I'm trying to push a file to a Amazon s3 filebucket.
I'm posting the file through an html form.
I try to generate a path to the file like this($file is a part of a foreach, because i need to support multiple files in a form-submit.)
$file['tmp_name'].'/'.$file['name'];
this outputs a filepath like this
/Applications/MAMP/tmp/php/phpZDcVQv/pdf.pdf
/Applications/MAMP/tmp/php/ exists, but nothing is inside it. I have set access read and write for everyone to that folder.
I use a library to post the images to Amazon: https://github.com/tpyo/amazon-s3-php-class It also complains that the filepath i have provided doesn't exist. It's running a check like:
if (!file_exists($file) || !is_file($file) || !is_readable($file))
How come the files aren't added?
Am I referencing the wrong folder? The file with the code is in /web/projectname/
Someone on the internet said something unclear about php removing the temp-file directly. Is this after the response has been run? Do I need to address this in some way?
The most simple code that generates the problem:
foreach ($_FILES as $file) {
$filepath = $file['tmp_name'].'/'.$file['name'];
if(file_exists($filepath)){
echo 'true <br />';
}else{
echo 'false <br />';
}
}
This echo:es false even if files have been uploaded.
$filepath contains the path i described above.
as the manual states:
$_FILES['userfile']['tmp_name']
The temporary filename of the file in which the uploaded file was stored on the server.
and
$_FILES['userfile']['name']
The original name of the file on the client machine.
this means that file_exists($file['tmp_name']) should be true.
The path $file['tmp_name'].'/'.$file['name'] is bogus, since $file['name'] is there only for informing you of the original name, but this name is not used while saving the uploaded file on the server.
So in your example /Applications/MAMP/tmp/php/phpZDcVQv is actually the uploaded file.
Related
My purpose is uploading a remote file create from my PC to specific folder, but I don't know whats wrong with my code below. It uploads the file with the name and the .jpg extension, but it is not moving the file to the specified folder.
if(isset($_POST["image"])){
define("SITE_NAME","project_name/"); //constant for project name
define("SITE_PATH",$_SERVER['DOCUMENT_ROOT']."/".SITE_NAME); //constant for project base directory
define("IMAGES_URL",SITE_URL."images/"); //constant for image directory
$upload_base_dir=IMAGES_URL;
$upload_time_dir=date('Y')."/".date('m')."/".date('d')."/"; // setup directory name
$upload_dir = $upload_base_dir.$upload_time_dir;
if (!file_exists($upload_dir)) {
mkdir($upload_dir, 0777, true); //create directory if not exist
}
$input = $_POST["image"];
$file = fopen(time()."image.jpg", 'wb');
fwrite($file, $input);
//$image_name=basename($_FILES['image']['name']);
$image=time().'_'.$image_name;
move_uploaded_file($file,$upload_dir.$image);
fclose($file);
}
Any suggestions? Thank you in advance.
move_uploaded_file($file,$upload_dir.$image) will only work for items within temp, that are accessable via $_FILES superglobal. If you are sending your file as a strieam within post, that wont work.
1) If file is a form upload make sure form is a multipart and access your file via $_FILES superglobal
move_uploaded_file($_FILES['userfile']['tmp_name'], $yourDirectory.$yourFilename);
2) If you post the file as a stream via post (keep in mind this will only work for small files as large ones will exceed request limit). Save the file directly to it's destiantion using fopen or to move it after you created it use rename() - http://php.net/manual/en/function.rename.php
rename($currentFilePath, $newFilePath)
P.S. sending files as post streams is a very very bad idea.
I am having a problem with move_uploaded_file().
I am trying to upload a image path to a database, which is working perfectly and everything is being uploaded and stored into the database correctly.
However, for some reason the move_uploaded_file is not working at all, it does not produce the file in the directory where I want it to, in fact it doesn't produce any file at all.
The file uploaded in the form has a name of leftfileToUpload and this is the current code I am using.
$filetemp = $_FILES['leftfileToUpload']['tmp_name'];
$filename = $_FILES['leftfileToUpload']['name'];
$filetype = $_FILES['leftfileToUpload']['type'];
$filepath = "business-ads/".$filename;
This is the code for moving the uploaded file.
move_uploaded_file($filetemp, $filepath);
Thanks in advance
Try this
$target_dir = "business-ads/";
$filepath = $target_dir . basename($_FILES["leftfileToUpload"]["name"]);
move_uploaded_file($_FILES["leftfileToUpload"]["tmp_name"], $filepath)
Reference - click here
Try using the real path to the directory you wish to upload to.
