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PHP - Help Making The Uploaded Image-path Be Displayed On The Page
I need help making the uploaded image file name, that's chosen to be uploaded, be displayed on the html page with the path /upload/ added to the beginning of the displayed file name like so:
../upload/test.png Any help/improvements will be appreciated.
<html>
<head>
<title>PHP Test</title>
</head>
<body>
<?php
if ($form_submitted == 'yes') {
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = strtolower( end($temp) );
if ( $_FILES["file"]["size"] < 200000
&& in_array($extension, $allowedExts) )
{
if ($_FILES["file"]["error"]!= 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
} else {
$length = 20;
move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $newfilename );
$file_location = '<a href="http://../upload/' . $newfilename . '">' . $newfilename . '</a>';
}
} else {
echo "Invalid upload file";
}
?>
<label for="file">Filename:</label>
<input type="file" name="file" id="file">
</body>
</html>
Similar TutorialsHi: I'm a newby regarding uploading files to MySQL, turning report output into an HREF, and getting MySQL data via a hyperlink. I have successfully uploaded, files to MySQL, I have also been able to display filename information as a hyperlink in report output, but when I click on the hyperlink, I get the following message format on a 404 page: The requested URL /current_dir_of_requesting_page/filename.filetype was not found on this server. My reporting page has teh following line of code in a report table to create the hyperlink: <td style="width:225px"><? echo "<a href=".$row['name'].">".$row['name']."</a>";?></td> Can anyone assist me with this? Hi! So I'm working for someone, and they want me to fix this error in a PHP file.. Here is the code: <?php
include_once('config.php');
$online = mysql_query("SELECT * FROM bots WHERE status LIKE 'Online'");
$offline = mysql_query("SELECT * FROM bots WHERE status LIKE 'Offline'");
$dead = mysql_query("SELECT * FROM bots WHERE status LIKE 'Dead'");
$admintrue = mysql_query("SELECT * FROM bots WHERE admin LIKE 'True'");
$adminfalse = mysql_query("SELECT * FROM bots WHERE admin LIKE 'False'");
$windows8 = mysql_query("SELECT * FROM bots WHERE so LIKE '%8%'");
$windows7 = mysql_query("SELECT * FROM bots WHERE so LIKE '%7%'");
$windowsvista = mysql_query("SELECT * FROM bots WHERE so LIKE '%vista%'");
$windowsxp = mysql_query("SELECT * FROM bots WHERE so LIKE '%xp%'");
$unknown = mysql_query("SELECT * FROM bots WHERE so LIKE 'Unknown'");
$totalbots = mysql_num_rows(mysql_query("SELECT * FROM bots"));
$onlinecount = 0;
$offlinecount = 0;
$deadcount = 0;
$admintruecount = 0;
$adminfalsecount = 0;
$windows8count = 0;
$windows7count = 0;
$windowsvistacount = 0;
$windowsxpcount = 0;
$unknowncount = 0;
while($row = mysql_fetch_array($online)){
$onlinecount++;
}
while($row = mysql_fetch_array($offline)){
$offlinecount++;
}
while($row = mysql_fetch_array($dead)){
$deadcount++;
}
while($row = mysql_fetch_array($admintrue)){
$admintruecount++;
}
while($row = mysql_fetch_array($adminfalse)){
$adminfalsecount++;
}
while($row = mysql_fetch_array($windows8)){
$windows8count++;
}
while($row = mysql_fetch_array($windows7)){
$windows7count++;
}
while($row = mysql_fetch_array($windowsvista)){
$windowsvistacount++;
}
while($row = mysql_fetch_array($windowsxp)){
$windowsxpcount++;
}
while($row = mysql_fetch_array($unknown)){
$unknowncount++;
}
$statustotal = $onlinecount + $offlinecount + $deadcount;
$admintotal = $admintruecount + $adminfalsecount;
$sototal = $windows7count + $windowsvistacount + $windowsxpcount + $unknowncount;
?>
