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  • PHP, MySQL: Security concern; Page loads in a weird way

    - by Devner
    Hi all, I am testing the security of my website. I am using the following URL to load a PHP page in my website, on localhost: http://localhost/domain/user/index.php/apple.php When I do this, the page is not loading normally; Instead the images, icons used in the page simply vanish/disappear from the page. Only text appears. And also on any link I click on this page, it brings me to this same page again without navigating to the required page. So if I have hyperlinks to other pages, such as "SEARCH", which points to search.php, instead of navigating to the search.php page, it refreshes the index.php page and just appends the page name of the destination page to the end of the URL. For example, say I used the link above. It then loads the index.php page minus the images at it's will. When I click on the "Search" link to navigate to the search page, I see the following in the URL: http://localhost/domain/user/index.php/search.php I have a redirection configured to a 404 error page in my .htaccess file, but the page does not redirect to the 404 error page. Notice the search.php towards the end of the URL above. Any other link that I click, reloads the index.php page and just appends the destination page name to the end of the URL like I have shown above. I was expecting to see a 404 Error but that does not happen. The URL should not even be able to load the page because I do NOT have a "index.php" folder in my website. What can I do to solve this? All help is appreciated. Thank you.

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  • PHP class_exists always returns true

    - by Ali
    I have a PHP class that needs some pre-defined globals before the file is included: File: includes/Product.inc.php if (class_exists('Product')) { return; } // This class requires some predefined globals if ( !isset($gLogger) || !isset($db) || !isset($glob) ) { return; } class Product { ... } The above is included in other PHP files that need to use Product using require_once. Anyone who wants to use Product must however ensure those globals are available, at least that's the idea. I recently debugged an issue in a function within the Product class which was caused because $gLogger was null. The code requiring the above Product.inc.php had not bothered to create the $gLogger. So The question is how was this class ever included if $gLogger was null? I tried to debug the code (xdebug in NetBeans), put a breakpoint at the start of Product.inc.php to find out and every time it came to the if (class_exists('Product')) clause it would simply step in and return thus never getting to the global checks. So how was it ever included the first time? This is PHP 5.1+ running under MAMP (Apache/MySQL). I don't have any auto loaders defined.

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  • PHP While loop seperating unique categories from multiple 'Joined' tables

    - by Hob
    I'm pretty new to Joins so hope this all makes sense. I'm joining 4 tables and want to create a while loop that spits out results nested under different categories. My Tables categories id | category_name pages id | page_name | category *page_content* id | page_id | image_id images id | thumb_path My current SQL join <?php $all_photos = mysql_query(" SELECT * FROM categories JOIN pages ON pages.category = categories.id JOIN image_pages ON image_pages.page_id = pages.id JOIN images ON images.id = image_pages.image_id ");?> The result I want from a while loop I would like to get something like this.... Category 1 page 1 Image 1, image 2, image 3 page 2 Image 2, image 4 Category 2 page 3 image 1 page 4 image 1, image 2, image 3 I hope that makes sense. Each image can fall under multiple pages and each page can fall under multiple categories. at the moment I have 2 solutions, one which lists each category several times according to the the amount of pages inside them: eg. category 1, page 1, image 1 - category 1, page 1, image 2 etc One that uses a while loop inside another while loop inside another while loop, resulting in 3 sql queries. <?php while($all_page = mysql_fetch_array($all_pages)) { ?> <p><?=$all_page['page_name']?></p> <?php $all_images = mysql_query("SELECT * FROM images JOIN image_pages ON image_pages.page_id = " . $all_page['id'] . " AND image_pages.image_id = images.id"); ?> <div class="admin-images-block clearfix"> <?php while($all_image = mysql_fetch_array($all_images)) { ?> <img src="<?=$all_image['thumb_path']?>" alt="<?=$all_image['title']?>"/> <?php } ?> </div> <?php } } ?

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  • Pass a variable from the source file to an included file in PHP

    - by Alpha1
    For my website I want to store the general format of the site in a single PHP file in a single location, and each of the different pages content in the local location of the page. I then want to pass the title and content address to the included file via a variable. However I can't get the included format file to read the variables storing the title and content data. AKA, the called file for the individual page would be: <?php $title = 'Some Title'; $source_file = 'content.php'; readfile('http:...../format.php'); ?> The format file would be: <html> ... <title> <?php echo $title; ?> </title> ... <?php include($source_file); ?> ... I recall reading somewhere I need to include something to get the variables at the start of the format file, however I can't remember what it is or find where I found that information.

