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PHP - Open Pdf File From Variable
Alright I am so close but can not figure this out, thanks in advance for any help.
So what I am trying to do is take the input variables from a form and have them load a specific PDF .
<html>
<body>
<?php
if( $_GET["startpoint"] || $_GET["endpoint"])
{
$a = $_GET["startpoint"];
$b = $_GET["endpoint"];
echo "Your Startpoint is:". $_GET["startpoint"]. "<br />";
echo "Your Endpoint is: ". $_GET["endpoint"]. "<br />";
echo $a,$b,".pdf";
}
?>
</body>
</html>
This is what I have and it works to display what the two points are what the final pdf file should be.
The final echo generates the PDF name
echo $a,$b,".pdf";How can I get it to load a file using this as a variable or the name it generates, thanks. Similar Tutorialshello can i open file with name in another language rather than english. For example торта.jpg. How can i open that kind of file. I've tryed that <meta content="text/html" charset="utf-8" /> <?php $fh = fopen("торта.jpg", 'r'); fclose($fh); ?> but it's don't work. Help me pls! Hello, I want to do, when someone click on button/link will open this: but not Save As, but Open. I know how to make simple form: <html> <body> <form action="edit.php" method="post" enctype="multipart/form-data"> <label for="file">File:</label> <input type="file" name="file" id="file" /> <br><br> <input type="submit" name="submit" value="Open"/> </form> </body> </html> But here You must click Browse, but I want to do like in photo: if You click on button, then at once will open form like photo. Many thanks, Tadas P.S. Sorry for my really bad English. This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=317065.0 This topic has been moved to HTML Help. http://www.phpfreaks.com/forums/index.php?topic=308625.0 Hello all, i am go9090go. Today i made a domains for a jar file people can upload from my website. I made this to make the jar file close source and its easy to update. Now i made a java classloader and everything i made works. The classloader call a php document with the password and username. The pass and name will be checked inside a databse and if its inside i use header() to load the jar file. But when i just go to my main domain i get the index of the site and people can easly download the jar file without have to walk thru the php pass checker. So i want to place the jar file inside a protected folder,and i want that only way you get acces to this jar is by the php file. How can i get a file from a protected folder? here is the php used when the jar file is not inside a protected folder: <?php $DBName = "name";//name database $DBUser = "name";//user $DBPassword = "pass"; //passs $DBHost = "host"; //might be different mysql_connect($DBHost, $DBUser, $DBPassword); mysql_select_db($DBName); $username = $_GET['username']; $password = $_GET['password']; $IP = $_SERVER['REMOTE_ADDR']; $string = "Java"; $pos = strpos($agent, $string); if (!strpos($_SERVER['HTTP_USER_AGENT'], "Java")) { echo("Your Auth has been banned for trying to breach security."); //mysql_query("delete from users where username='$username'"); exit(); } $query = "select * from users where name='$username' and pass='$password'"; mysql_query($query); $num = mysql_affected_rows(); if ($num > 0) { header('Location:script/Script.jar'); } ?> now i want to use the header to a file inside a folder that is protected : so how can i make the header() methode to open script.jar inside a protected folder. The folder haves name and pass: blabla,balbla for exempel thanks for help Hi guys <?php sleep(00); $handle = fopen("http://www.myurl.com/?bla=bla=bla" , "r"); fclose($handle); ?> How do i prevent this from outputing to a file on my server. I only need the url to be accessed to add a new row of data to my DB. Any Ideas? Regards Mark I have a page with links to images. Currently the images open in a blank browser window, and I think this is too plain. I don't want to create an HTML file for each image. I want to have an 'image display php file' - img_temp.php. Along with the reference to the image file, in the link, I want to also be able to send a <title> to the template file. I think the code below shows kinda' what I'm trying to do, and also shows that I don't quite have a grasp on how to do it. Any tips would be appreciated. Index Page (index.php) Code: [Select] <html> <head> <title>Gallery Links</title> </head> <body> <!