PHP Passing image from HTML form errors

Question

Planning a basic image up load to start, not done it before. Having a problem with the form not passing the picture to the PHP file. The errors i am recieving are;

Notice: Undefined index: file in C:\Users\PC\Documents\XAMPP\htdocs\php\addpic.php on line 2 Notice: Undefined index: file in C:\Users\PC\Documents\XAMPP\htdocs\php\addpic.php on line 8 Notice: Undefined index: file in C:\Users\PC\Documents\XAMPP\htdocs\php\addpic.php on line 9 Notice: Undefined index: file in C:\Users\PC\Documents\XAMPP\htdocs\php\addpic.php on line 10 Notice: Undefined index: file in C:\Users\PC\Documents\XAMPP\htdocs\php\addpic.php on line 11

HTML form

<div id="addpicdialog" title="Change Pic">
<p>Please Fill in out member information.</p>
<form name="changepic" method="POST" action="php/addpic.php?id=<?php echo $id ?>" >
<p>DISPLAY CURRENT PIC HERE</p>

<label for="file">Filename:</label>
<input type="file" name="file" id="file"><br>
<input name="submit" type="submit" value="Submit" id="submit"  /> 
</form>
</div>

addpic.php

<?php
if ($_FILES["file"]["error"] > 0)
{
echo "Error: " . $_FILES["file"]["error"] . "<br>";
}
else
{
echo "Upload: " . $_FILES["file"]["name"] . "<br>";
echo "Type: " . $_FILES["file"]["type"] . "<br>";
echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
echo "Stored in: " . $_FILES["file"]["tmp_name"];
}
?>

have you actually sent any data to the websit when you receive the error? and you should also add enctype="multipart/form-data" to your form. — Johannes H.
– Johannes H., Commented Feb 13, 2014 at 17:49

Korvo · Accepted Answer · 2014-02-20 13:34:03Z

2

Uploads requires multipart/form-data

Try this:

<form enctype="multipart/form-data" name="changepic" method="POST" action="php/addpic.php?id=<?php echo $id ?>" >

Tip:

Use isset(...): http://php.net/manual/en/function.isset.php

edited Feb 20, 2014 at 13:34

answered Feb 13, 2014 at 17:48

Korvo

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CodeBird · Accepted Answer · 2014-02-13 17:48:48Z

0

Replace this:

 <form name="changepic" method="POST" action="php/addpic.php?id=<?php echo $id ?>" >

By:

 <form name="changepic" method="POST" action="php/addpic.php?id=<?php echo $id ?>" enctype="multipart/form-data" >

answered Feb 13, 2014 at 17:48

CodeBird

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