JSON format in PHP

Question

{
  "required_items": [{
      "filename": "abcd",
      "no": "3"
    },
    {
      "filename": "abc",
      "no": "2"
    }
  ]
}

I am not getting the code of the JSON format in PHP - I want to insert the filename and no through a loop.

After your edit, your JSON string was still not valid. Also, please read meta.stackexchange.com/questions/22186/… — Marcel Korpel
– Marcel Korpel, Commented Jun 26, 2010 at 13:44
Yes, after I edited your source code. Before that, it wasn't. @Gordon: Nice tip, thanks! — Marcel Korpel
– Marcel Korpel, Commented Jun 26, 2010 at 14:27

Paul Dixon · Accepted Answer · 2010-06-26 13:31:15Z

2

Pass the JSON to json_decode and you'll end up with an regular PHP data structure you can operate on. Use var_dump to take a look at it. When you're done manipulating it, turn it back into JSON with json_encode

I don't know if it's typos on your part, but the format you pasted isn't valid JSON.

answered Jun 26, 2010 at 13:31

Paul Dixon

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Jon Benedicto · Accepted Answer · 2010-06-26 13:49:31Z

1

If you're asking how to generate the above JSON code in PHP, do something like this:

$object->required_items = array();

for( ... your loop here ... )
{
    $item->filename = 'filename';
    $item->no = 1;
    $object->required_items[] = $item;
}

$json = json_encode( $object );

answered Jun 26, 2010 at 13:49

Jon Benedicto

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1 Comment

JetPro Over a year ago

Good day, sir. I tried to use your code. But my output displays only one row from my database (that contains 3 rows). Can you please help me? Thanks. Here's my question. stackoverflow.com/questions/11806959/…

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