Where do I put my link in this php? - php

This is a really trivial question, but I can't figure it out:
I have this php scrip
<?php
// If you want to ignore the uploaded files,
// set $demo_mode to true;
$demo_mode = false;
$upload_dir = 'upload/';
$allowed_ext = array('jpg','jpeg','png','gif');
if(strtolower($_SERVER['REQUEST_METHOD']) != 'post'){
exit_status('Error! Wrong HTTP method!');
}
if(array_key_exists('pic',$_FILES) && $_FILES['pic']['error'] == 0 ){
$pic = $_FILES['pic'];
if(!in_array(get_extension($pic['name']),$allowed_ext)){
exit_status('Only '.implode(',',$allowed_ext).' files are allowed!');
}
if($demo_mode){
// File uploads are ignored. We only log them.
$line = implode(' ', array( date('r'), $_SERVER['REMOTE_ADDR'], $pic['size'], $pic['name']));
file_put_contents('log.txt', $line.PHP_EOL, FILE_APPEND);
exit_status('Uploads are ignored in demo mode.');
}
// Move the uploaded file from the temporary
// directory to the uploads folder:
if(move_uploaded_file($pic['tmp_name'], $upload_dir.'Bild.jpg')){
exit_status('File was uploaded successfuly!');
}
}
exit_status('Something went wrong with your upload!');
// Helper functions
function exit_status($str){
echo json_encode(array('status'=>$str));
exit;
}
function get_extension($file_name){
$ext = explode('.', $file_name);
$ext = array_pop($ext);
return strtolower($ext);
}
?>
After finishing the process, I want to redirect to another page, or even better, directly start another script.
I tried inserting
header('Location: start_conversion.php');
after the last if (File was uploaded sucessfully), but that didn't work.
Where is my mistake?

exit_status appears to terminate your script. So you'd need to put your redirect before that.
Perhaps instead you could use:
function exit_redirect($loc){
header("Location: $loc",TRUE,302);
exit;
}

Ok, the solution was an originally added javascript. I had to redirect from there, then everything worked.

Related

Upload fails "move uploaded file"

First off, the upload folder is given 777, and my old upload script works, so the server accepts files. How ever this is a new destination.
I use krajee bootstrap upload to send the files. And I receive a Jason response. The error seems to be around move uploaded file. I bet it's a simple error from my side, but I can't see it.
<?php
if (empty($_FILES['filer42'])) {
echo json_encode(['error'=>'No files found for upload.']);
// or you can throw an exception
return; // terminate
}
// get the files posted
$images = $_FILES['filer42'];
// a flag to see if everything is ok
$success = null;
// file paths to store
$paths= [];
// get file names
$filenames = $images['name'];
// loop and process files
for($i=0; $i < count($filenames); $i++){
$ext = explode('.', basename($filenames[$i]));
$target = "uploads" . DIRECTORY_SEPARATOR . md5(uniqid()) . "." . array_pop($ext);
if(move_uploaded_file($_FILES["filer42"]["tmp_name"][$i], $target)) {
$success = true;
$paths[] = $target;
} else {
$success = false;
break;
}
}
// check and process based on successful status
if ($success === true) {.
$output = [];
$output = ['uploaded' => $paths];
} elseif ($success === false) {
$output = ['error'=>'Error while uploading images. Contact the system administrator'];
// delete any uploaded files
foreach ($paths as $file) {
unlink($file);
}
} else {
$output = ['error'=>'No files were processed.'];
}
// return a json encoded response for plugin to process successfully
echo json_encode($output);
?>
I think field name is the issue. Because you are getting image name with filer42 and upload time, you are using pictures.
Please change
$_FILES["pictures"]["tmp_name"][$i]
to
$_FILES["filer42"]["tmp_name"][$i]
And check now, Hope it will work. Let me know if you still get issue.
The error is not in this script but in the post.
I was using <input id="filer42" name="filer42" type="file">
but it have to be <input id="filer42" name="filer42[]" type="file" multiple>
as the script seems to need an arrey.
It works just fine now.

