JSON Parsing issue with PHP

Question

I'm attempting to pull in data from WoWProgress API. I want to parse the data online and just decode them directly while posting them for view. I'm still learning, but I seem to be having issues with this array. Would really need some help.

$json = file_get_contents("http://www.wowprogress.com/guild/eu/twisting-nether/hellenic%20horde/json_rank");
if($json == false)
{
throw new Exception("Failed To load infomation. Check setup options");
}
$result = json_decode($json, true);
echo "<pre>";
foreach ($result["realm_rank"] as $value) {
    print_r($value);
}
echo "</pre>";

But all i receive is "Invalid argument supplied for foreach()"

Would really love some help. Thanks in Advance!

Sorry for my english. English is not my native language.

Looks like $result["realm_rank"] is an integer value, not an array. — Rajdeep Paul
– Rajdeep Paul, Commented Jan 28, 2016 at 10:17

Pupil · Accepted Answer · 2016-01-28 10:33:27Z

2

The $result['realm_rank'] is a string, not an array.

Therefore, foreach on it is causing error.

This error causes when either the array is not set, or is blank or not an array.

<?php
$json = file_get_contents("http://www.wowprogress.com/guild/eu/twisting-nether/hellenic%20horde/json_rank");
if ($json == false) {
 throw new Exception("Failed To load infomation. Check setup options");
}
$result = json_decode($json, true);
 if (isset($result['realm_rank'])) {
  if (is_array($result['realm_rank'])) {
   foreach ($result["realm_rank"] as $value) {
    echo "<pre>";
    print_r($value);
    echo "</pre>";
   }
  } 
  else {
   echo $result['realm_rank'];
  } 
}
else {
 echo 'Unknown error';
}
?>

edited Jan 28, 2016 at 10:33

answered Jan 28, 2016 at 10:18

Pupil

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Comments

Franz Gleichmann · Accepted Answer · 2016-01-28 10:18:12Z

0

Your $result["realm_rank"] is not an array, but an integer. you can't iterate over an integer.

Try simply printing it:

print_r($result["realm_rank"]);

answered Jan 28, 2016 at 10:18

Franz Gleichmann

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Comments

devpro · Accepted Answer · 2016-01-28 10:18:16Z

0

Result of $result is this

print_r($result);

Array
(
    [score] => 750000
    [world_rank] => 2117
    [area_rank] => 1031
    [realm_rank] => 44
)

Invalid argument supplied for foreach() means $result["realm_rank"] is not an array.

For solution you can use values as:

foreach($result as $value) {
    echo $value. "<br/>";
}

Or just use

echo $result["realm_rank"];

answered Jan 28, 2016 at 10:18

devpro

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Comments

P. Jairaj · Accepted Answer · 2016-01-28 10:22:28Z

0

$result["realm_rank"] is not an array but a value. So you cant pass it in foreach.

Just do :

print_r($result["realm_rank"]);

I tried this and its working:

<?php

$json = file_get_contents("http://www.wowprogress.com/guild/eu/twisting-nether/hellenic%20horde/json_rank");
if($json == false)
{
throw new Exception("Failed To load infomation. Check setup options");
}
$result = json_decode($json, true);
echo "<pre>";
    print_r($result["realm_rank"]);
echo "</pre>";
?>

edited Jan 28, 2016 at 10:22

answered Jan 28, 2016 at 10:17

P. Jairaj

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1 Comment

Alex Papadopoulos Over a year ago

but then i get the table empty.

Professor Abronsius · Accepted Answer · 2016-01-28 10:27:38Z

0

Unless you specifically need to convert it to an array to loop through the values you could keep it in the default object format and access the values by name directly.

define('NL',PHP_EOL);/* only to prettify output */

$json = file_get_contents("http://www.wowprogress.com/guild/eu/twisting-nether/hellenic%20horde/json_rank");

if( !$json ) {
    throw new Exception("Failed To load infomation. Check setup options");
}

$json = json_decode( $json );

echo '<pre>',
    $json->score . NL, 
    $json->world_rank . NL, 
    $json->area_rank . NL, 
    $json->realm_rank . NL,
    '<pre>';

answered Jan 28, 2016 at 10:27

Professor Abronsius

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