Not getting appropriate php string result from mysql_query()

Question

I'm looking to label a string based on database results. I have the following code:

$loc1 = "A";
$query  = "SELECT loc FROM User where nick='".$users[$j]."'";
    $loc = mysql_query($query);
    if($loc == 1)
        $loc1 = "A";
    if($loc== 2)
        $loc1 = "B";

The loc1 location is always "A". I've confirmed that the query works in MySQL and I'm having a difficult time understanding why this simple case is not working. I would appreciate your help!

$loc is mysql resource and not mysql data

Poonam
– Poonam

2012-02-22 13:18:00 +00:00
Commented Feb 22, 2012 at 13:18 — Poonam
– Poonam, Commented Feb 22, 2012 at 13:18
Check out the usage examples in the mysql_query() docs.

Marcus Adams
– Marcus Adams

2012-02-22 14:00:03 +00:00
Commented Feb 22, 2012 at 14:00 — Marcus Adams
– Marcus Adams, Commented Feb 22, 2012 at 14:00

codaddict · Accepted Answer · 2012-02-22 13:18:59Z

4

The function mysql_query does not return the value of the column fetched, instead it returns a MySQL resource which you use to extract the column value returned by the query.

Change:

$loc = mysql_query($query);

to:

$result = mysql_query($query);
if (!$result) {
    die('Invalid query: ' . mysql_error());
}
if ($row = mysql_fetch_assoc($result)) {
    $loc = $row["loc"];
}

answered Feb 22, 2012 at 13:18

codaddict

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Comments

Scott Rowley · Accepted Answer · 2014-06-13 19:17:23Z

2

You are storing mysql_query() return value in $loc which is a resource.

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

So, $loc is not 1 or neither equals to 2.

You need to use mysql_fetch_array / mysql_fetch_object, etc to manipulate the resource.

edited Jun 13, 2014 at 19:17

Scott Rowley

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answered Feb 22, 2012 at 13:22

Avisek Chakraborty

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Comments

corretge · Accepted Answer · 2012-02-22 14:16:58Z

1

You must fetch the recordset with, in example mysql_fetch_row

Example #1 Fetching one row with mysql_fetch_row()

<?php
$result = mysql_query("SELECT id,email FROM people WHERE id = '42'");
if (!$result) {
    echo 'Could not run query: ' . mysql_error();
    exit;
}
$row = mysql_fetch_row($result);

echo $row[0]; // 42
echo $row[1]; // the email value
?>

But, be careful with SQL injection attacks!

http://es.php.net/manual/en/security.database.sql-injection.php

edited Feb 22, 2012 at 14:16

answered Feb 22, 2012 at 13:23

corretge

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