How to get json output in php from url

Question

I see there is some more questions about this, but not useful for me.

Here i have the link and following code i am Link

<?php 
$json=file_get_contents("http://2strok.com/radio.json");
$details=json_decode($json);
if($details->Response=='True')
?>

<?php echo $details->name[3];?><br>

my question is how can i get the result from the json link in the php code? — Zabkas
– Zabkas, Commented Jan 26, 2018 at 4:50

Panos Kalatzantonakis · Accepted Answer · 2018-01-26 05:16:03Z

3

$json=file_get_contents("http://2strok.com/radio.json");
$details=json_decode($json,true);
foreach($details as $output) {
    echo $output['name'].'<br>';
}

Update - get specific name

$json=file_get_contents("http://2strok.com/radio.json");
$details=json_decode($json,true);
foreach($details as $output) {
    if ($output['name']=='FM 101 / Islamabad'){
        echo 'Specific name found: '.$output['name'];
    }
    // echo $output['name'].'<br>';
}

edited Jan 26, 2018 at 5:16

answered Jan 26, 2018 at 4:55

Panos Kalatzantonakis

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6 Comments

Zabkas Over a year ago

Thankyou so much ;)

Zabkas Over a year ago

Do u have also any idee how can i get specific name or url? from the json?

Zabkas Over a year ago

I am quite new here so i do not know so much about the rules here. do u want me to take it back?

Zabkas Over a year ago

The last code u update it is not giving me any respons?

Panos Kalatzantonakis Over a year ago

Glad I've helped

|

Phil · Accepted Answer · 2018-01-26 04:57:41Z

1

After json decode, it returns an array of objects, so try this:

$json=file_get_contents("http://2strok.com/radio.json");
$details=json_decode($json);

foreach ($details as $detail){
    echo $detail->name;
    echo "<br>";
} 
?>

answered Jan 26, 2018 at 4:57

Phil

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