CODEWITHSUNDEEP
phpjson
user123456789
user123456789

Reputation: 566

Printing JSON object using PHP

I have a JSON file and I would like to print that object in JSON:

JSON

[{"text": "Aachen, Germany - Aachen/Merzbruck (AAH)"}, {"text": "Aachen, Germany - Railway (ZIU)"}, {"text": "Aalborg, Denmark - Aalborg (AAL)"}, {"text": "Aalesund, Norway - Vigra (AES)"}, {"text": "Aarhus, Denmark - Aarhus Airport (AAR)"}, {"text": "Aarhus Limo, Denmark - Aarhus Limo (ZBU)"}, {"text": "Aasiaat, Greenland - Aasiaat (JEG)"}, {"text": "Abadan, Iran - Abadan (ABD)"}]

I have tried with following method,

<?php   
  $jsonurl='http://website.com/international.json'; 
  $json = file_get_contents($jsonurl,0,null,null);  
  $json_output = json_decode($json);        
  foreach ($json_output as $trend)  
  {         
   echo "{$trend->text}\n";     
  } 
?>

but it didn't work:

Fatal error: Call to undefined function var_dup() in /home/dddd.com/public_html/exp.php on line 5

Can anyone help me understand what I'm doing wrong?

Upvotes: 6

Views: 69877

Answers (5)

Vishal Mhatre
Vishal Mhatre

Reputation: 1

JSON_FORCE_OBJECT in your json call eg : 

$obj = json_decode($data);

Instead write like this:

$obj = json_decode($data, JSON_FORCE_OBJECT);

Upvotes: 0

Mortgy
Mortgy

Reputation: 563

<?php   

  $jsonurl='http://website.com/international.json'; 
  $json = file_get_contents($jsonurl,0,null,null);  
  $json_output = json_decode($json, JSON_PRETTY_PRINT); 
  echo $json_output;
?>

by using JSON_PRETTY_PRINT u transform your json to pretty formatting, using json_decode($json, true) doesn't reformat your json to PRETTY formatted output, also you don't have to run loop over all keys to export same JSON object again, you could use those constants also which could clean up your json object before exporting it.

json_encode($json, JSON_PRETTY_PRINT | JSON_UNESCAPED_UNICODE | JSON_UNESCAPED_SLASHES)

Upvotes: 7

Sandeep Kotian
Sandeep Kotian

Reputation: 1

$data=[{"text": "Aachen, Germany - Aachen/Merzbruck (AAH)"}, {"text": "Aachen, Germany - Railway (ZIU)"}, {"text": "Aalborg, Denmark - Aalborg (AAL)"}, {"text": "Aalesund, Norway - Vigra (AES)"}, {"text": "Aarhus, Denmark - Aarhus Airport (AAR)"}, {"text": "Aarhus Limo, Denmark - Aarhus Limo (ZBU)"}, {"text": "Aasiaat, Greenland - Aasiaat (JEG)"}, {"text": "Abadan, Iran - Abadan (ABD)"}]
$obj = json_decode($data);
$text = $obj[0]->text;

This will work.

Upvotes: 0

dino.keco
dino.keco

Reputation: 1401

Try this code:

<?php   
  $jsonurl='http://website.com/international.json'; 
  $json = file_get_contents($jsonurl,0,null,null);  
  $json_output = json_decode($json, true);        
  foreach ($json_output as $trend){         
   echo $trend['text']."\n";     
  } 
?>

Thanks, Dino

Upvotes: 0

Sumit Gupta
Sumit Gupta

Reputation: 2192

use

$json_output = json_decode($json, true);

by default json_decode give OBJECT type but you are trying to access it as Array, so passing true will return an array.

Read documentation : http://php.net/manual/en/function.json-decode.php

Upvotes: 4

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