-- This is the page with the links to the images --> <ul> <li><a href="img_temp.php?file=images/01.jpg<?php ?title = "This is image 1"?>">Image 01</a></li> </ul> </body> </html> Image Display Template (img_temp.php) Code: [Select] <html> <head> <title><?php echo $title ?></title> </head> <body> <!-- This is the page that displays the sent image and displays the appropriate doc title --> <?php //Output the image $file echo $file; ?> </body> </html> I'm trying to copy an existing file using a php script, run on the command line. I've tried copy() and also a script from the comments on copy() http://www.php.net/manual/en/function.copy.php#102320 No matter how I try to open the file I get " failed to open stream: No such file or directory". I tried changing the name of the file, and the ext, and both. The file does exist, and if I run cp old new, the file gets copied fine. Current code Code: [Select] <?php $source = '/www/documents/myfile.pdf'; $destination = '/www/documents/mypdf.pdf.bak'; copy($source, $destination); ?> Warning: copy(/www/documents/myfile.pdf): failed to open stream: No such file or directory in /www/cron/historic.php on line 3 Code: [Select] ~$ ls /www/hua/compucom113dev/client/documents/ myfile.pdf ~$ cp /www/documents/myfile.pdf /www/documents/myfile.pdf.bak ~$ ls /www/hua/compucom113dev/client/documents/ myfile.pdf myfile.pdf.bak I had chmod in the script at one point to make sure the directory had the permissions needed but it didn't make a difference so I removed it. I can add it back if someone can help explain what to set it to. I was doing 0777 then back to 0755 at the end. This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=320945.0 Hello, this is my first post I hope you guys can help me out here, basically whats going on is I declare a variable "$diree" then I insert it in a function but its not getting recognized, here is the code: $institutea "some directory"; $dire = str_replace(" ","-",$institutea); # SETTINGS $max_width = 110; $max_height = 130; $per_page = 5; $page = $_GET['page']; $has_previous = false; $has_next = false; function getPictures() { global $page, $per_page, $has_previous, $has_next; if ( $handle = opendir($diree.'/') ) { // done changes here $lightbox = rand(); ?><table border="1"><tr><td> <form action="pro.php" method="post"> <?php echo "<ul id='pictures'>"; $count = 1; $skip = $page * $per_page; if ( $skip != 0 ) $has_previous = true; while ( $count < $skip && ($file = readdir($handle)) !== false ) { if ( !is_dir($file) && ($type = getPictureType($file)) != '' ) $count++; } $count = 1; while ( $count < $per_page && ($file = readdir($handle)) !== false ) { if ( !is_dir($file) && ($type = getPictureType($file)) != '' ) { if ( ! is_dir($dire .'/') ) { // done changes here mkdir($dire .'/'); // done changes here } if ( ! file_exists($dire .'/'.$file) ) { makeThumb( $file, $type ); } echo '<li>'; echo '<img src="'.$dire .'/'.$file.'" alt="" / width="110" height="110"><br/><input type="checkbox" name="food[]" value='.$file.'>'; // done changes here echo '</li>'; $count++; } } echo '</ul>'; ?> </td></tr><tr><td><div align="center"><input type="submit" value="Add"></div></td></tr> </tr></table></form> <?php while ( ($file = readdir($handle)) !== false ) { if ( !is_dir($file) && ($type = getPictureType($file)) != '' ) { $has_next = true; break; } } } } function getPictureType($file) { $split = explode($dire .'/', $file); $ext = $split[count($split) - 1]; if ( preg_match('/jpg|jpeg/i', $ext) ) { return 'jpg'; } else if ( preg_match('/png/i', $ext) ) { return 'png'; } else if ( preg_match('/gif/i', $ext) ) { return 'gif'; } else { return ''; } } function makeThumb( $file, $type ) { global $max_width, $max_height; if ( $type == 'jpg' ) { $src = imagecreatefromjpeg($file); } else if ( $type == 'png' ) { $src = imagecreatefrompng($file); } else if ( $type == 'gif' ) { $src = imagecreatefromgif($file); } if ( ($oldW = imagesx($src)) < ($oldH = imagesy($src)) ) { $newW = $oldW * ($max_width / $oldH); $newH = $max_height; } else { $newW = $max_width; $newH = $oldH * ($max_height / $oldW); } $new = imagecreatetruecolor($newW, $newH); imagecopyresampled($new, $src, 0, 0, 0, 0, $newW, $newH, $oldW, $oldH); if ( $type == 'jpg' ) { imagejpeg($new, $dire .'/'.$file); } else if ( $type == 'png' ) { imagepng($new, $dire .'/'.$file); } else if ( $type == 'gif' ) { imagegif($new, $dire .'