Printing JSON object using PHP

Question

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?

did you var_dump($json_output); to see what you get? And to see whether json_decode makes an array out of your data? — herrhansen
– herrhansen, Commented Jun 25, 2013 at 8:55
@quidage meant var_dump($json_output) - or alternatively simply print_r($json_output) — Aleks G
– Aleks G, Commented Jun 25, 2013 at 8:56
These links might help you: stackoverflow.com/questions/9138625/how-to-process-json-in-php stackoverflow.com/questions/9597941/how-to-echo-json-in-php — JunM
– JunM, Commented Jun 25, 2013 at 8:56
Fatal error: Call to undefined function var_dup() in /home/dddd.com/public_html/exp.php on line 5 — user123456789
– user123456789, Commented Jun 25, 2013 at 8:58

Mortgy · Accepted Answer · 2016-04-17 11:32:27Z

7

<?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)

edited Apr 17, 2016 at 11:32

answered Mar 18, 2016 at 13:06

Mortgy

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3 Comments

Hatchet Over a year ago

While this answer is probably correct and useful, it is preferred if you include some explanation along with it to explain how it helps to solve the problem. This becomes especially useful in the future, if there is a change (possibly unrelated) that causes it to stop working and users need to understand how it once worked. Thanks!

Mortgy Over a year ago

thanks for your comment, i will update my answer with more explanation :)

Vlax Over a year ago

@Mortgy this doesn't work with json_decode directly, you need something like this: json_encode(json_decode($json_ugly), JSON_PRETTY_PRINT); taken from: link

Sumit Gupta · Accepted Answer · 2013-06-25 08:57:46Z

4

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

answered Jun 25, 2013 at 8:57

Sumit Gupta

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Comments

dino.keco · Accepted Answer · 2013-06-25 09:02:01Z

0

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

answered Jun 25, 2013 at 9:02

dino.keco

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3 Comments

user123456789 Over a year ago

Warning: Invalid argument supplied for foreach() in /home/website/public_html/exp.php on line 6

dino.keco Over a year ago

There is a problem in your JSON which you download from that URL. Please double check if JSON is valid.

dino.keco Over a year ago

This codes works without problems:<?php $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)"}]'; $json_output = json_decode($json, true); foreach ($json_output as $trend){ echo $trend['text']."\n"; } ?>

Sandeep Kotian · Accepted Answer · 2013-09-20 15:52:33Z

0

$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.

answered Sep 20, 2013 at 15:52

Sandeep Kotian

1

Comments

Algamest · Accepted Answer · 2020-02-19 14:01:09Z

0

JSON_FORCE_OBJECT in your json call eg :

$obj = json_decode($data);

Instead write like this:

$obj = json_decode($data, JSON_FORCE_OBJECT);

edited Feb 19, 2020 at 14:01

Algamest

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answered Feb 19, 2020 at 13:32

Vishal Mhatre

1

1 Comment

Algamest Over a year ago

Hi Vishal. I've made a suggested edit to your answer to format the code. If possible could you also provide details as to what the fix is doing? Thanks!

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