Resource id #4 Error on Binary Row

[Kriekie]

Hi

I'm trying to access an image from a binary blob row in mySQL, but keep getting the error "Resource id #4"

The code I'm using:

<?php
@require("scripts/connect.php");
$iid = $_REQUEST[iid];
$Query = mysql_query("Select image from image where image_id = $iid");
header("Content-type: image/jpeg");
$result = mysql_fetch_array($Query);
$image = $result["image"];
echo $image;
?>

Thanks for any help.

 

[sleipnir214]

You've missed a step. Before you can use mysql_fetch_array(), you have to invoke mysql_query().

Mysql_query(), when used with a successful SELECt query, will return a result handle that your code will pass to mysql_fetch_array().

See the PHP online manual entry for mysql_fetch_array() for example code.


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TANSTAAFL!!

 

[sleipnir214]

Also, I don't know the content of the connect.php file your script includes. Does it select the correct database within your MySQL server?


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TANSTAAFL!!

 

[Olavxxx]

Also, a tip:
When writing a query, it the mysql functions should be in uppercase, fieldnames, etc. should be in lowercase.

SELECT ... FROM ... WHERE ... = ...

// Quote variable to make safe
function quote_smart($value)
{
   // Stripslashes
   if (get_magic_quotes_gpc()) {
       $value = stripslashes($value);
   }
   // Quote if not integer
   if (!is_numeric($value)) {
       $value = "'" . mysql_real_escape_string($value) . "'";
   }
   return $value;
}

// Connect
$link = mysql_connect('mysql_host', 'mysql_user', 'mysql_password')
   OR die(mysql_error());

// Make a safe query
$query = sprintf("SELECT `image` FROM `image` WHERE `image_id` = '%s'",
           quote_smart($_POST['username']));

mysql_query($query);

That query is also improved, regarding sql injection

ref.:

http://no2.php.net/mysql_real_escape_string

Olav Alexander Mjelde

http://www.volvo-power.net

Admin & Webmaster