Can’t figure out what’s wrong with this array

[kentonator]

Ok, to be fair, I’m brand new to PHP and just trying to piece this together for a friends website. The idea of this code is if a value of 1 is given in the data entry it would return house and so on. But right now, even if it’s a 3 it’s just returning house instead of Mobile Home.  I’ve tried everything I know and googled for ever, but if you don’t know the right language to google, it’s impossible.  I have a feeling this is a simple fix and I just don’t know it.

function property_structure_type ($data){ $data = array("1", "2", "3", "4", "5", "6", "7");
if (in_array("1", $data)) {
    return "House";
}
if (in_array("2", $data)) {
    return "Cabin";
}
if (in_array("3", $data)) {
    return "Mobile Home";
}
if (in_array("4", $data)) {
    return "Apartment";
}
if (in_array("5", $data)) {
    return "Condominium";
}
if (in_array("6", $data)) {
    return "Townhome";
}
if (in_array("7", $data)) {
    return "None";
}

 

[benanamen]

It' simple really. You are providing an array, it matches the 1 and returns back to the caller so the rest of the if's are not processed.

Try this...

<?php declare(strict_types=1);

function property_structure_type ($data)
{
 if (in_array("1", $data)) {
    return "House";
}
if (in_array("2", $data)) {
    return "Cabin";
}
if (in_array("3", $data)) {
    return "Mobile Home";
}
if (in_array("4", $data)) {
    return "Apartment";
}
if (in_array("5", $data)) {
    return "Condominium";
}
if (in_array("6", $data)) {
    return "Townhome";
}
if (in_array("7", $data)) {
    return "None";
}
}

echo property_structure_type ([2]);

 

Edited April 17, 2019 by benanamen

[Barand]

Of course - no matter what value you pass, "1" is in the array so it returns "House". Functions exit as soon as a value is returned.

Perhaps

function property_structure_type ($data)
{
    $types = [ 1 => 'House',
               2 => 'Cabin',
               3 => 'Mobile Home',
               4 => 'Apartment',
               5 => 'Condominium',
               6 => 'Townhome'
             ];
    return isset($types[$data]) ? $types[$data] : 'None';
}

echo property_structure_type(3);    //-> Mobile home

 

[kentonator]

3 hours ago, Barand said:

Of course - no matter what value you pass, "1" is in the array so it returns "House". Functions exit as soon as a value is returned.

Perhaps

function property_structure_type ($data)
{
    $types = [ 1 => 'House',
               2 => 'Cabin',
               3 => 'Mobile Home',
               4 => 'Apartment',
               5 => 'Condominium',
               6 => 'Townhome'
             ];
    return isset($types[$data]) ? $types[$data] : 'None';
}

echo property_structure_type(3);    //-> Mobile home

 

This worked perfectly, thanks a ton!