PHP/MYSQL Query returning reference ID instead of INT value

Question

I've got a php script with this statement:

$query = "SELECT type 
            FROM users 
           WHERE username = '$username'";

$result = $database->query($query);
if($result == 1) {
  echo "whatever";
  continue;
}

The problem is that the if never runs and when I created a print statement to print $result before the if runs, it prints a Reference ID#. So result is never == 1 because it is being assigned the reference ID #.

What the heck am I doing wrong? How do I assign the value of 'type' which is an INT, instead of it's contents Reference ID#?

What is $database is it your own custom object, amysqli connection, something else? — Michael Mior
– Michael Mior, Commented Jul 13, 2011 at 18:10
$database is an object used to open a connection to the database — Jmichelsen
– Jmichelsen, Commented Jul 13, 2011 at 19:36

genesis · Accepted Answer · 2011-07-13 19:34:25Z

7

you have to fetch that line first ...

$query = "SELECT type FROM users WHERE username = '$username'";
$result = $database->query($query);
$row = mysql_fetch_assoc($result);
if($row['type'] == 1)
{
      echo "whatever";
}

edited Jul 13, 2011 at 19:34

answered Jul 13, 2011 at 18:09

genesis

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3 Comments

Jmichelsen Over a year ago

I tried this and got an error stating: PHP Fatal error: Cannot break/continue 1 level

genesis Over a year ago

it's not my fault, I thought you got it in for or something like that. Answer edited ;)

Jmichelsen Over a year ago

hehe didn't mean to point fingers, I thought that continue was supposed to be there. I'm giving this a try, thank you.

Wilson212 · Accepted Answer · 2011-07-13 18:10:37Z

3

You need to use mysql_fetch_assoc() or mysql_fetch_array() to get the result into an array:

$query = $database->query("SELECT type FROM users WHERE username = '$username'");
$result = mysql_fetch_assoc($query);

if($result['type'] == 1)
{
    echo "whatever";
    continue;
}

answered Jul 13, 2011 at 18:10

Wilson212

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Comments

JConstantine · Accepted Answer · 2011-07-13 18:10:22Z

2

try

<?php
$result = mysql_fetch_assoc($database->query($query));
$id = $result['type'];
if($type == 1)
{

answered Jul 13, 2011 at 18:10

JConstantine

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Comments

Rolice · Accepted Answer · 2011-07-13 18:18:04Z

0

Yes. You have to check it like:

$query = "SELECT type FROM users WHERE username = '$username'";
$result = $database->query($query);
if($result)
{
    echo "whatever";
    continue;
}

This is a Boolean check, but... In php we have weak types - everything is true, except on false, 0, '' (empty string), null, etc.

If you want to make a good Boolean check including type, then you have:

if(bool_val === true)
{
..
}
// or
if(bool_val && is_bool(bool_val))
{
..
}

This query object returns a result resource in php. And you probably receive #Resource ID, which is some kind of pointer in php code.

If you receive something like this (#Resource ID) - this means your query has passed correctly and no error occured.

answered Jul 13, 2011 at 18:18

Rolice

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