Hi trying to create a website that will let the user upload or take photo from the camera, and what I'm trying to use is the accept="image/*" capture that i found in the internet.
Here is my html file
<!DOCTYPE html>
<html>
<head>
<title>Take Photo</title>
</head>
<body>
<form action="upload.php" method="post" enctype="multipart/form-data">
<input type="file" name="uploadedfile" accept="image/*" capture>
<input type="submit" value="Upload">
</form>
</body>
</html>
and here is my upload.php file
<?php
$target_path = "filecheck/check/takephoto/img/";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
?>
The problem is that i'm getting this error
Warning: move_uploaded_file(takephoto/img/panda-010.jpg): failed to open stream: No such file or directory in /home1/filecheck/check/takephoto/upload.php on line 6
Warning: move_uploaded_file(): Unable to move '/tmp/phpz3tKo0' to 'filecheck/check/takephoto/img/panda-010.jpg' in /home1/filecheck/check/takephoto/upload.php on line 6
There was an error uploading the file, please try again!
How can I get the image uploaded to img folder.
I'm new at this.
Related
Hi there I'm trying to create an upload form in php. Containing, upload input, text input (name to be given to the file that was uploaded), and submit button. However I don't know much about php, so I don't know how to actually link that what I had typed on the <input type="text"/> becomes the name of the file when uploaded. If someone can help ? Thanks.
Here is my code:
<!DOCTYPE html>
<html>
<head>
<title>Tu peux uploader ici ta video.</title>
</head>
<body>
<form enctype="multipart/form-data" action="upload.php" method="POST">
<p>Upload your file</p>
<input type="file" name="uploaded_file"></input><br />
<input type="text" name="file-name"></input><br /> <!-- [ASK]: How to make this the file name of the uploaded file -->
<input type="submit" value="Upload"></input>
</form>
</body>
</html>
<?PHP
if(!empty($_FILES['uploaded_file']))
{
$path = "uploads/";
$path = $path . basename( $_FILES['uploaded_file']['name']);
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $path)) {
echo "The file ". basename( $_FILES['uploaded_file']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
}
?>
move_uploaded_file ( string $filename , string $destination ) takes filename which is the filename of the uploaded file and destination is the destination of the moved file. Note that destination directory must exist; move_uploaded_file() will not automatically create it. So lets get the name now...
$_FILES['uploaded_file']['tmp_name'] gives you the temporary filename of the file in which the uploaded file was stored on the server eg. C:\Users\USER\AppData\Local\Temp\phpD3C.tmp
while
$_FILES['uploaded_file']['name'] gives you the actual name which you need to extract the extension eg. myfile.jpg
To link that what you had typed on the , first get it via $_POST["file-name"], then concatenate with the extension. Use pathinfo to retrieve the original extension.
Change your code to become...
<!DOCTYPE html>
<html>
<head>
<title>Tu peux uploader ici ta video.</title>
</head>
<body>
<form enctype="multipart/form-data" action="upload.php" method="POST">
<p>Upload your file</p>
<input type="file" name="uploaded_file"></input><br />
<input type="text" name="file-name"></input><br /> <!-- [ASK]: How to make this the file name of the uploaded file -->
<input type="submit" value="Upload"></input>
</form>
</body>
</html>
<?php
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if(!empty($_FILES['uploaded_file']))
{
$file_parts = pathinfo(basename( $_FILES['uploaded_file']['name'])); //access the actual name instead of tmp_name
//just pick the extension of the file_parts and concatenate it to your path
$path = 'images/';
$path = $path . $_POST["file-name"] . "." . $file_parts['extension'] ;
if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $path)) {
echo "The file ". basename($path) ." has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
}
}
?>
This should get you the desired name. So if you upload file dog.php and in the text field you have cow, the resulting name should be cow.php.
Result:
I am hosting a file upload script on an digital ocean server. I have copied line for line tutorials from all over the internet and changed my permissions using CHMOD 777. I cannot for the life of me figure out why the file won't upload.
PHP
<?php
$target_path = "uploads/";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
?>
HTML
<form enctype="multipart/form-data" action="upload_manager.php" method="POST">
<input type="hidden" name="MAX_FILE_SIZE" value="100000" />
Choose a file to upload: <input name="uploadedfile" type="file" /><br />
<input type="submit" value="Upload File" />
</form>
The file I'm trying to upload is only a few hundred KB and is a PNG. I have tried JPEG too.
