return array from function by passing string

Question

i have following function where i have brands info in an array. i should get array containing those info when i pass brand name to this function.

function brand_info($brand)
{
    $brands_list=array ( 
    'lg'=>
    array(
    'name'           => 'LG Phone Company',
    'country'        =>  'country',
    'founded_year'   =>  '2001'
    ),
    'nokia'=>
    array(
    'name'           => 'Nokia Phone Company',
    'country'        =>  'country',
    'founded_year'   =>  '2001'
    )
    );


    if(in_array($brand,$brands_list))
    {
        // return array containg company info
    }
}

this should return an array by which i can show these info.

$brand_info=brand_info($brand_name);
echo $brand_info['name'];

what may be the best way to do this?

Do you mean return $brands_list[$brand];? Apparently you already know how to work with arrays, so what is the problem? — Felix Kling
– Felix Kling, Commented May 18, 2012 at 10:18
@felix yes exactly, but its returning Null, i don't know what i am missing — naeplus
– naeplus, Commented May 18, 2012 at 10:26
Well, if the brand is not in the array, it does not return anything... — Felix Kling
– Felix Kling, Commented May 18, 2012 at 10:27
@FelixKling i am passing 'nokia' to this function. you can see array contain 'nokia', but its not returning array. — naeplus
– naeplus, Commented May 18, 2012 at 10:33
Ah, in_array tests whether the value exists in the array. You are testing for a key, so it should be if(isset($brands_list[$brand])). — Felix Kling
– Felix Kling, Commented May 18, 2012 at 10:37

Nick · Accepted Answer · 2012-05-18 12:04:46Z

3

If you're passing in the brandname then this would suffice:

function brand_info($brand)
{
    $brands_list=array ( 
    'lg'=>
    array(
    'name'           => 'LG Phone Company',
    'country'        =>  'country',
    'founded_year'   =>  '2001'
    ),
    'Nokia'=>
    array(
    'name'           => 'Nokia Phone Company',
    'country'        =>  'country',
    'founded_year'   =>  '2001'
    )
    );

    if (array_key_exists($brand,$brands_list)) {
      return $brands_list[$brand];
    } else {
      return false;
    }
}

$brandinfo = brand_info('Nokia');
echo $brandinfo['name']; // will print "Nokia Phone Company"

edited May 18, 2012 at 12:04

answered May 18, 2012 at 10:19

Nick

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6 Comments

naeplus Over a year ago

but its returning Null, i used var_dump($brandinfo); to show.

Nick Over a year ago

The above returns correctly for me, and it's not using anything that would depend on your PHP config, so it should be working for you.

Robin Castlin Over a year ago

Make sure you've typed all variables correctly @NaeemX2. This code is valid.

naeplus Over a year ago

you are right... i am missing. its working now. thanks for your help.

edelgado Over a year ago

You may have problems if you do not test weather the array key exists.

|

Amarnasan · Accepted Answer · 2012-05-18 10:19:52Z

1

May seem trivial but...

return $brands_list[$brand]

answered May 18, 2012 at 10:19

Amarnasan

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Comments

chrvadala · Accepted Answer · 2012-05-18 10:23:29Z

1

function brand_info($brand)
{
    $brands_list=array ( 
    'lg'=>
    array(
    'name'           => 'LG Phone Company',
    'country'        =>  'country',
    'founded_year'   =>  '2001'
    ),
    'nokia'=>
    array(
    'name'           => 'Nokia Phone Company',
    'country'        =>  'country',
    'founded_year'   =>  '2001'
    )
    );


    if(in_array($brand,$brands_list))
    {
        return $brand_list[$brand];
    }else{
        return null;
    }
}

and then

$info = brand_info($my_brand);
if(!is_null($info)){ echo $info['name']; }

answered May 18, 2012 at 10:23

chrvadala

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Comments

edelgado · Accepted Answer · 2012-05-18 10:32:00Z

1

function brand_info($brand) {
    $brands_list=array (
        'lg'=>
            array(
                    'name'           => 'LG Phone Company',
                    'country'        =>  'country',
                    'founded_year'   =>  '2001'
            ),
        'Nokia'=>
            array(
                    'name'           => 'Nokia Phone Company',
                    'country'        =>  'country',
                    'founded_year'   =>  '2001'
            )
    );


    foreach ($brands_list as $brandname=>$info) {
        if($brandname==$brand) {
            return $info;
        }
    }
    return array();
}

answered May 18, 2012 at 10:32

edelgado

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