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  • How to avoid notice in php when one of the conditions is not true

    - by user225269
    I've notice that when one of the two conditions in a php if statement is not true. You get an undefined index notice for the statement that is not true. And the result in my case is a distorted web page. For example, this code: <?php session_start(); if (!isset($_SESSION['loginAdmin']) && ($_SESSION['loginAdmin'] != '')) { header ("Location: loginam.php"); } else { include('head2.php'); } if (!isset($_SESSION['login']) && ($_SESSION['login'] != '')) { header ("Location: login.php"); } else { include('head3.php'); } ?> If one of the if statements is not true. The one that is not true will give you a notice that it is undefined. In my case it says that the session 'login' is not defined. If session 'LoginAdmin' is used. What can you recommend that I would do in order to avoid these undefined index notice.

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  • Unable to Get values from Web Form to a PHP Class to Display

    - by kentrenholm
    I am having troubles getting the value from my variables submitted via a web form using a PHP class file. Here is my structure of the web page: Order Form Page Process.php Page Book.php Page I can easily get the user data entered (on Order Form Page), process, and display it on the Process.php page. The issue is that I must create a Book class and print the details of the data using the Book class. I have an empty constructor printing out "created" so I know my constructor is being called. I also am able to print the word "title" so I know I can print to the screen by using the Book class. My issue is that I can't get values in my variables in the Book class. Here is my variable declaration: private $title; Here is my printDetails function: public function printDetails () { echo "Title: " . $this->title . "<br />"; } Here is my new instance of the book class: $bookNow = new book; Here are my get and set functions: function __getTitle($title) { return $this->$title; } function __setTitle($title,$value) { $this->$title = $value; } I do have four other variables that I'm looking to display as well. Each of those have their own variable declaration, a line in printDetails, and their own setter and getter. Lastly, I also have a call to the Book class in my process PHP. It looks like this: function __autoload($book) { include $book . '.php'; } $bookNow = new book(); Any help, much appreciated. It must be something so very small (I'm hoping).

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  • JQuery to PHP function and back Ajaxed

    - by Xaris
    Hi all, i have a set of php function that i want to call on different events mostly onclick with jquery async (ajax). The first function is called on load $(document).ready(function() { $("#div2").hide('slow'); $("#div1").empty().html('<img src="ajax-loader.gif" />'); $.ajax( { type: "POST", url: "WebFunctions.php", data: {'func':'1'}, success: function(html) { $("#div1").show('slow').html(html) } }); The Data: {'func':'1'} -- is a switch statement on the php side switch($_POST['func']) { case '1': getParents(); break; case '2': getChilds(params); break; case '3': getChildObjects(params); break; default: } "This functions are calls to a soap server" <-- irrelevant. So when that function finishes i get an array which contains IDs and Names. I echo the names but i want the ID for reference so when i click on the echoed name i can call an other php function with parameter the ID of the name... How do i get rid of the switch statement?? How do i call properly php functions and pass params to it??? How can i save this IDs so when i click on an item with that id an other php function is called?? Plz feel free to ask any question, any answer is welcome :)

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  • How to load data on option box in php

    - by user225269
    I'm having trouble loading the mysql data on the html form option box using php. Here's my code: <?php $con = mysql_connect("localhost","myuname","mypassword"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("school", $con); $idnum= mysql_real_escape_string($_POST['idnum']); $result = mysql_query("SELECT * FROM student WHERE IDNO='$idnum'"); ?> <?php while ( $row = mysql_fetch_array($result) ) { ?> <tr> <td width="30" height="35"><font size="2">*I D Number:</td> <td width="30"><input name="idnum" type="text" maxlength="5" value="<?php echo $row["IDNO"]; ?>" readonly="readonly"></td> </tr> My problem is loading it here: <td><font size="2">Gender</td> <td> <select name="gender" id="gender"> <font size="2"> <option value="<?php echo $line['IDNO']; ?> "><?php $line['GENDER'] ; ?></option> </select></td></td> The table looks like this: IDNO | GENDER 123 | M 321 | F What am I supposed to do?To load the exact gender corresponding to the IDNO?

