How do I filter an array with only valid object in php

Question

i have an array like this which i got from parsing the html source:

Array
(
    [0] => 12345
    [1] => 54321
    [2] => 32546
    [3] => 98754
    [4] => 15867
    [5] => 75612
)

if I put it in a variable and run a foreach loop prefixing a url like so:

$x = Array;
foreach ($x as $y){
$r = 'http://example.com/' . $y;
echo $r . '<br>';
}

It will output like this:

http://example.com/12345
http://example.com/54321
http://example.com/32546
http://example.com/98754
http://example.com/15867
http://example.com/75612

and each of those output url if run in the browser will output an object EITHER like this:

{
   "key1": "value1",
   "key2": "value2",
   "key3": "value3"
}

OR like this:

{
   "error": {
      "errorMessage": "errorMessageValue",
      "errorType": "errorTypeValue"
   }
}

so my question is... how do I filter the Array in php so that it will only give me an Array that has a valid key/value pair and not the error object.

as suggested, I tried the following:

$x = Array;
foreach ($x as $y){
$link = 'http://example.com' . $y;
$json = file_get_contents($link);
$array = array_filter((array)json_decode($json), "is_scalar");
echo '<pre>';
echo json_encode($array) . '<br>';
echo '</pre>';
}

but it still output the same array and does not exclude the error object.

Please add addition information, where you produce that array? — user1646111
– user1646111, Commented Jul 27, 2013 at 0:35

bwoebi · Accepted Answer · 2013-07-27 00:42:20Z

4

$array = array_filter((array)json_decode($json), "is_scalar");

This will exclude all the objects from the array and you only have key => value pairs where value is nor an object nor an array.

edited Jul 27, 2013 at 0:42

answered Jul 27, 2013 at 0:36

bwoebi

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8 Comments

Martin Over a year ago

json_decode return an object unless the second argument is set to true

bwoebi Over a year ago

@Martin okay, added array cast ;-)

anagnam Over a year ago

@bwoebi using your code, i get Notice: Array to string conversion when trying to use file_get_contents to get $json to work then echoing $array

bwoebi Over a year ago

if you want to output my array / store it somewhere, you have to re-encode it with json_encode($array).

anagnam Over a year ago

@bwoebi it still output the same array as I had in the original. it did not exclude the error object.

|

Diego · Accepted Answer · 2013-07-27 01:21:15Z

0

Maybe I'm missing something, but can't you just check if the returned object contains an "error" entry and, if it does, skip it?

$x = Array;
foreach ($x as $y){
  $link = 'http://example.com' . $y;
  $json = file_get_contents($link);
  $returned_data = json_decode($json, true);

  // If returned data contains an "error" entry, just move to next one
  if(isset($returned_data['error'])) {
    continue;
  }
  echo '<pre>';
  echo json_encode($returned_data) . '<br>';
  echo '</pre>';
}

answered Jul 27, 2013 at 1:21

Diego

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