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  • PHP IIS problems downloading file says it is corrupt

    - by Matt
    Hi, I am running PHP on IIS 6 with mssql. I have uploaded a file to my webserver through a php script. Upon checking the file on the server the file is ok and not corrupt. However, when i then have a link on my website to try and download the file, it says the file is corrupt. I know the file isnt corrupt as i can view it perfectly if i look at the file on the server. Is seems like this is a common problem as a similar problem was posted here: http://www.bigresource.com/Tracker/Track-php-1pAakBhT/ Any help would be much appreciated. Thanks, M My download code is as follows: $filesize = $rows->filesize; $filepath = $rows->filepath; header("Content-Disposition: attachment; filename=$filename"); header("Content-length: $filesize"); header("Content-type: application/pdf"); header("Cache-control: must-revalidate"); header("Content-Description: PHP Generated Data"); readfile($filepath);

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  • Creating an HTML file by combining multiple PHP files via the command line?

    - by FishOrDie
    Is it possible to combine multiple PHP files via the command line and create an HTML file? For example, this will save the rendered version of a single PHP file as HTML: php /path/to/my/file/filename.php > /path/to/my/file/test.html I need it to combine multiple files, but I can't seem to get it it to work. Ideally, it would be something like this: php /path/to/my/file/filename.php + /path/to/my/file/filename2.php + /path/to/my/file/filename3.php > /path/to/my/file/test.html Is this possible? If so, how?

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  • Problem with checkboxes, sql select statements & php

    - by smokey20
    I am trying to display some rows from a database table based on choices submitted by the user. Here is my form code <form action="choice.php" method="POST" > <input type="checkbox" name="variable[]" value="Apple">Apple <input type="checkbox" name="variable[]" value="Banana">Banana <input type="checkbox" name="variable[]" value="Orange">Orange <input type="checkbox" name="variable[]" value="Melon">Melon <input type="checkbox" name="variable[]" value="Blackberry">Blackberry From what I understand I am placing the values of these into an array called variable. Two of my columns are called receipe name and ingredients(each field under ingredients can store a number of fruits). What I would like to do is, if a number of checkboxes are selected then the receipe name/s is displayed. Here is my php code. <?php // Make a MySQL Connection mysql_connect("localhost", "*****", "*****") or die(mysql_error()); mysql_select_db("****") or die(mysql_error()); $variable=$_POST['variable']; foreach ($variable as $variablename) { echo "$variablename is checked"; } $query = "SELECT receipename FROM fruit WHERE $variable like ingredients"; $row = mysql_fetch_assoc($result); foreach ($_POST['variabble'] as $ingredients) echo $row[$ingredients] . '<br/>'; ?> I am very new to php and just wish to display the data, I do not need to perform any actions on it. I have tried many select statements but I cannot get any results to display. My db connection is fine and it does print out what variables are checked. Many thanks in advance.

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  • Image generated with PHP dosn't show

    - by lolalola
    Hi, whats wrong with my code. image dosn't show in test2.php file File: test2.php: <img src = "test.php" /> File: test.php session_start(); $md5_hash = md5(rand(0,999)); $security_code = substr($md5_hash, 15, 5); $_SESSION["security_code"] = $security_code; $width = 100; $height = 20; header("Content-type: image/png"); $image = ImageCreate($width, $height); $white = ImageColorAllocate($image, 255, 255, 255); $black = ImageColorAllocate($image, 0, 0, 0); $grey = ImageColorAllocate($image, 204, 204, 204); ImageFill($image, 0, 0, $black); //Add randomly generated string in white to the image ImageString($image, 3, 30, 3, $security_code, $white); //Throw in some lines to make it a little bit harder for any bots to break imageRectangle($image,0,0,$width-1,$height-1,$grey); imageline($image, 0, $height/2, $width, $height/2, $grey); imageline($image, $width/2, 0, $width/2, $height, $grey); imagepng($image); imagedestroy($image);

