Ik krijg de volgende error.
Wat kan ik doen om dit te fixen?
lijn 16:
Code (php)
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// Create an Image from it so we can do the resize
$ext = end(explode('.', $uploaded_Name));
$ext = end(explode('.', $uploaded_Name));
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<?php
include '../config.php';
ini_set('display_errors', 1);
error_reporting(E_ALL | E_STRICT);
$target = "../images/preview/";
$uploaded_Name = $_FILES['previewe']['name'];
$uploaded_Temp = $_FILES['previewe']['tmp_name'];
$previewe = "http://www.xxx.com/images/preview/".$uploaded_Name ;
// This is the temporary file created by PHP
$uploadedfile = $_FILES['previewe']['tmp_name'];
// Create an Image from it so we can do the resize
$ext = end(explode('.', $uploaded_Name));
if($ext == 'jpg'){
$src = imagecreatefromjpeg($uploadedfile);
}
else if($ext == 'png'){
$src = imagecreatefrompng($uploadedfile);
}
else if($ext == 'gif'){
$src = imagecreatefromgif($uploadedfile);
}
else
{
echo "No preview uploaded";
}[/code]
include '../config.php';
ini_set('display_errors', 1);
error_reporting(E_ALL | E_STRICT);
$target = "../images/preview/";
$uploaded_Name = $_FILES['previewe']['name'];
$uploaded_Temp = $_FILES['previewe']['tmp_name'];
$previewe = "http://www.xxx.com/images/preview/".$uploaded_Name ;
// This is the temporary file created by PHP
$uploadedfile = $_FILES['previewe']['tmp_name'];
// Create an Image from it so we can do the resize
$ext = end(explode('.', $uploaded_Name));
if($ext == 'jpg'){
$src = imagecreatefromjpeg($uploadedfile);
}
else if($ext == 'png'){
$src = imagecreatefrompng($uploadedfile);
}
else if($ext == 'gif'){
$src = imagecreatefromgif($uploadedfile);
}
else
{
echo "No preview uploaded";
}[/code]
Jelle schreef op 25.01.2009 15:36:
Strict Standards: Only variables should be passed by reference
Lijkt mij duidelijk. Je hebt er nu een functie instaan.
Gewoon twee regels van maken.
Code (php)
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$ext = end(explode('.', $uploaded_Name));
dit maakt:
Code (php)
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$ext = explode('.', $uploaded_Name);
ik weet niet of het helpt