sargardoon
پنج شنبه 19 اسفند 1389, 07:25 صبح
نمونه کد
<html>
<head>
<title>Upload Files</title>
</head>
<body>
<form action="index.php" method="post" enctype="multipart/form-data">
File: <input type="file" name="image" /><input type="submit" value="Upload" />
</form>
<?php
// Connect to database
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("phptut") or die(mysql_error());
// File properties
$file = $_FILES['image']['tmp_name'];
if(!isset($file)){
echo "<br><b>Please select image to upload</b>";
} else {
$image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
$image_name = addslashes($_FILES['image']['name']);
$image_size = getimagesize($_FILES['image']['tmp_name']);
if($image_size == FALSE){
echo "This is not an image.";
} else {
if(!$sql = mysql_query("INSERT INTO imageupload VALUES ('','$image_name','$image')")){
echo "Problem in uploading image.";
} else {
$lastid = mysql_insert_id();
echo "Image uploaded successfully<p />Your image:<p /><img src=getimage.php?id=$lastid>";
}
}
}
?>
</body>
</html>
<?php
// Connect to database
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("phptut") or die(mysql_error());
$id = addslashes($_REQUEST['id']);
$dbimage = mysql_query("SELECT * FROM imageupload where id=$id");
$dbimage = mysql_fetch_assoc($dbimage);
$dbimage = $dbimage ['image'];
header("Content-type: image/jpeg");
echo $dbimage;
?>
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