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  • php, image dosn't show

    - by lolalola
    Hi, whats wrong with my code. image dosn't show in test2.php file File: test2.php: <img src = "test.php" /> File: test.php session_start(); $md5_hash = md5(rand(0,999)); $security_code = substr($md5_hash, 15, 5); $_SESSION["security_code"] = $security_code; $width = 100; $height = 20; header("Content-type: image/png"); $image = ImageCreate($width, $height); $white = ImageColorAllocate($image, 255, 255, 255); $black = ImageColorAllocate($image, 0, 0, 0); $grey = ImageColorAllocate($image, 204, 204, 204); ImageFill($image, 0, 0, $black); //Add randomly generated string in white to the image ImageString($image, 3, 30, 3, $security_code, $white); //Throw in some lines to make it a little bit harder for any bots to break imageRectangle($image,0,0,$width-1,$height-1,$grey); imageline($image, 0, $height/2, $width, $height/2, $grey); imageline($image, $width/2, 0, $width/2, $height, $grey); imagepng($image); imagedestroy($image);

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  • Any PHP MVC framework planning to use 5.3 features?

    - by alexandrul
    I would like to get started with PHP, and 5.3 release seems to bring many nice features (namespaces, lambda functions, and many others). I have found some MVC frameworks, and some of them support only PHP 5: PHP Frameworks PHP MVC Frameworks Model–view–controller on Wikipedia but can anyone recommend one of those MVC frameworks that plans to actively use PHP 5.3 features, not just being compatible with PHP 5.3? Update Results so far: Zend Framework 2.0 (in development) Lithium (in development, based on CakePHP) Symfony (in development) FLOW3 (in development, alpha)

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  • submittingthe form even if the form is in invalid state

    - by Abu Hamzah
    i am using asp.net web forms and i am using bassistance.de jquery validation.. i have bunch of fields in the form and its validatating how it suppose to but its still executing my code behind code, why is it doing that? here is my code: form: <script type="text/javascript"> $(document).ready(function() { $("#aspnetForm").validate({ rules: { <%=txtVisitName.UniqueID %>: { maxlength:1, //minlength: 12, required: true } ................. }, messages: { <%=txtVisitName.UniqueID %>: { required: "Enter visit name", minlength: jQuery.format("Enter at least {0} characters.") } ............ }, success: function(label) { label.html("&nbsp;").addClass("checked"); } }); $("#aspnetForm").validate(); }); </script> <div > <h1> Page</h1> <asp:Label runat="server" ID='Label1'>Visit Name:</asp:Label> <asp:TextBox ID="txtVisitName" runat='server'></asp:TextBox> ............ ............... <button id="btnSubmit" name="btnSubmit" type="submit"> Submit</button> </div> and i have a external .js file referencing to my web form page. $(document).ready(function() { $("#btnSubmit").click(function() { SavePage(); }); }); i have a WebMethod .cs and i have bookmark and it does hitting that line of code even thoug my page is in invalid state. how would i fix this? thanks.

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  • PHP - How can I check if return() was called from an include()'d file?

    - by John Himmelman
    How can I tell if return() was called from within the included file. The problem is that include() returns 'int 1', even if return() wasn't called. Here is an example... included_file_1.php <?php return 1; included_file_2.php <?php echo 'no return here, meep'; main.php <?php $ret = include('included_file_1.php'); // This file DID return a value, int 1, but include() returns this value even if return() wasn't called in the included file. if ($ret === 1) { echo 'file did not return anything'; } var_dump($ret); $ret = include('included_file_2.php'); // The included file DID NOT return a value, but include() returns 'int 1' if ($ret === 1) { echo 'file did not return anything'; } var_dump($ret);

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  • Adding with PHP to a MySQL database

    - by shinjuo
    I am pretty new to PHP and I am trying to make an inventory database. I have been trying to make it so that a user can enter a card ID and then amount the want to add to the inventory and have it update the inventory. For example someone could type in test and 2342 and it would update test. Here is what I have been trying with no success: add.html <body> <form action="add.php" method="post"> Card ID: <input type="text" name="CardID" /> Amount to Add: <input type="text" name="Add" /> <input type="submit" /> </form> </body> </html> add.php <?php $link = mysql_connect('tulsadir.ipowermysql.com', 'cbouwkamp', '!starman1'); if (!$link){ die('Could not connect: ' . mysql_error()); } mysql_select_db("tdm_inventory", $link); $add = $_POST[Add] mysql_query("UPDATE cardLists SET AmountLeft = '$add' WHERE cardID = 'Test'"); echo "test successful"; mysql_close($link); ?>

