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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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• 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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• 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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• 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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• Loop not working when I try to traverse a Simple XML Object

  - by Ben Shelock

  I'm trying to loop through the results from the Last.fm API but it's not working. function get_url($url){ $ch = curl_init(); curl_setopt($ch,CURLOPT_URL,$url); curl_setopt($ch,CURLOPT_RETURNTRANSFER,1); curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,5); $content = curl_exec($ch); curl_close($ch); return $content; } $xml = get_url('http://ws.audioscrobbler.com/2.0/?method=album.search&album=kid%20a&api_key=b25b959554ed76058ac220b7b2e0a026'); $doc = new SimpleXMLElement($xml); $albums = $doc->results->albummatches; foreach($albums as $album){ echo $album->album->name; } This just shows the first album. If I change the code within the foreach loop to echo $album->name; it shows nothing at all. What am I doing wrong?

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• Break a string into parts, returning all characters

  - by Benjamin

  I want to break a string according to the following rules: all consecutive alpha-numeric chars, plus the dot (.) must be treated as one part all other consecutive chars must be treated as one part consecutive combinations of 1 and 2 must be treated as different parts no whitespace must be returned For example this string: Method(hierarchy.of.properties) = ? Should return this array: Array ( [0] => Method [1] => ( [2] => hierarchy.of.properties [3] => ) [4] => = [5] => ? ) I was unsuccessful with preg_split(), as AFAIK it cannot treat the pattern as an element to be returned. Any idea for a simple way to do this?

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• PHP GET issue - it aint workin!

  - by benhowdle89

  I have a simple PHP form which displays inputs with values from a mysql DB and send the form results to another page which updates a db table based on the GET results: echo "<form method='get' action='updateprojects.php'>"; echo "<table>"; echo "<tr>"; echo " <th>Project No</th> <th>Customer Name</th> <th>Description</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td><input value=" . $row['project_no'] . "></input></td>"; echo "<td><input value='" . $row['cust_name'] . "'></input></td>"; echo "<td><input value='" . $row['description'] . "'></input></td>"; echo "</tr>"; } echo "</table>"; echo "<input type='submit' value='Update' />"; echo "</form>"; on updateprojects.php i have set up the GET variable and even echoed them to check but nothing comes through! Cannot see why!? This is the start of updateprojects.php: echo $_GET['project_no'].$_GET['cust_name'].$_GET['description'];

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• Display number of retweets of a blog post

  - by Clint

  Hi, I was wondering if anyone has a link to any information regarding building in Twitter Mention functionality into a website like here, http://paul.boagworld.com/?page=2, where each post dispays the number of times it has been mentioned on twitter. Many thanks, C

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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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• 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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• Help to comment a simple statement

  - by Woppie the Russian

  How does you call this statement?: $arr = array(); initialize an array or declare an array? In brief, how would you comment such a statement? (P.S. That array would then hold a bunch of values.) EDIT: I doubt because what I basically do is declare a variable of a particular type, don't I?

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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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• Connection to mysql works through php but not through a form submission?

  - by Legend

  I have a download.php file that connects to a remote mysql database. If I run it using php download.php it works fine. But if I create another php file form.php and then submit the form to this download.php, it complains the following: Can't connect to MySQL server on '<IP_ADDRESS>' (13) Does anyone know why this might be happening? I can't see a reason why this works directly but fails to work upon form submission...

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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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• 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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• How to give new life into a five years old, simple but reliable PHP form?

  - by Sam

  Hi all. I have a script in php 5.2. I want to use a simple form. I found something a programmer made for me about 5 years ago. When I use it, PHP outputs an error now unless I set register_long_arrays = On, then it works fine. On the PHP website, however, it says: Warning This feature has been DEPRECATED as of PHP 5.3.0. Relying on this feature is highly discouraged. It's recommended to turn them off, for performance reasons. Instead, use the superglobal arrays, like $_GET. Should I listen to PHP's warning, or just enable the option and keep using my old form happily? If the former, then how/where do I change this simple form, so it does not rely on the deprecated setting? Your answer is much appreciated. form.htm <html><body> <form method="POST" action="form_sent.php"> ... </form> </body></html> form_sent.php <html><body> <?php $email = $HTTP_POST_VARS[email]; $mailto = "[email protected]"; $mailsubj = "A Form was Sent from Website!"; $mailhead = "From: $email\n"; reset ($HTTP_POST_VARS); $mailbody = "Values submitted from web site form:\n"; while (list($key, $val) = each ($HTTP_POST_VARS)){$mailbody .= "$key : $val\n";} if (!eregi("\n",$HTTP_POST_VARS[email])) { mail($mailto, $mailsubj, $mailbody, $mailhead); } ?> <b>Form Sent. Thank you.</b> </body></html>

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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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• 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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• 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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• using a href (html)tag along with PHP

  - by dexter

  i have tried: <?php include("delete.php") ?> <?php .... .... .... if($result=mysql_query($sql)) { echo "<table><th>Id</th><th>Name</th><th>Description</th><th>Unit Price</th>"; while($row = mysql_fetch_array($result)) { echo "<tr><td>".$row['Id']."</td><td>".$row['Name']."</td><td>".$row['Description']."</td><td>".$row['UnitPrice']."</td> <td><a href='delproduct($row[Id])' onclick = 'return MsgOkCancel()'>Delete</a></td></tr>"; echo "<br/>"; } } ?> following javascript is in the same page: <script type="text/javascript" language="javascript"> function MsgOkCancel() { if (confirm("Are You Sure You Want to Delete?")) { return true } else {return false} } </script> where delproduct is a javascript function in delete.php written like: <script type="javascript"> function delproduct(Id) { alert('Id '+ Id); } <script> ** after ** clicking Delete a okcancel message-box appear asking conformation ** but ** after clicking 'ok' it should execute statements inside delproduct function but it doesn't it gives error like: Object Not Found :The requested URL was not found on this server. what would be the problem? pls help, thanks

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