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  • Include php code within echo from a random text

    - by lisa
    I want to display a php code at random and so for I have <?php // load the file that contain thecode $adfile = "code.txt"; $ads = array(); // one line per code $fh = fopen($adfile, "r"); while(!feof($fh)) { $line = fgets($fh, 10240); $line = trim($line); if($line != "") { $ads[] = $line; } } // randomly pick an code $num = count($ads); $idx = rand(0, $num-1); echo $ads[$idx]; ?> The code.txt has lines like <?php print insert_proplayer( array( "width" => "600", "height" => "400" ), "http://www.youtube.com/watch?v=xnPCpCVepCg"); ?> Proplayer is a wordpress plugin that displays a video. The codes in code.txt work well, but not when I use the pick line from code.txt. Instead of the full php line I get: "width" => "600", "height" => "400" ), "http://www.youtube.com/watch?v=xnPCpCVepCg"); ?> How can I make the echo show the php code, rather than a txt version of the php code?

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  • creating new instance fails PHP

    - by as3isolib
    I am relatively new to PHP and having some decent success however I am running into this issue: If I try to create a new instance of the class GenericEntryVO, I get a 500 error with little to no helpful error information. However, if I use a generic object as the result, I get no errors. I'd like to be able to cast this object as a GenericEntryVO as I am using AMFPHP to communicate serialize data with a Flex client. I've read a few different ways to create constructors in PHP but the typical 'public function Foo()' for a class Foo was recommended for PHP 5.4.4 //in my EntryService.php class public function getEntryByID($id) { $link = mysqli_connect("localhost", "root", "root", "BabyTrackingAppDB"); if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit(); } $query = "SELECT * FROM Entries WHERE id = '$id' LIMIT 1"; if ($result = mysqli_query($link, $query)) { // $entry = new GenericEntryVO(); this is where the problem lies! while ($row = mysqli_fetch_row($result)) { $entry->id = $row[0]; $entry->entryType = $row[1]; $entry->title = $row[2]; $entry->description = $row[3]; $entry->value = $row[4]; $entry->created = $row[5]; $entry->updated = $row[6]; } } mysqli_free_result($result); mysqli_close($link); return $entry; } //my GenericEntryVO.php class <?php class GenericEntryVO { public function __construct() { } public $id; public $title; public $entryType; public $description; public $value; public $created; public $updated; // public $properties; } ?>

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  • php session_start() error

    - by tooepic
    Hi, i've used a sample found on online and applied it to my code: <?php session_start(); if (isset($_REQUEST["email"])) { $_SESSION["name"] = true; $host = $_SERVER["HTTP_HOST"]; $path = dirname($_SERVER["PHP_SELF"]); $sid = session_name() . "=" . session_id(); header("Location: index.php?$sid"); exit; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> ... ... and rest of the html code When I open this page, I got an error: Warning: session_start() [function.session-start]: Cannot send session cookie - headers already sent by (output started at /data/server/user/directory/sub-directory/login.php:1) in /data/server/user/directory/sub-directory/login.php on line 2 Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /data/server/user/directory/sub-directory/login.php:1) in /data/server/user/directory/sub-directory/login.php on line 2 I looked around to resolve this issue and saw few posts about this in this site also, but I just can't get a good grip on this...can't find the answer. Please help. Thanks.

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  • php search database for row

    - by Brenden Morley
    Okay I got code the code to pull data based on a users account number well here is what im using (And yes I know it isnt safe now that is the reason for my post) <?php include('config.php'); $user_info = fetch_user_info($_GET['AccountNumber']); ?> <html> <body> <div> <?php if ($user_info === false){ $Output = 'http://www.MyDomain.Com/'; echo '<META HTTP-EQUIV=Refresh CONTENT="0; URL='.$Output.'">'; }else{ ?> <center> <title><?php echo $user_info['FirstName'], ' ', $user_info['LastName'], ' - ', $user_info['City'], ', ', $user_info['State']; ?> - Name of site</title> So basically what this code is allowing me to do is have a file called Profile.php And when a user visits this this page it will return the data Like this http://MyDomain.com/Profile.php?AccountNumber=50b9c965b7c3b How can I do this securely cause right now its using a get method really unsafe to retive the account number from the url bar.

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  • PHP object help -> unexpected T_OBJECT_OPERATOR

    - by php-b-grader
    Please help me understand: print gettype(new CustomerObject()) prints: "object" (so it is an object) BUT print gettype((new CustomerObject())->get_customer()); prints: unexpected T_OBJECT_OPERATOR If I do it over two lines it works fine $object = new Customer($order->customer_id); print gettype($object); prints: object $customer = $object->get_customer(); print gettype($customer); prints: array It appears that the lines cannot be joined into a single call. Is this correct? and what is the logic behind that?

