PHP Scripting
Upload File
PHP script for upload image and file
By Geekboots
6/9/2020
0 views
upload-file.phpPHP
<html>
<head>
<title>Upload File</title>
</head>
<body>
/* Encoded form data for uploading a file */
<form method="post" enctype="multipart/form-data" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
<input type="file" name="up_file" id="up_file" />
<input type="submit" value="upload" />
</form>
<?php
if(isset( $_FILES['up_file'])){
echo "Name: ". $_FILES['up_file']['name'] ."<br />";
echo "Size: ". $_FILES['up_file']['size'] ." bytes<br />";
echo "Temp name: ".$_FILES['up_file']['tmp_name'] ."<br />";
echo "Type: ". $_FILES['up_file']['type'] ."<br />";
echo "Error: ". $_FILES['up_file']['error'] ."<br />";
/* Upload file if it's a images */
if($_FILES['up_file']['type'] == "image/jpeg" ) {
$source = $_FILES['up_file']['tmp_name'];
$target = "upload/".$_FILES['up_file']['name'];
/* Moves an uploaded file to a new location */
if(move_uploaded_file($source, $target)) {
$size = getImageSize( $target );
echo '<img width="'.$size[0].'" height="'.$size[1].'" src="'.$target.'" alt="uploaded image" /><br />';
echo "Image Uploaded Successfully!";
}
else
echo "Fail to upload!";
}
else
echo "It's not a images!";
}
?>
</body>
</html>
/* Output */
Input:
Choose a image file
Name: order05.jpg
Size: 197978 bytes
Temp name: /tmp/phphWY5Zn
Type: image/jpeg
Error: 0
Image Uploaded Successfully!PHPupload file in phpupload file PHP script
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