wrong after my else statement. It is also probable that there is an sql injection vulnerability too. Can you help??4 <?php include '../login/dbc.php'; page_protect(); $customerid = $_SESSION['user_id']; $checkid = "SELECT customerid FROM content WHERE customerid = $customerid"; if ($checkid = $customerid) {echo 'You cannot post any more entries, you have already created one';} else $sql="INSERT INTO content (customerid, weburl, title, description) VALUES ('$_POST[customerid]','$_POST[webaddress]','$_POST[pagetitle]','$_POST[pagedescription]')"; if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } echo "1 record added"; ?>

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  • ajax how to read out $_post variable

    - by alex
    Hi, I am trying to filter/search a database with ajax $.ajax({ type: "POST", url: "filterSearch.php", queryString: qry, success: function(data){ alert( "Data Saved: " + data ); $('#searchResult').html(data); // Fill the search results box } }); Now in filterSearch.php i have the following test codes if(isset($_POST['queryString'])) { echo "TEST"; } if($_POST['runquery']==1) { $sql = "SELECT * FROM fs_vacatures WHERE here-the-like-query?"; $msg = $sql; echo $msg; die(); } die(); But nor TEST or the $sql is return in the alert??

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  • Secure method for linking to image files uploaded via custom CMS

    - by letseatfood
    How does one provide a direct URL for an image file (JPEG) that is secure? For example, if a PHP script writes uploaded images to directory http://www.somehost.com/images, is it okay to provide http://www.somehost.com/images/someimage.jpg as the URL, or is there a more secure way to do this? Should I look into something like the permalink feature that is a part of Wordpress? I am not sure if that is related. The main reason I ask is that I have a custom PHP/MySQL CMS for managing images. I would like for the client to be able to copy a link to the image they want and then include it in a TinyMCE text editor for inserting the image in their website. Thanks!

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