Question

I have a script that is importing data from a csv source. Many of the fields/columns are in “code.” For example, the school field is a numeric number apposed to the actual school name. I need to convert the code into the actual name before it imports into the db.

My question is, is there a quicker way to do this conversion other than using a “if” statement. Some of the fields that need to be converted have 20+ options so i am wondering if there is another way to write the conversion instead of having 20 if statments. Right now I have:

if($data['school'] = 0001) {$school = "school1"; } 
if($data['school'] = 0002) {$school = "school2"; }
if($data['school'] = 0003) {$school = "school3"; }...

or i could use if else statements

So my question is: Is there a quicker way either through some sort of array or loop or other type of php statement that I could speed up this process?

Answer 1

Do you have all of this in a file, or do you have to type it into PHP code by hand either way?

You can put the code mappings into an associative array, then a single statement will get the right name.

$schoolCodeToName = array(
    '0001' => 'school1',
    '0002' => 'school2',
    '0003' => 'school3'
    // ...
);

$school = $schoolCodeToName[ $data['school'] ];

If you want to save yourself from the typing, then you need some code to load a data file and build the $schoolCodeToName array.