Nu moet je een afbeelding selecteren en eigenlijk wil ik het zo hebben dat hij al een vast plaatje heeft wat in het script vast gesteld is.
Ik had de select file al op text gezet en fupload al vervangen door een xxx.jpeg, maar dan gaat het niet goed.
Wat moet ik veranderen, en waarom ? (zodat ik er wat van kan leren)
Verder vraag ik me af of hij niet automatisch kan submitten, maar ik kon daar niets over vinden op het www.
Gr. Jack
<html>
<form enctype="multipart/form-data"
action="<?print $_SERVER['PHP_SELF']?>" method ="post">
<tr><td><input type = "hidden" name="MAX_FILE_SIZE" value = "102400"></td></tr>
<tr><td>Select File: </td><td><input type = "file" name = "fupload"><t/d></tr>
<tr><td><input type = "submit" value = "Upload!"></td></tr>
</table>
<?php
//checking and uploading file-----------------------------------------------------------
if (isset ($_FILES['fupload'])){
$filename = $_FILES['fupload']['name'];
$randomdigit = rand(0000,9999);//create random digit
$newfilename = $randomdigit.$filename;//make new file name with random digit
//printing file information
print "<table>";
print "<tr><td>Original Name:</td><td> ". $_FILES['fupload']['name']."</td></tr>";
print "<tr><td>New Name:</td><td> ".$newfilename."</td></tr>";
print "<tr><td>Size: </td><td>". $_FILES['fupload']['size']."</td></tr>";
print "<tr><td>Temp Name: </td><td>". $_FILES['fupload']['tmp_name']."</td></tr>";
print "<tr><td>Type: </td><td>". $_FILES['fupload']['type']. "</td></tr>";
print "<tr><td>Error: </td><td>". $_FILES['fupload']['error']. "</td></tr>";
print "</table>";
//checking the type of file, if it is image it will display it
if ($_FILES['fupload']['type'] == "image/jpeg"){
$source = $_FILES['fupload']['tmp_name'];
$target = "upload/".$_SESSION['username']."/".$newfilename;
move_uploaded_file($source, $target); // or die ("Couldnt copy");
//displaying the image
$imagesize = getImageSize($target);
$imgstr = "<p><img width=\"$size[0]\" height=\"$size[1]\" ";
$imgstr .= "src=\"$target\" alt=\"uploaded image\" ></p>";
$imagepath = $yoursite.$target;
print $imgstr;
print "The link to your image is: ".$yoursite.$target;//link to the image
}
}
//-------------------------------------------------------------------------------------
?>
</html>