Using Variable Variables in PHP on Classes (CodeIgniter)

Question

I am running PHP with CodeIgniter. My application needs to output formatted addresses throughout the site, pulled from MySQL. In some cases, address_1 may be blank or may be filled, in other cases postcode may or may not be there, etc. I have written a helper function that formats these addresses neatly, when passed a MySQL result set.

The issue is that due to the size and complexity of the site, in some places I need to use this function with multiple addresses in the same result set, and in other places I may pass the results as $query->row('address_1') or as $row->address_1.

I have written my function out like this:

// Get every element from the MySQL Resultset       
            if(method_exists($address, 'row'))
            {
                $a['abode']         = ($address->row($prefix.'abode')         ? $address->row($prefix.'abode') : '');
                $a['building_no']   = ($address->row($prefix.'building_no')   ? $address->row($prefix.'building_no') : ''); 
                $a['building_name'] = ($address->row($prefix.'building_name') ? $address->row($prefix.'building_name') : '');
                $a['office_number'] = ($address->row($prefix.'office_number') ? $address->row($prefix.'office_number') : '');
                $a['address_1']     = ($address->row($prefix.'address_1')     ? $address->row($prefix.'address_1') : '');
                $a['address_2']     = ($address->row($prefix.'address_2')     ? $address->row($prefix.'address_2') : '');
                $a['town']          = ($address->row($prefix.'town')          ? $address->row($prefix.'town') : '');
                $a['county']        = ($address->row($prefix.'county')        ? $address->row($prefix.'county') : '');
                $a['postcode']      = ($address->row($prefix.'postcode')      ? $address->row($prefix.'postcode') : '');
                $a['country']       = ($address->row($prefix.'country')       ? $address->row($prefix.'country') : '');
            }
            else
            {
                // PHP Variable Variables
                $a['abode']         = (isset($address->{$prefix.'abode'})           ? $address->{$prefix.'abode'}: '');     
                $a['building_no']   = (isset($address->{$prefix.'building_no'})     ? $address->{$prefix.'building_no'} : '');
                $a['building_name'] = (isset($address->{$prefix.'building_name'})   ? $address->{$prefix.'building_name'} : '');
                $a['office_number'] = (isset($address->{$prefix.'office_number'})   ? $address->{$prefix.'office_number'} : '');
                $a['address_1']     = (isset($address->${$prefix.'address_1'})      ? $address->${$prefix.'address_1'} : '');
                $a['address_2']     = (isset($address->{$prefix.'address_2'})       ? $address->{$prefix.'address_2'} : ''); 
                $a['town']          = (isset($address->{$prefix.'town'})            ? $address->{$prefix.'town'} : ''); 
                $a['county']        = (isset($address->{$prefix.'county'})          ? $address->{$prefix.'county'} : ''); 
                $a['postcode']      = (isset($address->{$prefix.'postcode'})        ? $address->{$prefix.'postcode'} : ''); 
                $a['country']       = (isset($address->{$prefix.'country'})         ? $address->{$prefix.'country'} : '');
            }

As you can see I was 'experimenting' a little on $a['address_1'] of the else statement trying to find a combination that worked.

I am trying to get the else statement case to work, but PHP throws no errors or anything except for the $a['address_1'] line, in which it says:

Message: Undefined variable: app_address_1, Filename: helpers/address_helper.php.

My $prefix variable is set to app_, but this isn't working. Why?

$address->{$prefix.'address_1'} should work just fine. Maybe it just doesn't have a address_1 property. Try to use var_dump($address) to see what's in it. — gen_Eric
– gen_Eric, Commented Dec 7, 2011 at 19:55

Francis Avila · Accepted Answer · 2011-12-07 20:07:07Z

1

Look carefully at that line:

$a['address_1']     = (isset($address->${$prefix.'address_1'})      ? $address->${$prefix.'address_1'} : '');

Do you see what I see?

$address->${$prefix.'address_1'} 
----------^

You have an excess $ in there! The error is because it's trying to find the value of a variable named app_address_1, which is the result of $prefix.'address_1'.

Let's evaluate step-by-step:

$address->${$prefix.'address_1'}
$address->${'app_address_1'}
$address->{$app_address_1}
$address->{NULL}

answered Dec 7, 2011 at 20:07

Francis Avila

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