当数值存在时，php in array返回false

Question

Simple one here, but hurting ones head.

echo in_array('275', $searchMe);

is returning false. But if I print the array out and then search it manually using my web browser I am able to see that the value exists in the array.

[0] => ExtrasInfoType Object
                (
                    [Criteria] => 
                    [Code] => 275
                    [Type] => 15
                    [Name] => Pen
                )

Extra information. The array has been coverted from an object to an array using

$searchMe = (array) $object;

Would it be because the values are not quoted? I have tried using the following with the in_array function:

echo in_array('275', $searchMe); // returns false
echo in_array(275, $searchMe); // returns error (Notice: Object of class Extras could not be converted to int in)

var_dump of $searchMe

array
  'Extras' => 
    object(Extras)[6]
      public 'Extra' => 
        array
          0 => 
            object(ExtrasInfoType)[7]
              ...
          1 => 
            object(ExtrasInfoType)[17]
              ...
          2 => 
            object(ExtrasInfoType)[27]
              ...

Answer 1

One scenario I think that would be helpful.

If the array index that contains the value is 0, the returning value of in_array would be 0. If in our logic, 0 is considered as false, then there will be issues. Avoid this pattern in your code.

Answer 2

in_array can't see inside the ExtrasInfoType Object . Basically its comparing ExtrasInfoType Object to 275 , which in this case returns false .

Answer 3

You have an object there, use...

// Setup similar object
$searchMe = new stdClass;
$searchMe->Criteria = '';
$searchMe->Code = 275;
$searchMe->Type = 15;
$searchMe->Name = 'Pen';

var_dump(in_array('275', (array) $searchMe)); // bool(true)

CodePad .

When I cast ( (array) ) to an array, I got...

array(4) {
  ["Criteria"]=>
  string(0) ""
  ["Code"]=>
  int(275)
  ["Type"]=>
  int(15)
  ["Name"]=>
}