PHP SSH move file to another directory [duplicate]

I am uploading files to a server using php and while the move_uploaded_file function returns no errors, the file is not in the destination folder. As you can see I am using the exact path from root, and the files being uploaded are lower than the max size.
$target = "/data/array1/users/ultimate/public_html/Uploads/2010/";
//Write the info to the bioHold xml file.
$xml = new DOMDocument();
$xml->load('bioHold.xml');
$xml->formatOutput = true;
$root = $xml->firstChild;
$player = $xml->createElement("player");
$image = $xml->createElement("image");
$image->setAttribute("loc", $target.basename($_FILES['image']['name']));
$player->appendChild($image);
$name = $xml->createElement("name", $_POST['name']);
$player->appendChild($name);
$number = $xml->createElement("number", $_POST['number']);
$player->appendChild($number);
$ghettoYear = $xml->createElement("ghettoYear", $_POST['ghetto']);
$player->appendChild($ghettoYear);
$schoolYear = $xml->createElement("schoolYear", $_POST['school']);
$player->appendChild($schoolYear);
$bio = $xml->createElement("bio", $_POST['bio']);
$player->appendChild($bio);
$root->appendChild($player);
$xml->save("bioHold.xml");
//Save the image to the server.
$target = $target.basename($_FILES['image']['name']);
if(is_uploaded_file($_FILES['image']['tmp_name']))
echo 'It is a file <br />';
if(!(move_uploaded_file($_FILES['image']['tmp_name'], $target))) {
echo $_FILES['image']['error']."<br />";
}
else {
echo $_FILES['image']['error']."<br />";
echo $target;
}
Any help is appreciated.
Eric R.
Most like this is a permissions issue. I'm going to assume you don't have any kind of direct shell access to check this stuff directly, so here's how to do it from within the script:
Check if the $target directory exists:
$target = '/data/etc....';
if (!is_dir($target)) {
die("Directory $target is not a directory");
}
Check if it's writeable:
if (!is_writable($target)) {
die("Directory $target is not writeable");
}
Check if the full target filename exists/is writable - maybe it exists but can't be overwritten:
$target = $target . basename($_FILES['image']['name']);
if (!is_writeable($target)) {
die("File $target isn't writeable");
}
Beyond that:
if(!(move_uploaded_file($_FILES['image']['tmp_name'], $target))) {
echo $_FILES['image']['error']."<br />";
}
Echoing out the error parameter here is of no use, it refers purely to the upload process. If the file was uploaded correctly, but could not be moved, this will still only echo out a 0 (e.g. the UPLOAD_ERR_OK constant). The proper way of checking for errors goes something like this:
if ($_FILES['images']['error'] === UPLOAD_ERR_OK) {
// file was properly uploaded
if (!is_uploaded_File(...)) {
die("Something done goofed - not uploaded file");
}
if (!move_uploaded_file(...)) {
echo "Couldn't move file, possible diagnostic information:"
print_r(error_get_last());
die();
}
} else {
die("Upload failed with error {$_FILES['images']['error']}");
}
You need to make sure that whoever is hosting your pages has the settings configured to allow you to upload and move files. Most will disable these functions as it's a sercurity risk.
Just email them and ask whether they are enabled.
Hope this helps.
your calls to is_uploaded_file and move_uploaded_file vary. for is_uploaded_file you are checking the 'name' and for move_uploaded_file you are passing in 'tmp_name'. try changing your call to move_uploaded_file to use 'name'

How to copy an image from one folder to another using php

I'm having difficulty in copying an image from one folder to another, now i have seen many articles and questions regarding this, none of them makes sense or work, i have also used copy function but its giving me an error. " failed to open stream: No such file or directory" i think the copy function is only for files. The image i wanna copy is present in the root directory. Can anybody help me please. What i am doing wrong here or is there any other way???
<?php
$pic="somepic.jpg";
copy($pic,'test/Uploads');
?>
You should write your code same as below :
<?php
$imagePath = "/var/www/projectName/Images/somepic.jpg";
$newPath = "/test/Uploads/";
$ext = '.jpg';
$newName = $newPath."a".$ext;
$copied = copy($imagePath , $newName);
if ((!$copied))
{
echo "Error : Not Copied";
}
else
{
echo "Copied Successful";
}
?>
You should have file name in destination like:
copy($pic,'test/Uploads/'.$pic);
For your code, it must be like this:
$pic="somepic.jpg";
copy($pic,'test/Uploads/'.$pic);
Or use function, like this:
$pic="somepic.jpg";
copy_files($pic,'test/Uploads');
function copy_files($file_path, $dest_path){
if (strpos($file_path, '/') !== false) {
$pathinfo = pathinfo($file_path);
$dest_path = str_replace($pathinfo['dirname'], $dest_path, $file_path);
}else{
$dest_path = $dest_path.'/'.$file_path;
}
return copy($pic, $dest_path);
}