Can anyone tell me the error here, can how to fix it? This topic has been moved to mod_rewrite. http://www.phpfreaks.com/forums/index.php?topic=351137.0 Hi. I have a script here that will let users upload an image to my website but I just can't figure out how to save the uploaded image as "upload/logo.png" so that it will replace the already existing "upload/logo.png". Help would be greatly appreciated. Code: [Select] <html> <body> <form action="" method="post" enctype="multipart/form-data"> <label for="file">Filename:</label> <input type="file" name="file" id="file" /> <br /> <input type="submit" name="submit" value="Submit" /> </form> </body> </html> <?php if(isset($_POST['submit']) && !empty($_FILES["file"]["name"])) { $timestamp = time(); $target = "upload/"; $target = $target . basename($_FILES['uploaded']['name']) ; $ok=1; $allowed_types = array("image/gif","image/jpeg","image/pjpeg","image/png","image/bmp"); $allowed_extensions = array("gif","png","jpg","bmp"); if ($_FILES['file']['size'] > 350000) { $max_size = round(350000 / 1024); echo "Your file is too large. Maximum $max_size Kb is allowed. <br>"; $ok=0; } if ($_FILES["file"]["error"] > 0) { echo "Error: " . $_FILES["file"]["error"] . "<br />"; $ok=0; } else { $path_parts = pathinfo(strtolower($_FILES["file"]["name"])); if(in_array($_FILES["file"]["type"],$allowed_types) && in_array($path_parts["extension"],$allowed_extensions)){ $filename = $timestamp."-".$_FILES["file"]["name"]; echo "Name: " . $filename . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; $path_parts = pathinfo($_FILES["file"]["name"]); echo "Extension: " . $path_parts["extension"] . "<br />"; echo "Size: " . round($_FILES["file"]["size"] / 1024) . " Kb<br />"; //echo "Stored in: " . $_FILES["file"]["tmp_name"]. " <br />"; } else { echo "Type " . $_FILES["file"]["type"] . " with extension " . $path_parts["extension"] . " not allowed <br />"; $ok=0; } } if($ok == 1){ @move_uploaded_file($_FILES["file"]["tmp_name"], $target . $filename); $file_location = $target . $filename; if(file_exists($file_location)){ echo "Uploaded to <a href='$file_location'>$filename</a> <br />"; } else { echo "There was a problem saving the file. <br />"; } } } else { echo "Select your file to upload."; } ?> Thanks! I am trying to run a php code on my site but it kept on saying page cannot be found. Any idea why its saying that??? Not sure if this is the right place to post this. I have PHP form that I use to upload a document, PDF or Word Doc, I would also like the form to create a thumbnail of the document when it is uploaded, is this possible? Hi, I have a script that uploads an image to a directory and then saves the fill path to a field in a table for later use. The only problem is people are uploading huge images and then when I produce a catalogue of my listings it takes forever to load because the images are so big. What I am after is an add in script to create an ADDITIONAL image 100 x 75px, I don't really want to change my upload script. Any ideas? Thanks in advace. Here is what I have: Code: [Select] <?php $idir = "../fleet/"; // Path To Images Directory if (isset ($_FILES['fupload'])){ //upload the image to tmp directory $url = $_FILES['fupload']['name']; // Set $url To Equal The Filename For Later Use if ($_FILES['fupload']['type'] == "image/jpg" || $_FILES['fupload']['type'] == "image/jpeg" || $_FILES['fupload']['type'] == "image/pjpeg") { $file_ext = strrchr($_FILES['fupload']['name'], '$account.'); // Get The File Extention In The Format Of , For Instance, .jpg, .gif or .php $copy = copy($_FILES['fupload']['tmp_name'], "$idir" . $_FILES['fupload']['name']); // Move Image From Temporary Location To Perm } } $fleetimage1 = mysql_real_escape_string("$idir" . $_FILES['fupload']['name']); //then insert sql code below... ?> Hi everyone, I have a script below, which uplads an image, however, the image name always starts with a capital letter, I want all letters to be small, how to adjust this please, thank you
$target_dir = "";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$target_file = "$get_current_user.jpg";
$uploadOk = 1;
$imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
// Check if image file is a actual image
if(isset($_POST["submit"])) {
$check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
if($check !== false) {
echo ""; /// was File is an image
$uploadOk = 1;
} else {
echo "File is not an image<br>";
$uploadOk = 0;
}
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 10000000) {
echo "Sorry, this image is too large, please resize using Paint<br>";
$uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg" && $imageFileType != "gif" ) {
echo "Only JPG, JPEG, PNG & GIF files are allowed<br>";
$uploadOk = 0;
}
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
echo "There was an error<br>";
} else {
if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {
echo "Uploaded successfully (You may need to clear Cache to see the new image)";
mysql_query("UPDATE user SET user_image = 'https://.../images/users/$get_current_user.jpg' WHERE user_name = '$get_current_user' ");
} else {
echo ""; /// was Select a suitable image, file not uploaded yet
}
}
Hello I have uploaded images from a server onto a website, they run through a mysql database (database holds the image file names). The problem I am having is resizing the images. Can anyone help? Thank you in advance GWG I'm beginning with simple codes to understand how things work and work my way up with what I need. My upload form looks like this: <form name="upload about" action="display.php" enctype="multipart/form-data" method="post"> Header: <input type="text" name="header" /> Body: <input type="text" name="body" /> <input type="hidden" name="MAX_FILE_SIZE" value="30000" /> Image: <input type="file" name="pic" /> <input type="submit" /> </form> And my display page is just as simple. <?php echo $_POST["header"]; ?> <?php echo $_POST["body"]; ?> <?php echo $_FILES['pic']['name']; ?> The problem is, whenever I submit the data, everything is fine, except the image does not display and just posts the filename of the image I uploaded. According to my research setting the enctype to multipart/formdata should display the image, but it does not. Can anyone tell me what's wrong? Hello. I'm using an Amazon S3 class to uploaded to S3. I have 2 upload boxes - The first uploads once and the second needs to upload twice - 1 full size, 1 thumb. The issue i'm having is that the 2nd image (the thumb) seems to be failing, although if I don't save the first full sized image I am able to upload the thumb. So I think the issue is with using the temp file twice? This is my code: //retreive post variables $fileName = $randomString . "_" . $_FILES['theFile']['name']; $fileTempName = $_FILES['theFile']['tmp_name']; $fileName2 = $randomString . "_" . $_FILES['theFile2']['name']; $fileTempName2 = $_FILES['theFile2']['tmp_name']; //move the file if ($s3->putObjectFile($fileTempName, "containerhere", $fileName, S3::ACL_PUBLIC_READ)) { echo "<strong>Uploaded Image</strong>"; }else{ echo "<strong>Something went wrong while uploading your file... sorry.</strong>"; } //move the file if ($s3->putObjectFile($fileTempName2, "containerhere", $fileName2, S3::ACL_PUBLIC_READ)) { echo "<strong>Uploaded Image</strong>"; }else{ echo "<strong>Something went wrong while uploading your file... sorry.</strong>"; } include('simpleImage.php'); $image = new SimpleImage(); $image->load($_FILES['theFile2']['tmp_name']); $image->resizeToWidth(100); $image->save($_FILES['theFile2']['tmp_name']); $fileName3 = $randomString . "_" . $_FILES['theFile2']['name']; $fileTempName3 = $_FILES['theFile2']['tmp_name']; //move the file if ($s3->putObjectFile($fileTempName3, "containerhere", "thumbs/" . $fileName3, S3::ACL_PUBLIC_READ)) { echo "<strong>Uploaded Image</strong>"; }else{ echo "<strong>Something went wrong while uploading your file... sorry.</strong>"; } Can anyone offer any advice? Thanks. I have a very simple page, where I use the include statement that includes my connection string. The connection string and a variables are not being passed to my test server - (WAMP). My include file is below
$server = "servername";
$username = "username";
$password = "";
$database = "dbname";
$portNumber = 3308;
$link = mysqli_connect($server,$username,$password,$database,$portNumber);
if(!$link)
{
echo "cannot connet to the server";
}
else
{
echo "This works";
}
echo "Hi";
I am including this file with the following code:
include ("includes/connect.inc");
My site is not connecting to the database, and is not displaying "This works", or "Hi". Do you have any pointers on how to get this connect?