When I print out the value of
$_FILES['uploadedfile']['tmp_name']
It is empty
The value of
$_FILES['uploadedfile']['name']
is the file name.
I had a look at my php ini and everything related to uploading files is correct.
What else can I try to get this to work?
UPDATE
I have now solved it, the problem was that I did not have the preciding '/' and I did not have the '/' at the end, here is the final syntax of the folder path: $folder = '/home/dl/www/uploads/';
UPDATE
I have updated the code so that is can output some more detailed debugging information, and this is what is output:
Warning: move_uploaded_file(home/dl/www/uploads/test.txt): failed to open stream: No such file or directory in /home/dl/public_html/file-upload/upload.php on line 9
I have the below script which is attempting to upload a selected file to a specified directory on my web server. However, it is not uploading; I have checked that the permissions on the directory is at '777' as per numerous tutorials have suggested, but it is still throwing the 'File is not uploaded' message.
Does anyone have any suggestions as to why this may not be working?
Thanks!
<?php
$folder = "/home/dl/www/uploads";
if (is_uploaded_file($HTTP_POST_FILES['filename']['tmp_name'])) {
if (move_uploaded_file($HTTP_POST_FILES['filename']['tmp_name'], $folder.$HTTP_POST_FILES['filename']['name'])) {
Echo "File uploaded";
} else {
Echo "File not moved to destination folder. Check permissions";
};
} else {
Echo "File is not uploaded.";
};
?>
The HTML form:
<html>
<head>
<title>File Uploading Form</title>
</head>
<body>
<h3>File Upload:</h3>
Select a file to upload: <br />
<form action="upload.php" method="post"
enctype="multipart/form-data">
<input type="file" name="file" size="50" />
<br />
<input type="submit" value="Upload File" />
</form>
</body>
</html>
Try
$folder = "/home/dl/www/uploads/";
instead of
$folder = "/home/dl/www/uploads";
That is, a slash at the end.
i'ma beginner and i'm trying to upload a file to uploads/ folder from this tutorial but as i run my program after submit it's showing error There was an error uploading the file, please try again! can you tell me what's wrong here??
form.php:
<html>
<head>
<title>Form image</title>
</head>
<body>
<form enctype="multipart/form-data" action="uploader.php" method="post" name="changer">
<input type="hidden" name="MAX_FILE_SIZE" value="100000">
<input type="file" name="uploadedfile" accept="image/jpeg">
<input type="submit" value="Submit">
</form>
</body>
</html>
uploader.php
<?php
//where the file is going to be placed..
$target_path="uploads/";
$target_path=$target_path.basename($_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['name'], $target_path)){
echo "The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
The correct code is
move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)
You are using
move_uploaded_file($_FILES['uploadedfile']['name'], $target_path)
This will not work since the file is stored in the temp folder after the upload is initiated and while moving the file it must know the temp name so that the file could be moved from temp folder to the server.
I'm attempting to build my first form/file uploader (I'm a newb fyi).
I am testing on a local server on my mac, both the form, file handler, and the uploads folder are in the same file directory with one another.
When I select a file using the submit form (test file is 'testFilego.txt' and 3 bytes in size'), i get the following error: http://localhost/PhP_exercises/__tizag/280-php-fileupload-test.php?MAX_FILE_SIZE=2500000&uploadedfile=testFilego.txt. The submit form doesn't seem to connect to the handler (?). I thought the test file would appear in my uploads folder. Help.
This is the submit form:
<form>
<form enctype="multipart/form-data" action="280-php-uploader.php" method="POST">
<input type="hidden" name="MAX_FILE_SIZE" value="2500000" />
Choose a file to upload: <input name="uploadedfile" type="file" /><br />
<input type="submit" value="Upload File" />
</form>
</form>
This is the file hander the form ought to be contacting
<?php
280-php-uploader.php<?php
//This is '280-php-uploader.php'
// Where the file is going to be placed
$target_path = "uploads/";
// Add the original filename to our target path.
//Result is "uploads/filename.extension"
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded";
} else{
echo "There was an error uploading the file, please try again!";
}
?>
As Marco Acierno observed, i'd posted the form inside a form and that seems to have caused the problem.