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  • I'm trying to make a php Curl call to formstack api, but I get nothing

    - by Oskar Calvo
    This is one of my first curls code, so it can have mistakes I'm trying to be able to make calls to form/:id/submissions https://www.formstack.com/developers/api/resources/submission#form/:id/submission_GET If I load: https://www.formstack.com/api/v2/form/1311091/submission.json?oauth_token=44eedc50d015b95164897f5e408670f0&min_time=2012-09-01%2000:01:01&max_time=2012-10-27%2000:01:01 If works great. If I try this code: <?php $host = 'https://www.formstack.com/api/v2/'; // TODO this should manage dinamics values or build an action in every method. $action = 'form/1311091/submission.json'; $url = $host . $action; // TODO this values will arrive like an array with values $postData['oauth_token']= '44eedc50d015b95164897f5e408670f0'; $postData['min_time'] ='2012-09-01 00:01:01'; $postData['max_time'] ='2012-10-27 00:01:01'; // TODO make a method with this action function getElements($postData) { $elements = array(); foreach ($postData as $name=>$value) { $elements[] = "{$name}=".urlencode($value); } } $curl = curl_init(); curl_setopt($curl, CURLOPT_URL, $url); curl_setopt($curl, CURLOPT_HTTPGET, true); curl_setopt($curl, CURLOPT_POSTFIELDS, $elements); $result = curl_exec($curl) ; curl_close($curl); var_dump($result); ?>

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  • Unzip .zip file uploaded from Android via PHP to IIS 7

    - by HaOx
    I'm sending compressed file in .zip extension from Android via PHP to IIS server. Almost is working everything, but I cannot achieve unzip files with php. I've this code: <?php $target_path1 = "C:/Windows/Temp/"; $target_path1 = $target_path1 . basename( $_FILES['uploaded_file']['name']); /* I'm making the folder */ $directorio = substr($target_path1, 0, 32); if (!is_dir($directorio)) { mkdir($directorio); } /* I declare a path */ $barra = "/"; $target_path1 = $directorio . $barra . basename( $_FILES['uploaded_file']['name']); $target_path2 = $directorio . $barra; if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $target_path1)) { /* Here I want to unzip the uploaded file */ } else{ echo "There was an error uploading the file, please try again!"; echo "filename: " . basename( $_FILES['uploaded_file']['name']); echo "target_path: " .$target_path1; } ?> So, how can I unzip uploaded file? I tried so many methods but no one worked. I'd be appreciated if someone could tell what I have to do to unzip the uploaded file. And if I have to configure some parameters in php.ini or IIS server. Thanks in advance.

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  • how to close a windows form in c# when that form transfer control to an other form?

    - by moon
    i have encountered a problem in my application i have two windows forms one that is loaded when my application is started and ask for a password and other is shown when click log in with correct password , problem is how can i close the first form that is log in form when user proceed to next from that is actual application, currently i hide the first form but the requirement is to close that form to prevent extra processing when i close that log in form my application is closed and when i close the actual application form that log in form remain open. my application is not fully closed, log in form is running in background how can in fix this....

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  • php script randomly hangs up

    - by sergdev
    I install php 5 (more precisely 5.3.1) as apache module. After this one of my application becomes randomly hang up on mysql_connect - sometimes works, sometimes no, sometimes reload of page helps. How can this be fixed? I use Windows Vista, Apache/2.2.14 (Win32) PHP/5.3.1 with php module, MySql 5.0.67-community-nt. After a minute I obtain the error message: Fatal error: Maximum execution time of 60 seconds exceeded in path\to\mysqlcon.php on line 9 I run MySql locally and heavy load could not be a reason: SHOW PROCESSLIST shows about 3 process SHOW VARIABLES LIKE 'MAX_CONNECTIONS' is 100. UPDATE: At first I thought that this is connected with mysql_connect. But now I can't say for certain. More difficult thing is when I insert the line to debug: $fh = fopen("E://_debugLog", 'a'); fwrite($fh, __FILE__ . " : " . __LINE__ . "\n"); fclose($fh); script start working near that location as a rule.