/'.$file); } imagedestroy($new); imagedestroy($src); } ?> This is probably an easy one, but I can't figure it out and it's pretty much not searchable. on a linux-machine i have installed filezilla the filezilla runs pretty well and all is ok. now i need to have the passwd that i have stored years ago. The passprhase is stored in a plain in a file called sitemanager.xmlfile I want to find that file and open it with a terminal command. find . -name *.sitemanagerwell i thought that this will return the file I'm looking for. Now how do I open it automatically, without typing the name? find . -name *sitemanager.xm | openThis doesn't work. It says it doesn't found the open command. question: why it does not work on opensuse? should i use any other command - eg the following: find . -name *xyz | xargs openor find . -name *sitemanager.xml | xargs openor
find . -name *.xyz -exec open {} \;
and
find . -name *.xyz -exec open {} \; .
any and all help will be greatly appreciateagain: what is wanted and needet is to find out the passphrase in the filezilla-configuration Hello, I am new to the forum and somewhat new to php, nice to meet you all. This is the first time I have ever really scripted with PHP so I'm still learning about all the tools I have and what I have to call things in PHP. I have a list of urls and as I loop through each one, I'd like to be able to get information from the webpage. The <title> would be a good start. I also want to know the best way for me to compare data I have. I'll show the basic code below, but I successfully go through each url in this text file. I put it in a <ul><li> list just fine. So if $url == http://www.youtube.com/file how is the normal way to check and see if the word "youtube" is in $url? I found preg_match() but I think I'm approaching the whole thing wrong because I get no output. I am an intermediate to somewhat advanced scripter in other languages similar to php, I just need to learn how you do the normal things in PHP. So I'd like to compare a string "youtube" to a variable '$url'. And I would like to be able to grab the title or other info from the file $url. Here is what I have so far. (Recent research showed me how I should do this with an XML file so I will probably change the .txt to .xml) Can you please tell me what to look for as I have been searching and can't really find a comprehensive answer. I changed the whole page to an echo trying to fix something last night. Before it was written like.. Code: [Select] <?php if ($true) { $var = value ?> <html code>The value is <?php $var ?> .</html code> <?php } ?> Code: [Select] /index.php <?php include 'include/header.html'; echo "<div id='wrapper'> <div id='left'> <div class='article'> <br /> <p>"; echo "Today is " . date("l") . ", the " . date("jS") . " of " . date("F") . "."; $lines = file('data/news.txt'); if ($lines){ foreach ($lines as $line_num => $line) { $url = htmlspecialchars($line); //Now I have url. I want to check the url and get the <title> & misc. data. //if youtube is in $url {html code to embed youtube}; //my attempt was $x = file($url); but I got a lot of 404 and 403 errors. //now I fill html. echo "<ul id='menu1' class='auroramenu'> <li><a href='#'>Story ".$line_num."</a> <a style='display: none;' class='aurorashow' href='#'></a> <a style='display: inline;' class='aurorahide' href='#'></a> <ul> <br /> <p>".$url."</p><br /> <li style='text-align:right;'><a href='".$url."' target='_blank'>Read the story.</a> </li> </ul> </li> </ul>"; } } echo "</p><br /> </div> </div> <div id='right'>"; include 'include/sidebar.html'; echo "</div><br class='clr' /></div><br />"; include 'include/footer.html'; ?> Thank you for your help. Need Suggestion for my codes, I want msg1.php to be open in the frame bottomframe. if($username=="" || $password=="") { echo "<form method=\"post\" action=\"msg1.php\" target=\"bottomframe\" >"; } Not sure if this relates more to PHP or Javascript/jQuery, but is there a way to attach a file to an email client (i.e Outlook), when user click on a link/button? I have a link that generates a PDF and I want it so that, when someone click on a link, it will open up their email client with the PDF already attached, and subject already fill in. I know, you can use <a href="mailto:..." but that only opens up the mail client. I also need to attach a file with a subject fill in. Is this doable with PHP or Javascript? I have a