Php showing resource id and errors

Question

Whenever I run this code, I get a resource id. How do I fix this:

function ct()
{
    $result = mysql_query("SELECT `Quantity` FROM `shopping cart` WHERE `Customer_id`=1") or die(mysql_error());
    mysql_fetch_array($result);   
    echo "$result";
}

PTEM (peruse the excellent manual) for how to use the mysql extension: http://php.net/manual/en/mysql.examples-basic.php Also don't miss http://php.net/manual/en/mysqlinfo.api.choosing.php. — deceze, Apr 19 '13 at 05:45
because it return resource id by which you can fetch data. you need to pass this variable in mysql_fetch_* to get data. — Code Lღver, Apr 19 '13 at 05:46
go to [php.net](http://php.net/manual/en/function.mysql-query.php) and check the what this function returns. — Mithun Satheesh, Apr 19 '13 at 05:46
since you are using mysql_fetch_array it will return array so specify the field name — NullPoiиteя, Apr 19 '13 at 05:48

score 4 · Accepted Answer · edited Jun 20 '20 at 09:12

You need to fetch result by mysql_fetch_array or mysql_fetch_assoc

$row = mysql_fetch_array($result);

and

what mysql_query return is

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

For other type of SQL statements, INSERT, UPDATE, DELETE, DROP, etc, mysql_query() returns TRUE on success or FALSE on error.

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO, or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.

After edit in question

This mysql_fetch_array($result); Returns an array that corresponds to the fetched row and moves the internal data pointer ahead. So you must assigned that array to any variable to use like

while($row = mysql_fetch_array($result)){   
 echo $row ['field_name'];
}

Or

$row = mysql_fetch_array($result); // want to fetch only one row 
 echo $row ['field_name'];

success i did not place a field name – Apr 19 '13 at 05:59 — , Apr 19 '13 at 05:59

score -1 · Answer 2 · edited Apr 19 '13 at 05:58

-1

$link = mysqli_connect("HOSTNAME", "USERNAME", "PASS", "DBNAME");    
$result = mysqli_query($link,"SELECT `Quantity` FROM `shopping cart` WHERE `Customer_id`=1") or die(mysql_error());
while($row = mysqli_fetch_array($result))
{
    echo $row['Quantity'];
    echo "<br />";
}
mysqli_close($con);

edited Apr 19 '13 at 05:58

mrjimoy_05

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answered Apr 19 '13 at 05:45

Praveen kalal

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Php showing resource id and errors

2 Answers2