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  • PHP class to C# class?

    - by LordSauron
    I work for a company that makes application's in C#. recently we got a customer asking us to look in to rebuilding an application written in PHP. This application receives GPS data from car mounted boxes and processes that into workable information. The manufacturer for the GPS device has a PHP class that parses the received information and extracts coordinates. We were looking in to rewriting the PHP class to a C# class so we can use it and adapt it. And here it comes, on the manufacturers website there is a singel line of text that got my skin krawling: "The encoding format and contents of the transmitted data are subject to constant changes. This is caused by implementations of additional features by new module firmware versions which makes it virtually impossible to document it and for you to properly decode it yourself." So i am now looking for a option to use the "constantly changing" PHP class and access it in C#. Some thing link a shell only exposing some function's i need. Except i have no idea how i can do this. Can any one help me find a solution for this.

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  • Post valuse and upload Image to php server in android

    - by lawat
    I am trying to upload image from android phone to php server with additional values,the method is post my php file look like this if($_POST['val1']){ if($_POST['val2']){ if($_FILE['image']){ ...... } } }else{ echo "Value not found"; } I am doing is URL url=new URL("http://www/......../myfile.php"); HttpURLConnection con=(HttpURLConnection) url.openConnection(); con.setDoInput(true); con.setDoOutput(true); con.setUseCaches(false); con.setRequestMethod("POST");//Enable http POST con.setRequestProperty("Connection", "Keep-Alive"); con.setRequestProperty("Content-Type", "multipart/form-data;boundary="+"****"); connection.setRequestProperty("uploaded_file", imagefilePath); DataOutputStream ostream = new DataOutputStream( con.getOutputStream()); String res=("Content-Disposition: form-data; name=\"val1\""+val1+"****"+ "Content-Disposition: form-data; name=\"val2\""+val2+"****" "Content-Disposition: form-data; name=\"image\";filename=\"" + imagefilePath +"\""+"****"); outputStream.writeBytes(res); my actual problem is values are not posting so first if condition get false and else section is executed that is it give value not found please help me

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  • how to pass session_id() throught out the php pages?

    - by Piyush
    when user clicks on login button(index.php) I am calling chechlogin.php where I am checking loginId an password as- if($count==1) { // Register $myusername, $mypassword and redirect to file "login_success.php" session_register("myusername"); session_register("mypassword"); $_SESSION['UserId'] = $myusername; $_session['SessionId'] = session_id(); header("location:LoggedUser.php"); } in LiggedUser.php <?php session_start(); //starting session if (!isset($_SESSION['SessionId']) || $_SESSION['SessionId'] == '') { header("location:index.php"); } ? Problem: It is always going back to index.php page although I am entering right userid and password.I think session_id() is not working properly or ??

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  • JQuery-AJAX: No further request after timeout and delay in form post

    - by Nogga
    I got a form containing multiple checkboxes. This form shall be sent to the server to receive appropriate results from a server side script. This is already working. What I would achieve now: 1) Implementing a timeout: This is already working, but as soon as a timeout occurs, a new request is not working anymore. 2) Implementing a delay in requesting results: A delay shall be implemented so that not every checkbox is resulting in a POST request. This is what I have right now: function update_listing() { // remove postings from table $('.tbl tbody').children('tr').remove(); // get the results through AJAX var request = $.ajax({ type: "POST", url: "http://localhost/hr/index.php/listing/ajax_csv", data: $("#listing_form").serialize(), timeout: 5000, success: function(data) { $(".tbl tbody").append(data); }, error: function(objAJAXRequest, strError) { $(".tbl tbody").append("<tr><td>failed " + strError + "</td></tr>"); } }); return true; } Results are for now passed as HTML table rows - I will transform them to CSV/JSON in the next step. Thanks so much for your advice.