Hi all, I want to get the current page name as displayed in the URL, not neccesarily the name of the PHP script that is running. E.g. I have a PHP file 'home.php' that contains iframes that contain other PHP files such as 'frame1.php'. What I want is a script that can be run from frame1.php but that will return 'home.php' as the current page (as displayed in the URL in the browser). How do I do this? Thanks! Hi Basically I've built a CMS where by my clients can upload a number of images. On the success page I want to display the images they uploaded by file name. The issue is the number of images can vary. They may upload 2 or 10 or 50 etc. So far I've come up with this: Code: [Select] // number of files $UN = 3; //I've set this to 3 for now, but this is passed from the upload page! // server directories and directory names $dir = '../properties'; $images = glob($dir.'/*.{jpg}', GLOB_BRACE); //formats to look for $num_of_files = $UN; //number of images to display from number of uploaded files foreach($images as $image) { $num_of_files--; $newest_mtime = 0; $image = 'BROKEN'; if ($handle = @opendir($dir)) { while (false !== ($file = readdir($handle))) { if (($file != '.') && ($file != '..')) { $mtime = filemtime("$dir/$file"); if ($mtime > $newest_mtime) { $newest_mtime = $mtime; $image = "$file"; } } } } if($num_of_files > -1) //this made me laugh when I wrote it echo $trimmed = ltrim($image, "../properties").'<br />'; //display images else break; } Without this piece of code: Code: [Select] $newest_mtime = 0; $image = 'BROKEN'; if ($handle = @opendir($dir)) { while (false !== ($file = readdir($handle))) { if (($file != '.') && ($file != '..')) { $mtime = filemtime("$dir/$file"); if ($mtime > $newest_mtime) { $newest_mtime = $mtime; $image = "$file"; } } } } It shows the first 3 files alphabetically. I want to view the last number of images added. With the above code it simply shows the last image added 3 times! So I need to get the time each image was added and then order by the newest added and limit to the number of images uploaded. Any suggestions please? Kindest regards Glynn OK, When the user fills the info in the form out it goes in the DB fine. I can then array them on the "showroom page" fine. When they upload a picture it goes into the /images/ folder fine. Problem is... On each array on the showroom page I need the image they uploaded to be displayed. Cant work it out. Help would be GREAT!!!!! hi to all.Im currently displaying the images from my upload directory but the image does not display.please help thanks Hey guys! I have the following php code that grabs variables (and the browsed image) from Flash. //FLASH VARIABLES $Name = $_POST['Name']; $itemNumber = $_POST['itemNumber']; $filename = $_FILES['Filedata']['name']; $filetmpname = $_FILES['Filedata']['tmp_name']; $fileType = $_FILES["Filedata"]["type"]; $fileSizeMB = ($_FILES["Filedata"]["size"] / 1024 / 1000); list($filename, $extension) = explode('.', basename($_FILES['Filedata']['name'])); $filename = $Name; $target = $filename . $itemNumber . "." . $extension; // Place file on server, into the images folder move_uploaded_file($_FILES['Filedata']['tmp_name'], "images/".$target); This works perfect, but what I want to change is the width and height of the uploaded image. Any ideas/suggestions on how this could be done? Thanks in advance!! Cheers! Hey Everyone, I have these 3 scripts to upload an image but I'm having an issue because the images uploaded are going to the same directory as the pages. What do I need to change to make the uploaded images go to a folder path called "pictures". Thanks in advance for the help. Script 1 <form name="form1" method="post" action="adminpicturebrowse.php"> <p align="center">How many pictures for this dog? Max is 9</p> <p align="center"> <input name="uploadNeed" type="text" id="uploadNeed" maxlength="1"> <input type="submit" name="Submit" value="Submit"> </p> </form> Script 2 <form name="form1" enctype="multipart/form-data" method="post" action="adminaddupload.php"> <p align="center"> <? // start of dynamic form $uploadNeed = $_POST['uploadNeed']; for($x=0;$x<$uploadNeed;$x++){ ?> <input name="uploadFile<? echo $x;?>" type="file" id="uploadFile<? echo $x;?>"> </p> <div align="center"> <? // end of for loop } ?> </div> <p align="center"><input name="uploadNeed" type="hidden" value="<? echo $uploadNeed;?>"> <input type="submit" name="Submit" value="Submit"> </p> </form> Script 3 <? $uploadNeed = $_POST['uploadNeed']; // start for loop for($x=0;$x<$uploadNeed;$x++){ $file_name = $_FILES['uploadFile'. $x]['name']; // strip file_name of slashes $file_name = stripslashes($file_name); $file_name = str_replace("'","",$file_name); $copy = copy($_FILES['uploadFile'. $x]['tmp_name'],$file_name); // check if successfully copied if($copy){ echo "$file_name<br>"; }else{ echo "$file_name<br>"; } } // end of loop ?> I have the folder structure like:
root
application
system
assets
uploads
folder assets contains all css, img, and js.
uploads contains user uploaded file.
I set a "helper/assets_helper.php" file to define:
define ('ASSETS_PATH', base_url().'assets/');
define ('UPLOAD_URL', base_url().'uploads/');
For all the css, img, and js, it works well like
href="<?php echo ASSETS_PATH; ?>css/mycss.css"But when I display the uploaded images, it couldn't display image with <a href="<?php echo UPLOAD_URL;?>images/myupload01.jpg" ><img src="<?php echo UPLOAD_URL;?>images/myupload01.jpg" /></a>This uploaded image actually works fine with my localhost with the link like: http://localhost:900.../myupload01.jpg. But it couldn't display on my hosting server with like: http://users.mywebsi.../myupload01.jpg Can anyone shed some light on it. Thanks! Edited by TFT2012, 20 October 2014 - 10:43 AM. Hi, How can i show image using absolute path instead of virtual path?? Help please |
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