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up vote 0 down vote favorite

I have an array of arrays as follows (for example)

$array = (
    [0] => array (
        [a_key]     => 123456,
        [b_key]     => First Name,
        [c_key]     => Last Name,
        [this_key ] => Institution,
        ),
    [1] => array (
        [a_key]     => 123456,
        [b_key]     => First Name,
        [c_key]     => Last Name,
        [this_key ] => Institution,
        ),
    [2] => array (
        [a_key]     => 123456,
        [b_key]     => First Name,
        [c_key]     => Last Name,
        [this_key ] => Institution,
        ),
)

And I want to check if $array[$key]['this_key'] is not null or does not equal '' when

$key = (any number)

How do I set $key to match any number so that the following returns true when at least one second-dimension-array in the array has [this_key] set?

if( $array[$key]['this_key'] != '' )
    echo "true";
else 
    echo "false";

So, if $array[0][this_key] is not set, $array[1][this_key] IS set, and $array[2][this_key] is not set, the IF statement will return true because at least ONE of all of the arrays has [this_key] set.

Edit:

To give some context to the situation and provide a specific scenario to which this solution could be applied:

I have a list of people in a database that I need to display. For each person, I have a first and last name, and optionally an institution/degree/job title held by the individual. These need to display like this:

Without a title:
John Smith, Jane Doe, Fred Nerk

With a title:
John Smith, Penologist; Jane Doe, Professor; Fred Nerk, Astrophysicist

With mixed cases of a title:
John Smith; Jane Doe, Professor; Fred Nerk

Note the punctuation used in each case. If there is no optional information for ANY record, a comma separates the individuals. If there is optional information for ONE OR MORE records, a semicolon separates the individuals.

With the if statement, I would be able to determine if one of the individuals has optional information, even if every other individual does not.

By the way:

This PHP Sandbox Tool was incredibly useful for testing PHP code before integrating it into my file. It supports PHP versions from 4.4.9 up to 5.5.5.

share|improve this question
    
array_column() and array_filter() might be helpful. Else use a loop and a state variable. –  mario Aug 25 '14 at 15:21
    
Thanks, this looks like it will work, just need to update my version of PHP before I can use these. –  Nicky Aug 26 '14 at 16:10
    
There's a userland fallback for array_column in the manual, which becomes redundant when you upgrade, but can benefit code readability already now. –  mario Aug 26 '14 at 16:38
    
Thank you! I did not see that link. –  Nicky Aug 26 '14 at 19:03

2 Answers 2

up vote 0 down vote

You could do something like this:

foreach($array as $key=>$array){
    if(!empty($array['this_key'])){
        //found one, do something here           
    }
}
share|improve this answer
up vote 0 down vote accepted

For PHP >= 5.5.0, I used:

if(array_filter(array_column($array, 'this_key'))){
    // true, do something
}

When this returns true, I know that at least one array has information in this_key, and so I can do something unique for this case.

My server runs PHP 5.4.13, so I used the recommended userland implementation for PHP lower than 5.5. It was simple to implement, and it works as expected.

share|improve this answer

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