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  • Strange bug with PHP on Windows 7

    - by chessweb
    This is the configuration: Windows 7 Home Premium, XAMPP 1.7.3 (Apache 2.2.14 , PHP 5.3.1), Firefox 3.6 This is the PHP-code in a file named 'test.php' in htdocs: <?php echo('04556-8978765'); ?> On http://localhost/test.php I would expect to see the string 04556-8978765 in the browser. This is not what happens, though. The string appears for a short time and then it disappears altogether. Firebug shows an empty body-tag. However, when I look at page source, the string is there alright. When I change the string in the echo-statement to e.g. 4556-8978765, everything is fine. Internet Explorer 8 does not show this strange behavior. I could not reproduce this with the same Apache/PHP/Firefox configuration on Windows XP. '04556-8978765' is by no means unique. The couple '02065-96047' and '02065-9604' behave exactly the same. Can anybody reproduce this and offer an explanation as to what is going on? PS: If you can not see the string '04556-8978765' in the echo-statement above, look at this post with IE8.

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  • Session in php are not enough clear to me

    - by Lulzim
    I find sessions in php kind of confusing, can anybody of you explain those to me. I have an example which is not working in my case: I register sessions this way, would you please tell me is this the right way of registering sessions //this is the page from where i register myusername in sessions if($count==1){ session_start(); $_SESSION['myusername'] = $_POST['myusername']; include("enterpincover.php"); } else { echo "Wrong Pin"; } here i check first whether the username is registered in sessions in oder to open his account , otherwise open again login. It works, if user is not loged in, it will show login page which is right, if user is loged it shows welcome message but not the Welcome the name of the user as I want. for ex: Welcome David <?php session_start(); if(isset($_SESSION['myusername'])) { echo 'Welcome '.$_SESSION['myusername']; } else { include("leftmodules.php"); include("rightmodules.php"); include("login.php"); } ?>

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  • extract associative array within an array in php

    - by I Like PHP
    i have an array like this Array ( [0] => Array ( [s_id] => 4 [si_id] => sec_1 [d_id] => dep_4 [s_name] => sec1 [s_location] => LA [s_visibility] => yes [s_created_date] => 1273639343 [s_last_updated_date] => 1273639343 [s_created_by] => someone [s_last_updated_by] => everyone ) ) now i want to extract array[0] into an array... means i want this Array ( [s_id] => 4 [si_id] => sec_1 [d_id] => dep_4 [s_name] => sec1 [s_location] => LA [s_visibility] => yes [s_created_date] => 1273639343 [s_last_updated_date] => 1273639343 [s_created_by] => someone [s_last_updated_by] => everyone ) how do i get above results?

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  • please suggest good variable names that represent theirself ( php)

    - by I Like PHP
    Hi all, it may be wrong place to ask this question, but i hope u all programmer must have the interesting naming convention of variables. i have seen many places that some variable names are very good n effective like common variable names $link $db $connect $query $stmt $sql $qry $output $result $list so please suggest me some good names for variable , bcoz all time i have to write $x, $y etc.. if i want to save something instantly on page...that are even not relevant, so please suggest me good variable names

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  • (php) regexto remove comments but ignore occurances within strings

    - by David
    Hi there, I am writing a comment-stripper and trying to accommodate for all needs here. I have the below stack of code which removes pretty much all comments, but it actually goes too far. A lot of time was spent trying and testing and researching the regex patterns to match, but I don't claim that they are the best at each. My problem is that I also have situation where I have 'PHP comments' (that aren't really comments' in standard code, or even in PHP strings, that I don't actually want to have removed. Example: <?php $Var = "Blah blah //this must not comment"; // this must comment. ?> What ends up happening is that it strips out religiously, which is fine, but it leaves certain problems: <?php $Var = "Blah blah ?> Also: will also cause problems, as the comment removes the rest of the line, including the ending ? See the problem? So this is what I need... Comment characters within '' or "" need to be ignored PHP Comments on the same line, that use double-slashes, should remove perhaps only the comment itself, or should remove the entire php codeblock. Here's the patterns I use at the moment, feel free to tell me if there's improvement I can make in my existing patterns? :) $CompressedData = $OriginalData; $CompressedData = preg_replace('!/\*.*?\*/!s', '', $CompressedData); // removes /* comments */ $CompressedData = preg_replace('!//.*?\n!', '', $CompressedData); // removes //comments $CompressedData = preg_replace('!#.*?\n!', '', $CompressedData); // removes # comments $CompressedData = preg_replace('/<!--(.*?)-->/', '', $CompressedData); // removes HTML comments Any help that you can give me would be greatly appreciated! :)

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  • How can I secure my $_GETs in PHP?