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  • php lampp permissions of fopen function

    - by marmoushismail
    hi i'm programming php using: netbeans 6.8 lampp for ubuntu (xampp) apache which came with xampp $fh = fopen("testfile2.txt", 'w') or die("Failed to create file"); $text ="hello man cool good"; fwrite($fh, $text) or die("Could not write to file"); fclose($fh); echo "File 'testfile.txt' written successfully"; //i get the next error: Warning: fopen(testfile2.txt) [function.fopen]: failed to open stream: Permission denied in /home/marmoush/allprojects/phpprojects/myindex.php on line 91 Failed to create file anyway i know what this error is; it's about folder and files permissions; i looked into the folder permission tab made access available for "others" group ( to read and write) the program worked result was a file (test.txt) so i looked at the created file permission it appears to be that (php , xampp or whoever) creates file with (nobody permission) I have 2 QUESTIONS: 1- what if i need the file created by (php code and xampp ) to have the "root or user or myname" permissions ?? where to set this setting 2-also my concern (what if i send this files to actual web server will it make nobody permissions also nobody ? when they create files

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  • Collect all fields in form to an array

    - by Industrial
    Hi everyone, Each div with the class "row" is added upon request from the user, to be able to add multiple items at once. So now is the question how I'll collect all the forms in to an array that PHP can read (like JSON for instance). I'll guess that there's already some easy and effective way of doing this? <div class="container"> <div class="row"> <input type="text" name="value1" id="textfield" /> <input type="text" name="value2" id="textfield" /> <input type="text" name="value3" id="textfield" /> </div> </div> Here's what I would like to achieve out of the shown example: array( array ('value1' => '', 'value2' => '', 'value3' => '') ); Thanks! Update: The form will be handled with PHP and it would be super to be able to do something like a foreach loop on the specific container-div content.

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  • PHP Foreach statement issue. Multiple rows are returned

    - by Daniel Patilea
    I'm a PHP beginner and lately i've been having a problem with my source code. Here it is: <html> <head> <title> Bot </title> <link type="text/css" rel="stylesheet" href="main.css" /> </head> <body> <form action="bot.php "method="post"> <lable>You:<input type="text" name="intrebare"></lable> <input type="submit" name="introdu" value="Send"> </form> </body> </html> <?php //error_reporting(E_ALL & ~E_NOTICE); mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("robo") or die(mysql_error()); $intrebare=$_POST['intrebare']; $query = "SELECT * FROM dialog where intrebare like '%$intrebare%'"; $result = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_array($result) or die(mysql_error()); ?> <div id="history"> <?php foreach($row as $rows){ echo "<b>The robot says: </b><br />"; echo $row['raspuns']; } ?> </div> It returns me the result x6 times. This problem appeared when I've made that foreach because I wanted the results to stuck on the page one by one after every sql querry. Can you please tell me what seems to be the problem? Thanks!

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  • PHP: parse $_FILES[] data in multidimesional array

    - by superUntitled
    I having been looking around for an answer to this and have not found an answer anywhere, I am hoping someone has done this before! I have a form that allows for dynamic duplication of the form fields. The form allows for file uploads and text input, so the data is sent in both $_POST and $_FILES arrays. The the initial set of inputs look like this: <input type="text" name="primary[1][text]" /> <input type="file" name="primary[1][file]" /> <input type="text" class="a" name="secondary[1][text][]" /> <input type="file" name="secondary[1][file][]" /> When duplicated the fields are incremented, they look like this: <input type="text" name="primary[2][text]" /> <input type="file" name="primary[2][file]" /> <input type="text" class="a" name="secondary[2][text][]" /> <input type="file" name="secondary[2][file][]" /> To complicate matters, the "secondary" form fields can also be duplicated (thus the [] at the end of the secondary name array. How can I parse the posted $_FILES array? I have tried something like this: foreach ($_FILES['question'] as $f_num) { echo $f['file']['name']; } but I get an "Undefined index: file... " error.

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  • Write to PDF using PHP from android device

    - by Brent Mitchell
    I am trying to write to a pdf file on php server. I have sent variables to the server, create the pdf document, then have my phone download the document to view on device. The variables seem not to write on the php file. I have my code below public void postData() { try { Calculate calc = new Calculate(); HttpClient mClient = new DefaultHttpClient(); StringBuilder sb=new StringBuilder("myurl.com/pdf.php"); HttpPost mpost = new HttpPost(sb.toString()); String value = "1234"; List nameValuepairs = new ArrayList(1); nameValuepairs.add(new BasicNameValuePair("id",value)); mpost.setEntity(new UrlEncodedFormEntity(nameValuepairs)); } catch (UnsupportedEncodingException e) { Log.w(" error ", e.toString()); } catch (Exception e) { Log.w(" error ", e.toString()); } } And my php code to write the variable "value" onto the pdf document: //code to reverse the string if($_POST[] != null) { $reversed = strrev($_POST["value"]); $this->SetFont('Arial','u',50); $this->Text(52,68,$reversed); } I am just trying to write the variable in a random spot, but the variable the if statement is always null and I do not know why. Thanks. Sorry if it is a little sloppy.

