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upload the resume and store in database

100+
P: 118
hi,

i wrote a code to upload a resume in mysql database and download that one .
But in my database name,type,size of the file stored but content of the file not stored. So in my database content filed is empty.plz tell that what's the problem in my code.


[PHP]

<?
include("db.php");
if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0)
{
$fileName = $_FILES['userfile']['name'];
$tmpName = $_FILES['userfile']['tmpname'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];

$fp = fopen($tmpName, 'rb');
$content = fread($fp, filesize($tmpName));
$content = addslashes($content);
fclose($fp);

if(!get_magic_quotes_gpc())
{
$fileName = addslashes($fileName);
}

$query = "INSERT INTO upload (name, size, type, content ) ".
"VALUES ('$fileName', '$fileSize', '$fileType', '$content')";

mysql_query($query) or die('Error, query failed');

echo "<br>File $fileName uploaded<br>";
}
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>New Page 2</title>
</head>

<body>
<form method="post" enctype="multipart/form-data">
<table width="350" border="0" cellpadding="1" cellspacing="1" class="box">
<tr>
<td width="246">
<input type="hidden" name="MAX_FILE_SIZE" value="2000000">
<input name="userfile" type="file" id="userfile">
</td>
<td width="80"><input name="upload" type="submit" class="box" id="upload" value=" Upload "></td>
</tr>
</table>
</form>

</body>

</html>
[/PHP]

And i got the errors as

Warning: fread(): supplied argument is not a valid stream resource in e:\project\graphic\docu.php on line 11

Warning: fclose(): supplied argument is not a valid stream resource in e:\project\graphic\docu.php on line 13
Sep 18 '08 #1
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2 Replies


Dormilich
Expert Mod 5K+
P: 8,639
hm, what's the difference between $filename and $tmpname?
what are the values of the vars (line 5 to 8) (var_dump())? maybe this gives a hint...

regards
Sep 18 '08 #2

P: 1
You have done One silly mistake at


$tmpName = $_FILES['userfile']['tmpname'];
here you have written [tmpname] instead if ['tmp_name']
so you should have to write

$tmpName = $_FILES['userfile']['tmp_name'];
Dec 1 '13 #3

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