For instance "/var/www/html/website/business-ads/".$filename
Also make sure the web server has write access to the folder.
You need to check following details :
1) Check your directory "business-ads" exist or not.
2) Check your directory "business-ads" has permission to write files.
You need to give permission to write in that folder.
make sure that your given path is correct in respect to your current file path.
you may use.
if (is_dir("business-ads"))
{
move_uploaded_file($filetemp, $filepath);
} else {
die('directory not found.');
}
How can i verify a file name when uploading e.g. xyz.xlsx for this specific file and no other? Thus the file is uploaded if and only if its name is xyz with extension xlsx "xyz.xlsx".
If I understand what your looking for, just grab the filename and test it.
Example:
if($_FILES['uploadedfile']['name'] == "xyz.xlsx")
http://us2.php.net/manual/en/features.file-upload.post-method.php
When you do an upload with PHP, you get a $_FILES array (kind of like $_POST). You'll find the original filename in $_FILES['userfile']['name']. You can verify that as MyGlass suggests in his answer.
if($_FILES['uploadedfile']['name'] == "xyz.xlsx")
Additionally, codeignitor provides a file upload class that can be used to eliminate some of the boiler plate of file uploads here: http://ellislab.com/codeigniter/user-guide/libraries/file_uploading.html
Use $this->upload->data() to get the array with the filename info (the key is 'file_name').
$data = $this->upload->data();
if($data['file_name'] == "xyz.xlsx")
I have searched far and wide on this one, but haven't really found a solution.
Got a client that wants music on their site (yea yea, I know..). The flash player grabs the single file called song.mp3 and plays it.
Well, I am trying to get functionality as to be able to have the client upload their own new song if they ever want to change it.
So basically, the script needs to allow them to upload the file, THEN overwrite the old file with the new one. Basically, making sure the filename of song.mp3 stays intact.
I am thinking I will need to use PHP to
1) upload the file
2) delete the original song.mp3
3) rename the new file upload to song.mp3
Does that seem right? Or is there a simpler way of doing this? Thanks in advance!
EDIT: I impimented UPLOADIFY and am able to use
'onAllComplete' : function(event,data) {
alert(data.filesUploaded + ' files uploaded successfully!');
}
I am just not sure how to point THAT to a PHP file....
'onAllComplete' : function() {
'aphpfile.php'
}
???? lol
a standard form will suffice for the upload just remember to include the mime in the form. then you can use $_FILES[''] to reference the file.
then you can check for the filename provided and see if it exists in the file system using file_exists() check for the file name OR if you don't need to keep the old file, you can use perform the file move and overwrite the old one with the new from the temporary directory
<?PHP
// this assumes that the upload form calls the form file field "myupload"
$name = $_FILES['myupload']['name'];
$type = $_FILES['myupload']['type'];
$size = $_FILES['myupload']['size'];
$tmp = $_FILES['myupload']['tmp_name'];
$error = $_FILES['myupload']['error'];
$savepath = '/yourserverpath/';
$filelocation = $svaepath.$name.".".$type;
// This won't upload if there was an error or if the file exists, hence the check
if (!file_exists($filelocation) && $error == 0) {
// echo "The file $filename exists";
// This will overwrite even if the file exists
move_uploaded_file($tmp, $filelocation);
}
// OR just leave out the "file_exists()" and check for the error,
// an if statement either way
?>
try this piece of code for upload and replace file
if(file_exists($newfilename)){
unlink($newfilename);
}
move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $newfilename);
If I upload a text file via a form, is it possible to output its contents directly from the $_FILES variable rather than saving it onto the server first? I know this is a security risk, but it will only be run on a local machine.
Doing
file_get_contents($_FILES['uploadedfile']['tmp_name']);
is valid however you should also check to make sure that the file was uploaded through a form and that no errors occurred during upload:
if ($_FILES['uploadedfile']['error'] == UPLOAD_ERR_OK //checks for errors
&& is_uploaded_file($_FILES['uploadedfile']['tmp_name'])) { //checks that file is uploaded
echo file_get_contents($_FILES['uploadedfile']['tmp_name']);
}
A helpful link is http://us2.php.net/manual/en/features.file-upload.php
The file is saved to temp directory the moment it's uploaded, but you can use $_FILES['uploadedfile']['tmp_name'] to read it without having to save in a permanent place.
Unfortunately, no. At least not through the $_FILES variable. Sorry.
EDIT: It is always saved as the temp file in $_FILES and you'll always have to use that one for content.