how to check the required data in this php array

Question

I have a response in below format. I am using laravel. I want to check if there is street_number present or not despite of their index and if yes just print the long_name similar I want to check if there is route in types of this array or not. if yes print route long_name and so on.

dd($dataArray); gives this response

array:3 [
  "html_attributions" => []
  "result" => array:15 [
    "address_components" => array:7 [
      0 => array:3 [
        "long_name" => "19"
        "short_name" => "19"
        "types" => array:1 [
          0 => "street_number"
        ]
      ]
      1 => array:3 [
        "long_name" => "Becket Street South"
        "short_name" => "Becket St S"
        "types" => array:1 [
          0 => "route"
        ]
      ]
      2 => array:3 [
        "long_name" => "senroy"
        "short_name" => "Slenroy"
        "types" => array:2 [
          0 => "locality"
          1 => "political"
        ]
      ]
]

I am looping like this

$newArray=[];
foreach($dataArray['result'] as $data){
//    check if street_number in array or not if yes push it value to new array
//same for other types as well

}

just check inside types via in_array so that you don't need to use a hardcoded index, it'll just check if street_address exists — Kevin
– Kevin, Commented Oct 20, 2021 at 2:00
i did this if (in_array('street_number', $data[$key]['types'])){ // it gives error "Undefined index: types" } — David
– David, Commented Oct 20, 2021 at 2:04
but this is fine if (in_array('street_number', $data[0]['types'])). but i don't want to hardcode — David
– David, Commented Oct 20, 2021 at 2:05

Farhan Ibn Wahid · Accepted Answer · 2021-10-20 04:18:17Z

2

the inside of your foreach is problematic: should be like this:

foreach($dataArray as $key=> $data){
    if (in_array('street_number',$data['types'])) {
        
    }
}

because your types is already in $data inside the foreach. So you don't need to say $key again.

Edit 1: found the problem. remove isset($data[$key]) because it is always false. because the $key is returning 0,1,.... but $data keys are these three : 'long_name' , 'short_name' , 'types' . You can check by dd($key)

Edit 2: After your latest update of the question, your foreach should be like this:

foreach($dataArray['result']['address_components'] as $data){
    if (in_array('street_number',$data['types'])) {
        dd('should be here');
    }
}

edited Oct 20, 2021 at 4:18

answered Oct 20, 2021 at 2:21

Farhan Ibn Wahid

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11 Comments

David Over a year ago

not working: if (isset($data[$key]) && in_array('street_number',$data['types'])) { dd('yes'); }I have not got the yes answer which means it was skipped

Farhan Ibn Wahid Over a year ago

problem found and updated in the answer @David

David Over a year ago

message: "Undefined index: types" got error

Pablo Over a year ago

Share var_nump($data) from the loop

Pablo Over a year ago

Honestly as long you do dd($variableGivingWarning) you should be able to figure out which array keys to use. Your top-level loop should start with $dataArray['result']['address_components'] instead of $dataArray['result']

|

Dharman · Accepted Answer · 2021-10-20 04:31:04Z

0

May this can help you

foreach ($data['result']['address_components'] as $val) {
    if(\Illuminate\Support\Arr::has($val, 'types') && in_array('street_number', $val['types'])) {
        echo "For Street Nu.";
    }else if(\Illuminate\Support\Arr::has($val, 'types') && in_array('route', $val['types'])) {
        echo "For Route";
    }
}

edited Oct 20, 2021 at 4:31

Dharman♦

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answered Oct 20, 2021 at 4:25

Gautam Patadiya

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