Loop and echo specified JSON content

Question

How can I show artistName only in "wrapperType":"track"?

Here's the JSON example:

{  
   "resultCount":5,
   "results":[  
      {  
         "wrapperType":"collection",
         "artistName":"Liam Payne"
      },
      {  
         "wrapperType":"track",
         "artistName":"Liam Payne & French Montana"
      }
      {  
         "wrapperType":"track",
         "artistName":"Liam Payne & French Montana"
      }
      {  
         "wrapperType":"track",
         "artistName":"French Montana"
      }
      {  
         "wrapperType":"track",
         "artistName":"French Montana"
      }
   ]
}

and the php code:

but this will display all values

foreach ($obj->results as $row){
    echo $row->artistName;
}

Manish Chauhan · Accepted Answer · 2018-08-28 11:39:37Z

4

Put if($row->wrapperType == "track") condition in your code.

Try below code:

foreach ($obj->results as $row){
    if($row->wrapperType == "track"){
        echo $row->artistName;
    }
}

answered Aug 28, 2018 at 11:39

Manish Chauhan

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Comments

user8034901 · Accepted Answer · 2018-08-28 11:41:34Z

0

foreach ($obj->results as $row){
    echo $row->artistName;
}

loops through all available items. You should check wrapperType:

foreach ($obj->results as $row){
  if ($row->wrapperType == 'track') {
    echo $row->artistName;
  }
}

answered Aug 28, 2018 at 11:41

user8034901

Comments

Bilal Ahmed · Accepted Answer · 2018-08-28 11:52:46Z

0

first your json is not valid (missing ,) , you can check your desired output here

in loop used $data['results'] instead of $obj->results

Used if condition like if($row['wrapperType']=='track'){// if condition for what type of data you want

$data='{
    "resultCount": 5,
    "results": [{
            "wrapperType": "collection",
            "artistName": "Liam Payne"
        },
        {
            "wrapperType": "track",
            "artistName": "Liam Payne & French Montana"
        }, {
            "wrapperType": "track",
            "artistName": "Liam Payne & French Montana"
        }, {
            "wrapperType": "track",
            "artistName": "French Montana"
        }, {
            "wrapperType": "track",
            "artistName": "French Montana"
        }
    ]
}';
$data=json_decode($data, true);
print_r($data['results']);
foreach ($data['results'] as $row){
    if($row['wrapperType']=='track'){// if condition for what type of data you want
       echo "Artist Name: ".$row['artistName'];
       echo "Wrapper Type: ".$row['wrapperType'];
    }
}

edited Aug 28, 2018 at 11:52

answered Aug 28, 2018 at 11:46

Bilal Ahmed

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1 Comment

Theresia Vidi Over a year ago

what is the difference between $data['results'] and $obj->results?

Ketan Yekale · Accepted Answer · 2018-08-28 13:46:15Z

0

You need to add a condition in the foreach loop to check if wrapperType is track

Like:

foreach ($obj->results as $row){
    if($row->wrapperType == "track"){
      echo $row->artistName;
   }
}

edited Aug 28, 2018 at 13:46

answered Aug 28, 2018 at 11:41

Ketan Yekale

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1 Comment

starbeamrainbowlabs Over a year ago

Hey, thanks for answering this question! An answer that contains only code is not very helpful without some additional detail. For example, why doe your answer work? Why not do it a different way? Including these things in your answers helps the question-asker - and others - to better understand your solution.

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