conditional in_array trouble with PHP mySQL

Question

I have firePHP so i know exactly what the variables are, but I can't figure out why this code doesn't change it.

I receive from a mySQL call $query (which if returned produces [{"type":"2"}]) I have 4 potential types, and things can be multiple types (i.e. [{"type":"1"},{"type":"2"}])

Now I want to read this data and run various other functions based on the type it has, that is: if it's only type 2, call function TWO, if it's type 1 and 2 call function ONE and function TWO. I thought this would be easiest if i moved all the data into another array.

Here is the code I currently have:

$result = array('message'=>false, 'money'=>false, 'glasses'=>false, 'exclamation'=>false);
    if (in_array('1',$query)) {$result['message'] = true;}
    if (in_array('2',$query)) {$result['money'] = true;}
    if (in_array('3',$query)) {$result['glasses']=true;}
    if (in_array('4',$query)) {$result['exclamation']=true;}
    echo json_encode($result);

This however does not update the $result array (as I can tell all of the values of $message are false in firePHP.... Thus I assume something is wrong with my if statements, but what?

Can you please put the complete output of $query, using print_r for instance ? — Rodolphe
– Rodolphe, Commented May 23, 2011 at 15:15

jeroen · Accepted Answer · 2011-05-23 15:18:17Z

1

I´m not sure about the value of $query, but if it is something like:

array [0 => '{"type":"2"}']

You would have to use:

in_array('{"type":"2"}',$query)

as that is the value of your variable.

answered May 23, 2011 at 15:18

jeroen

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3 Comments

mazlix Over a year ago

Oh, wow I knew it was going to be something simple...I had tried in_array('2', $query[0]) but that gave an error, why is that different?

martynthewolf Over a year ago

@mazlix It's different because 2 is only a part of the value held in array(0 => '{"type":"2"}') The only way your code would have worked would have been if the array looked like this array(0 => 2)

jeroen Over a year ago

@mazlix Your value is not an array in php, it is a string. It´s the object notation for javascript, but php only sees a string.

Bob Baddeley · Accepted Answer · 2011-05-23 15:21:35Z

0

Is it because the results returned in $query are arrays of arrays, and thus in_array is only searching at the top level and not sub-levels? It seems like what you want is to recursively search $query.

answered May 23, 2011 at 15:21

Bob Baddeley

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