hi all,
i am having a page where i can upload images.so now my problem is i hav to view the image that is uploaded and that should display on the web page.
My code is uploading an image in the images folder. so now i need to display that image below browse button.
thank u..
<html>
<body>
<form name="newad" method="post" enctype="multipart/form-data" action="test.php">
<table>
<tr><td><input type="file" name="image"></td></tr>
<tr><td><input name="Submit" type="submit" value="View image"></td></tr>
</table>
</form>
</body>
</html>
<?php
define ("MAX_SIZE","1000");
function getExtension($str)
{
$i = strrpos($str,".");
if (!$i)
{
return "";
}
$l = strlen($str) - $i;
$ext = substr($str,$i+1,$l);
return $ext;
}
$errors=0;
if(isset($_POST['Submit']))
{
$image=$_FILES['image']['name'];
if ($image)
{
$filename = stripslashes($_FILES['image']['name']);
$extension = getExtension($filename);
$extension = strtolower($extension);
if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif"))
{
echo '<h1>Unknown extension!</h1>';
$errors=1;
}
else
{
$size=filesize($_FILES['image']['tmp_name']);
if ($size > MAX_SIZE*1024)
{
echo '<h1>You have exceeded the size limit!</h1>';
$errors=1;
}
$image_name=time().'.'.$extension;
$newname="images/".$image_name;
$copied = copy($_FILES['image']['tmp_name'], $newname);
if (!$copied)
{
echo '<h1>Copy unsuccessfull!</h1>';
$errors=1;
}
}
}
}
if(isset($_POST['Submit']) && !$errors)
{
echo "<h1>File Uploaded Successfully! Try again!</h1>";
}
?>