PHP file that is executed via batch file very frequently for live updating. This is a sort of "sync" file for reading/writing to a MySQL database. I am able to get it functioning absolutely fine when executed manually via a web browser (and my echo debug lines output the expected information), yet when I run the same PHP file via the command line, it seems to just not be capable of opening any file for reading, so the equivalent variables that are correct when executed on a web browser are blank when echoed through the command line. Is there a different way of handling reading of files when running a php script via the command line, or should it function exactly the same as when run via a browser? For instance:
$k = "0";
$line = file("examplefile.log")[$k];
echo $line;
Hey all, I viewed this tutorial and the guy used $_SERVER['DOCUMENT_ROOT'] to reference a view layout for his specific index file. So I tried to imitate and did exactly what he did, but for me I got the following error: Code: [Select] Warning: include(/Users/jmerlino/Sites/mark/public/diet/views/layouts/shop.php) [function.include]: failed to open stream: No such file or directory in /Users/jmerlino/Sites/diet/index.php on line 22 Warning: include() [function.include]: Failed opening '/Users/jmerlino/Sites/mark/public/diet/views/layouts/shop.php' for inclusion (include_path='.:/usr/lib/php') in /Users/jmerlino/Sites/diet/index.php on line 22 This is the php code that is causing this error: include($_SERVER['DOCUMENT_ROOT'].'/'.'diet/views/layouts/'.$controller.'.php'); I'm not sure why it's going to: /Users/jmerlino/Sites/mark/public/diet/views/layouts/shop.php instead of: /Users/jmerlino/Sites/diet/views/layouts/shop.php Thanks for any response Hi there, Can anybody help me to write php/javascript code which will allow users to open files directly from web browser into desktop application? Here is the specification: I have a photo editing business with many people working in Photoshop. I am currently developing a web based application (joblist) using Javascript and PHP which should allow the photoshop designers to browse and open files/images directly from joblist/web browser into photoshop. The reason I want this instead of browsing folder is that I have a database where I store who worked on which file, when and how long it took. The concept is that, designers will select a file and click on start, as soon as they click on start the original file will open in Photoshop and there will be an entry into database (using PHP). Once they finish the task they will close the file and click on Finish button. My joblist application will be published in a local server and the file will be open on a local network, so when they save the file it will be saved where the source file is located in (local server). The application should work in both PC and Mac. I have already done all other part of the application except file opening directly from browser to desktop application functionality. Anybody can help me to write the code (PHP or Javascript) which can open the file from browser (local server) directly into desktop application e.g. PHotoshop or Illustrator? Thank you very much I look forward to someone's real help! Best regards Mr. Sumon My images generator comprises
an array of image names extracted form an images table from a database using a select statement
a random number generator,
and a string that builds the correct pathname for the selected file.
To select one of the images for display, i generate a random number between 1 and the length of the array.
Though the generator is working, I noticed one of the random numbers is throwing up this error:
Warning: getimagesize(images/): failed to open stream: No such file or directory in
An inspection of the array reveals 2 array elements (representing my number of images) but one array element is NULL ( the first entry in the banner table
image_generator.php
require_once('connection.inc.php');
$sql = 'SELECT `filename` FROM banner';
$result = $mysqli->query($sql, MYSQLI_STORE_RESULT) or die(mysqli_error());
$row = $result->fetch_array(MYSQLI_ASSOC);//an array of image names
$count = $result->num_rows;
for ($i = 1; $i <= $count; ++$i)
{
$row[$i] = $result->fetch_array(MYSQLI_ASSOC);
}
$i = rand(1, $count); //a random number generator,
//The random number is used in the final line to build the correct pathname for the selected file.