    - by ggfan
    My profile.php displays all the user's postings,comments,pictures. If the user wants to delete, it sends the posting's id to the remove.php so it's like remove.php?action=removeposting&posting_id=2. If they want to remove a picture, it's remove.php?action=removepicture&picture_id=1. Using the get data, I do a query to the database to display the info they want to delete and if they want to delete it, they click "yes". So the data is deleted via $POST NOT $GET to prevent cross-site request forgery. My question is how do I make sure the GETs are not some javascript code, sql injection that will mess me up. here is my remove.php //how do I make $action safe? //should I use mysqli_real_escape_string? //use strip_tags()? $action=trim($_GET['action']); if (($action != 'removeposting') && ($action != 'removefriend') && ($action != 'removecomment')) { echo "please don't change the action. go back and refresh"; header("Location: index.php"); exit(); } if ($action == 'removeposting') { //get the info and display it in a form. if user clicks "yes", deletes } if ($action =='removepicture') { //remove pic } I know I can't be 100% safe, but what are some common defenses I can use. EDIT Do this to prevent xss $action=trim($_GET['action']); htmlspecialchars(strip_tags($action)); Then when I am 'recalling' the data back via POST, I would use $posting_id = mysqli_real_escape_string($dbc, trim($_POST['posting_id']));

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  • JavaScript two-dimensional Array to PHP

    - by vi
    Hi I have to send a two-dimensional JavaScript Array to a PHP page. Indeed, I'm working on a form-builder, in which the user can add or remove fields. These fields are added (or removed) using JavaScript (jQuery). When the user is done and hit a 'publish' button, I have to get all the fields concerned and send them to a PHP page which would build a real form with it. I found a way to do it but I'm pretty sure it's not very clean : addedFields = new Array(); $("#add-info .field").each(function() { addedFields.push(new Array($(this).find('.name').val(), $(this).find('.type').val(), $(this).find('.size').val())); }); Basically, the ".field" class objects are <tr> and the ".name", ".type" and ".size" objects are inputs. So I get an array of [name, type, size], then I convert it into a string using addedFields = addedFields.join(";"); Finally, I go to the PHP form that way ; document.location.href = "create.php?addedfields=" + addedFields; Concerning the PHP code, I create a PHP array using the explode() function: $addedFields = explode(";", $_GET['addedfields']); and then I use it again for each element in the array: foreach ($addedFields as $field) { $field = explode(",", $field); echo "<li>Field with name : '$field[0]', of '$field[1]' type and with a size of $field[2]"; }

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  • Show image on hover with PHP

    - by Sorosh
    I have a small problem with my PHP code and It would be very nice if someone could help me. I want to display an image when hovering over a link. This is the link with the PHP code that I have now: <a href="<?php the_permalink(); ?>"><?php if ( has_post_thumbnail() ) {the_post_thumbnail();} else if ( has_post_video() ) {the_post_video_image();}?></a> This code shows a image, but I want to execute this code when hovering over the link with the image: <?php echo print_image_function(); ?> The code also shows a image that belongs to a category. I don't want the initial image to disappear I simply want to show the second image on top off the first image when hovering over the first image. I don't know if it is helpful but I use Wordpress and I am not a PHP expert. I even don't know if this is going to work. Thats why I am asking if somebody can help me with this. Thanks in advance

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  • Textbox auto generate by php and html

    - by user2892997
    i have few field of data such as product , amount and barcode.The system just show 1 row of data insert form that contain the 3 field.when i completed insert the first row, then second row of textbox will auto generate, i can do it by microsoft access, can i do so for php ? <?php $i=0; ?> <form method="post" action=""> <input type="text" name="<?php echo $i; ?>" /> </form> <?php if(isset($_POST[$i])){ $i++; ?> <form method="post" action=""> <input type="text" name="<?php echo $i; ?>" /> </form> <?php }?> it work for the first and second textbox, but how can i continue to create more textbox accordingly?

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