PHP Display array element

Question

Have array

Array ( 
    [3] => 
    stdClass Object ( 
        [term_id] => 3 
        [name] => Lietuviu 
        [slug] => lietuviu 
        [term_group] => 0 
        [term_taxonomy_id] => 3 
        [taxonomy] => kalba 
        [description] => 
        [parent] => 0 
        [count] => 7 
        [object_id] => 135 
    ) 
)

want display: [name] => Lietuviu , try $var[3][name], but this not work

What is the output you're getting?

Aziz
– Aziz

2012-11-14 15:05:37 +00:00
Commented Nov 14, 2012 at 15:05 — Aziz
– Aziz, Commented Nov 14, 2012 at 15:05
$var[3] is an object, so $var[3]->name.

Ja͢ck
– Ja͢ck

2012-11-14 15:05:54 +00:00
Commented Nov 14, 2012 at 15:05 — Ja͢ck
– Ja͢ck, Commented Nov 14, 2012 at 15:05

svenbravo7 · Accepted Answer · 2012-11-14 15:06:48Z

1

That is because that Array is actually an object. Put $var into this function:

function object_to_array($data) 
    {
      if(is_array($data) || is_object($data))
      {
        $result = array(); 
        foreach($data as $key => $value)
        { 
          $result[$key] = object_to_array($value); 
        }
        return $result;
      }
      return $data;
    }

Like so:

$realArray = object_to_array($var);

answered Nov 14, 2012 at 15:06

svenbravo7

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2 Comments

Wizard Over a year ago

1 more question how to search values in this array, can I use function in_array ?

svenbravo7 Over a year ago

Yes you can, it just makes an array of the object you gave it. But I believe in_array doesnt work well with multi-dimensional arrays.

complex857 · Accepted Answer · 2012-11-14 15:07:34Z

1

The value under the index 3 is a stdClass object, you will have to use the arrow operator -> to get its values:

print $var[3]->name;

answered Nov 14, 2012 at 15:07

complex857

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Comments

Nelson Benítez León · Accepted Answer · 2012-11-14 15:09:09Z

0

As your $var[3] element is an object, you have to access its properties like this:

echo $var[3]->name;

answered Nov 14, 2012 at 15:09

Nelson Benítez León

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Comments

user1647708 · Accepted Answer · 2012-11-14 15:12:12Z

0

You could use object_to_array as defined or access the value in following way: $var[3]->name

answered Nov 14, 2012 at 15:12

user1647708

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