$selectedImage = "images/{$row[$i]['filename']}";
if (file_exists($selectedImage) && is_readable($selectedImage))
{
$imageSize = getimagesize($selectedImage);
}
var_dump($row)
array (size=1)
'filename' => string 'ginsomin2.jpg' (length=13)
null
RANDON IMAGE DISPLAY
require_once 'image_generator.php';
<div id="banner" class="wrapper clearfix">
<img src="<?php echo $selectedImage; ?>" alt="banner">
</div>
Kindly advice how i may proceed from here?
Thanks.
I'm new to php and i have a html form with 2 forms and 2 submit buttons, 1 of the forms is solely for uploading an image i downloaded the php script as i don't understand the code well enough to write it myself. The second form asks for your name and email then when you click submit it writes a new html file with the included information. I have managed to upload a file successfully and write the html document with the information successfully but i can't get the image to write into the html document. My file_upload.php script contains the variable for the image name, now how would i get the variable string from file_upload.php and use that name in compile.php. Here's my code. -------------- Form.html -------------- <!DOCTYPE html> <html lang="en"> <head> <title></title> </head> <body> <form action="compile.php" method="post"> <p>input name:<input type="text" name="name"/></p><br /> <p>email:<input type="text" name="email" /></p><br /> </form> <br /> <form enctype="multipart/form-data" method="post" action="file_upload.php"> Choose your file he <input name="file1" type="file" /><br /><br /> <input type="submit" value="Upload It" /> </form> </body> </html> ------------- compile.php ------------- <?php $name = $_POST['name']; $price = $_POST['price']; $desc = $_POST['desc']; $file = "new.html"; $handle = fopen($file,'w'); $data = "<!DOCTYPE html> <html> <head> <title></title> </head> <body> <table style='border-radius:8px;'> <tr> <td style='border-radius:8px;'> <h1>$name</h1> </td> </tr> </table> <table> <tr> <td> <ul> <lh>Product name: $name</lh> <br><br> <li>Price: $price</li><br> <li>Description: $desc</li><br> </ul> </td> <td> <img src='uploads/$fileName' width='' height='' alt='' title='' /> </td> </tr> </table> </body> </html>"; fwrite($handle, $data); print "data written"; fclose($handle); ?> ------------ file_upload.php ------------ <?php // Set local PHP vars from the POST vars sent from our form using the array // of data that the $_FILES global variable contains for this uploaded file $fileName = $_FILES["file1"]["name"]; // The file name $fileTmpLoc = $_FILES["file1"]["tmp_name"]; // File in the PHP tmp folder $fileType = $_FILES["file1"]["type"]; // The type of file it is $fileSize = $_FILES["file1"]["size"]; // File size in bytes $fileErrorMsg = $_FILES["file1"]["error"]; // 0 for false... and 1 for true // Specific Error Handling if you need to run error checking if (!$fileTmpLoc) { // if file not chosen echo "ERROR: Please browse for a file before clicking the upload button."; exit(); } else if($fileSize > 5000000000) { // if file is larger than we want to allow echo "ERROR: Your file was larger than 5mb in file size."; unlink($fileTmpLoc); exit(); } else if (!preg_match("/.(gif|jpg|png)$/i", $fileName) ) { // This condition is only if you wish to allow uploading of specific file types echo "ERROR: Your image was not .gif, .jpg, or .png."; unlink($fileTmpLoc); exit(); } // Place it into your "uploads" folder mow using the move_uploaded_file() function move_uploaded_file($fileTmpLoc, "uploads/$fileName"); // Check to make sure the uploaded file is in place where you want it if (!file_exists("uploads/$fileName")) { echo "ERROR: File not uploaded<br /><br />"; echo "Check folder permissions on the target uploads folder is 0755 or looser.<br /><br />"; echo "Check that your php.ini settings are set to allow over 2 MB files, they are 2MB by default."; exit(); } // Display things to the page so you can see what is happening for testing purposes echo "The file named <strong>$fileName</strong> uploaded successfuly.<br /><br />"; echo "It is <strong>$fileSize</strong> bytes in size.<br /><br />"; echo "It is a <strong>$fileType</strong> type of file.<br /><br />"; echo "The Error Message output for this upload is: <br />$fileErrorMsg"; ?> |
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