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  • Upload to a PHP Server, using Ajax ( XMLHttp POST)

    - by Krishnanunni
    Right now i'm using the below method to Upload a file to PHP <form enctype="multipart/form-data" action="http://sserver.com/fileupload.php" method="POST"> <input type="hidden" name="MAX_FILE_SIZE" value="30000000" /> <input type="hidden" name="filename" value="file_uploaded.gif" /> <input type="hidden" name="username" value="foobar"/> Please choose a file: <input name="uploaded" type="file" /><br /> <input type="submit" value="Upload" /> </form> I read the $_POST and $_FILE in php to complete upload like this. $target = $_SERVER['DOCUMENT_ROOT']."/test/upload/"; $target = $target . basename( $_FILES['uploaded']['name']) ; echo $target; $ok=1; if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) { echo "The file ". basename( $_FILES['uploaded']['name']). " has been uploaded"; } else { echo "Sorry, there was a problem uploading your file."; } My questions is , can i change the above said code (HTML) to an Ajax XMLHttpRequest without changes in PHP.

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  • PHP classes, parse syntax errors when using 'var' to declare variables

    - by jon
    I am a C# guy trying to translate some of my OOP understanding over to php. I'm trying to make my first class object, and are hitting a few hitches. Here is the beginning of the class: <?php require("Database/UserDB.php"); class User { private var $uid; private var $username; private var $password; private var $realname; private var $email; private var $address; private var $phone; private var $projectArray; public function _construct($username) { $userArray = UserDB::GetUserArray($username); $uid = $userArray['uid']; $username = $userArray['username']; $realname = $userArray['realname']; $email = $userArray['email']; $phone = $userArray['phone']; $i = 1; $projectArray = UserDB::GetUserProjects($this->GetID()); while($projectArray[$i] != null) { $projectArray[$i] = new Project($projectArray[$i]); } UserDB.php is where I have all my static functions interacting with the Database for this User Class. I am getting errors using when I use var, and I'm getting confused. I know I don't HAVE to use var, or declare the variables at all, but I feel it is a better practice to do so. the error is "unexpected T_VAR, expecting T_VARIABLE" When I simply remove var from the declarations it works. Why is this?

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  • Increasing understanding of validating a string with PHP string functions

    - by user1554264
    I've just started attempts to validate data in PHP and I'm trying to understand this concept better. I was expecting the string passed as an argument to the $data parameter for the test_input() function to be formatted by the following PHP functions. trim() to remove white space from the end of the string stripslashes() to return a string with backslashes stripped off htmlspecialchars() to convert special characters to HTML entities The issue is that the string that I am echoing at the end of the function is not being formatted in the way I desire at all. In fact it looks exactly the same when I run this code on my server - no white space removed, the backslash is not stripped and no special characters converted to HTML entities. My question is have I gone about this in the wrong approach? Should I be creating the variable called $santised_input on 3 separate lines with each of the functions trim(), stripslashes() and htmlspecialchars()? By my understanding surely I am overwriting the value of the $santised_input variable each time I recreate it on a new line of code. Therefore the trim() and stripslashes() string functions will never be executed. What I am trying to achieve is using the "$santised_input" variable to run all of these PHP string functions when the $data argument is passed to my test_input() function. In other words can these string functions be chained together so that I only need to create $santised_input once? <?php function test_input($data) { $santised_input = trim($data); $santised_input = stripslashes($data); $santised_input = htmlspecialchars($data); echo $santised_input; } test_input("%22%3E%3Cscript%3Ealert('hacked')%3C/script%3E\ "); //Does not output desired result "&quot;&gt;&lt;script&gt;alert('hacked')&lt;/script&gt;" ?>

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