Why is the key used in the WHERE condition not found as an identifier in php?

Question

I have an SQL statement that goes like this:

SELECT 
    ISO_id, country_name, gold, silver, bronze, total 
FROM 
    Country 
WHERE 
    ISO_id='GBR';

And I use it to receive information from a database via PHP and PEAR. The database returns GBR, UK, 29, 17, 19, 65 , meaning that all the information was correctly returned.

Successfully calling and storing the information via PEAR:

$countryResult =& $db->query($sql);
if(PEAR::isError($countryResult)){
    die($countryResult->getMessage());
}
$row =& $countryResult->fetchRow();

Now, if I use $row['country_name'], $row['gold'], or anything else except $row['ISO_id'], then it returns the information correctly as a normal associative array. However, when I use $row['ISO_id'] I get the error:

Notice: Undefined index: ISO_id in /disks/olympics/view.php on line 100 Call Stack: 0.0004 662648 1. {main}() /disks/olympics/view.php:0

My temporary workaround is calling ISO_id twice in the original sql statement and renaming one, which works like so:

SELECT 
    ISO_id, ISO_id as iso2, country_name, gold, silver, bronze, total 
FROM 
    Country 
WHERE 
    ISO_id='GBR';

//GBR, GBR, UK, 29, 17, 19, 65

Now I can call $row['iso2'] successfully for what should already work as $row['ISO_id'].

For some reason, it looks like using the original ISO_id in the WHERE clause in the SQL statement messes how I can use it later in the program. Why is that?

Answer 1

If you print out the $row during your fetch loop you will see the actual key for the ISO_id is being returned by the sql server as all lower case. So when you try to reference it as ISO_id instead of iso_id PHP can't find it.

You simply need to change the key to 'iso_id' , all lowercase and you will be good to go.