Copy images from one folder to another

My Web application stored in directory of XAMPP/htdocs/projectname/. And I have images(source) & img(destination) folders in above directory.I am writing following line of code to get the copy images from one folder to another. But I get the following Warnnigs: (Warning: copy(Resource id #3/image1.jpg): failed to open stream: No such file or directory in C:\xampp\htdocs) and images are not copied into destination.
<?php
$src = opendir('../images/');
$dest = opendir('../img/');
while($readFile = readdir($src)){
if($readFile != '.' && $readFile != '..'){
if(!file_exists($readFile)){
if(copy($src.$readFile, $dest.$readFile)){
echo "Copy file";
}else{
echo "Canot Copy file";
}
}
}
}
?>
Just a guess (sorry) but I don't believe you can use $src = opendir(...) and $src.$readFile like that. Try doing this:
$srcPath = '../images/';
$destPath = '../img/';
$srcDir = opendir($srcPath);
while($readFile = readdir($srcDir))
{
if($readFile != '.' && $readFile != '..')
{
/* this check doesn't really make sense to me,
you might want !file_exists($destPath . $readFile) */
if (!file_exists($readFile))
{
if(copy($srcPath . $readFile, $destPath . $readFile))
{
echo "Copy file";
}
else
{
echo "Canot Copy file";
}
}
}
}
closedir($srcDir); // good idea to always close your handles
Replace this line in your code, this will work definitely.
if(copy("../images/".$readFile, "../img/".$readFile))
you are giving wrong path ,if path of your file say script.php is "XAMPP/htdocs/projectname/script.php" and images and img both are in "projectname" folder than you should use following values for $srcPath and $destPath,change their values to
$srcPath = 'images/';
$destPath = 'img/';
public function getImage()
{
$Path='image/'; //complete image directory path
$destPath = '/edit_image/';
// makes new folder, if not exists.
if(!file_exists($destPath) || file_exists($destPath))
{
rmdir($destPath);
mkdir($destPath, 0777);
}
$imageName='abc.jpg';
$Path=$Path.$imageName;
$dest=$destPath.$imageName;
if(copy($Path, $dest));
}

PHP Uploading images and getting the url

I've been working on a premade image upload template for fun and I'm not too good with php so I can't seem to get around this problem.
I need to be able to extract the direct link to my image (the $upload_dir variable + file extension) so I can use it in my index.php page. I've tried to include my post_file.php file in my index.php but I am getting the Wrong HTTP-method message.
My website is in html5 and it's using the drag-and-drop to initiate the uploading if that is something that messes with my upload file.
post_file.php:
$demo_mode = false;
$upload_dir = 'uploads/'.date(d).rand(0000,9999);
$allowed_ext = array('jpg','jpeg','png','gif');
if (strtolower($_SERVER['REQUEST_METHOD']) != 'post') {
exit_status('Error! Wrong HTTP method!');
}
if (array_key_exists('pic',$_FILES) && $_FILES['pic']['error'] == 0 ) {
$pic = $_FILES['pic'];
if (!in_array(get_extension($pic['name']),$allowed_ext)) {
exit_status('Only '.implode(',',$allowed_ext).' files are allowed!');
}
// Uploads are being ignored, but logged.
if ($demo_mode) {
$line = implode(' ', array( date('r'), $_SERVER['REMOTE_ADDR'], $pic['size'], $pic['name']));
file_put_contents('log.txt', $line.PHP_EOL, FILE_APPEND);
exit_status('Uploads are ignored in demo mode.');
}
// Move image,rename it and log it all
if (move_uploaded_file($pic['tmp_name'], $upload_dir.'.'.get_extension($pic['name']))) {
$line = implode(' ', array( date(r), 'http://website.com/'.$upload_dir.'.'.get_extension($pic['name'])));
file_put_contents('log.txt', $line.PHP_EOL, FILE_APPEND);
exit_status('Upload was successful');
}
}
exit_status('Something went wrong with your upload!');
Functions
function exit_status($str) {
echo json_encode(array('status'=>$str));
exit;
}
function get_extension($file_name) {
$ext = explode('.', $file_name);
$ext = array_pop($ext);
return strtolower($ext);
}
When I use the include('post_file.php') is when I receive the error message, how can I get the new image name + path and extension without including the post_file.php in my index.php file and then echo it in a variable?
Thanks in advance!

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