Undefined offset error in PHP from json response

Question

The code I wrote supposed to do two things: 1- shows the latest post from a twitter user 2- searches for a specific keyword. For the first part, it works perfectly but it fails in case of 'search' option by dumping this error message:

PHP Notice: Undefined offset:

I run the code with this format:

$ php script.php search keyword number

Here is the code:

    $result = file_get_contents("http://search.twitter.com/search.json?q=$argv[2]&rpp=$argv[3]&include_entities=true&result_type=mixed");
    $decoded = json_decode($result,true);
    for($j=0; $j<$argv[3]; $j++)
    {

    echo ($decoded[$j]['text']);



   }

that error is just a notice it wont stop the code from running, what it means though is that ur trying to use probably an array that hasnt been set, best thing would be print_r(); whatever arrays your using and see if the contain what they're suppose to — Breezer
– Breezer, Commented Apr 15, 2013 at 1:30
Probably the array doesn't contain anything. You should check with isset or something. What exact line is the error on? — edwardmp
– edwardmp, Commented Apr 15, 2013 at 1:30

edwardmp · Accepted Answer · 2013-04-15 01:52:50Z

3

I've checked the Twitter response, it doesn't match the format of the array you trying to access. You can't access the data through $decoded[$j]['text'] simply because that doesn't exist in the response.

What you mean to do is:

$decoded[results][$j]['text']

I've checked it and that appears to works fine.

Excerpt from the response (print_r format), you see why it happens.

Array (
  [completed_in] => 0.015
  [max_id] => 322818055665225728
  [max_id_str] => 322818055665225728
  [page] => 1
  [query] => testsearch
  [refresh_url] => ?since_id=322818055665225728&q=testsearch&result_type=mixed&include_entities=1
  [results] => Array
    (
      [0] => Array
        (
          [created_at] => Fri, 12 Apr 2013 21:06:36 +0000
          [entities] => Array

edited Apr 15, 2013 at 1:52

answered Apr 15, 2013 at 1:43

edwardmp

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Comments

Joseph Middleton · Accepted Answer · 2013-04-15 01:36:10Z

0

If the user inputs '10' but the query only returns 9 results, the 10th iteration of the for loop will return that notice (as $decoded[9] doesn't exist).

Perhaps try something like this

$limit = $argv[3];

if (count($decoded) < $limit)
{
    $limit = count($decoded);
}

for ($j = 0; $j < $limit; $j++)
{
...
}

answered Apr 15, 2013 at 1:36

Joseph Middleton

793 bronze badges

1 Comment

edwardmp Over a year ago

Probably true, but it is not the root cause of the problem here. He should implement something like this, though.

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