PHP simple file upload form doesn't work

Question

I'm exploring the file upload and text parsing capabilities of PHP, but my first step is buggy, and I can't figure it out. The goal of the code is to display a form to upload a text, and then display the value of the corresponding $_FILES array. However, it doesn't work - it runs without error but doesn't display print_r($_FILES['upload']). What am I missing?

This is my Index_txtParsing file:

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
    <head>
        <title>test</title>
    </head>
    <body>
        <form enctype="multipart/form-data" class="Txt_upload" method='POST'  action='index_txtParsing.php'>
            <label for="file">Enter upload here:</label>
            <input type='file' name='upload'/>
            <input type='submit' name='submit' value="upload here"/>
        </form>
    </body>
</html>

<?php
    if(isset($_POST['upload'])){
    $upload=$_POST['upload'];
    print_r($_FILES['upload']);
    }else{
    $upload="unknown";
    }
?>

EDIT: After incorporating the recommendations below, this code works:

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
    <head>
        <title>test</title>
    </head>
    <body>
        <form enctype="multipart/form-data" class="Txt_upload" method='POST'  action='index_txtParsing.php'>
            <label for="file">Enter upload here:</label>
            <input type='file' name='upload'/>
            <input type='submit' name='submit' value="upload here"/>
        </form>
    </body>
</html>

<?php
    if(isset($_FILES['upload'])){
    echo "Value of FILES['upload'] ";
    print_r($_FILES['upload']);
    echo "<br/>";
    }else{
    }
?>

The PHP manual has a section on file uploads and a good example — Pekka
– Pekka, Commented Feb 11, 2012 at 19:14

Madara's Ghost · Accepted Answer · 2012-02-11 20:08:02Z

3

To get the uploaded file use $_FILES instead of $_POST:

<?php
    if(isset($_FILES['upload'])){
    $upload=$_FILES['upload'];
    print_r($_FILES['upload']);
    }else{
    $upload="unknown";
    }
?>

Also please read the manual on File Uploads.

edited Feb 11, 2012 at 20:08

Madara's Ghost

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answered Feb 11, 2012 at 19:19

Markel Mairs

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1 Comment

Maxim Strutinskiy Over a year ago

Don't forget set 777 permission for your upload/ directory. For ubuntu $sudo chmod -R 777 uploads/

Giannis · Accepted Answer · 2012-02-11 19:24:17Z

2

Your if statement should look like that:

if ($_FILES['upload']['name']!="")

answered Feb 11, 2012 at 19:24

Giannis

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Comments

topscoder · Accepted Answer · 2012-02-11 19:19:38Z

1

Try to clear the value of action="" and check for the submit element in _POST instead upload in _POST so you get:

<form enctype="multipart/form-data" method="POST" action="">
        <input type="file" name="upload" />
        <input type="submit" name="submit" value="upload here"/>
</form>

<?php if ( isset( $_POST['submit'] ) ) { print_r( $_FILES ); } ?>

This should work.

answered Feb 11, 2012 at 19:19

topscoder

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Comments

Kypros · Accepted Answer · 2012-02-11 19:41:06Z

1

First of all you are checking for $_POST['upload'] when you should be checking for $_POST['submit'] if you want to see if the form has been submitted, or fot $_FILES[upload']['name'] if you are checking for the filename and
when you are trying to show $_POST['upload'] that is not available as it's a file so try something like this if you want the name of the file:
```
if(isset($_POST['submit'])){
    print_r($_FILES['upload']);
    echo $_FILES['upload']['name'];
}else{
    $upload="unknown";
}
```

Update:

As i told you in my first point if you need to check if a file has been chosen for upload, try changing your if statement as such so you first check whether the form has been submitted, and afterwards if a file has been chosen:

if(isset($_POST['submit'])){
    if($_FILES['upload']['name']!="")
            $upload = $_FILES['upload']['name'];
    else
        $upload = "no file";
}else{
    $upload="form not submitted";
}

echo $upload;

This of course is just an example and not a complete solution. In real case scenarios you'd probably want to check file sizes, file type etc...

edited Feb 11, 2012 at 19:41

answered Feb 11, 2012 at 19:22

Kypros

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2 Comments

JDelage Over a year ago

That works but it only tests whether the button submit has been clicked, not whether a file has indeed been uploaded.

Kypros Over a year ago

@JDelage Check my update as i had already mentioned that in my first point

Ben D · Accepted Answer · 2012-02-11 20:03:46Z

Don't look for the file in the POST array, look for it in the FILES array:

if(isset($_FILES['upload']['tmp_name']) && $_FILES['upload']['size']>0){ //make sure the file has been uploaded correctly
    $upload=$_FILES['upload'];
    print_r($_FILES['upload']);
}else{
    $upload="unknown";
}

$_FILES['name of field'] is an array that can be accessed as follows (from PHP.net):

$_FILES['userfile']['name']

The original name of the file on the client machine.

$_FILES['userfile']['type']

The mime type of the file, if the browser provided this information. An example would be "image/gif". This mime type is however not checked on the PHP side and therefore don't take its value for granted.

$_FILES['userfile']['size']

The size, in bytes, of the uploaded file.

$_FILES['userfile']['tmp_name']

The temporary filename of the file in which the uploaded file was stored on the server.

$_FILES['userfile']['error']

Eugen Rieck · Accepted Answer · 2012-02-11 19:18:05Z

0

you need

<input type="hidden" name="MAX_FILE_SIZE" value="same_or_lower_than_in_your_php.ini">

in your form

answered Feb 11, 2012 at 19:18

Eugen Rieck

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1 Comment

JDelage Over a year ago

I don't think it's indispensable, but in any case I tried with it and still doesn't work.

PandemoniumSyndicate · Accepted Answer · 2012-02-11 19:24:23Z

I believe this is the direction you're looking to go:

// Check if you uploaded a file previously
if(isset($_POST['upload']))
{
    // Load the new file
    if (!empty($HTTP_POST_FILES['ufile']['tmp_name']))
    {
            // Extension? 
        $temp_name = basename($_FILES['ufile']['name']);
        $extension = explode(".", $temp_name);
        $extension = end($extension);
        $filename .= $extension;

        $path = $uploaddir . $filename;

        if ($ufile != none)
        {
            if (copy($HTTP_POST_FILES['ufile']['tmp_name'], $path))
            {
           // It copied to where ever you want it to go!    
            } 
            else
            //error 
        }
        else
        //error
    }
    else
    //error

}

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