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up vote 1 down vote favorite 
<?php
    $a = $monitoring->getMonitoringServers();
    if (in_array("192.168.100.253", $a))
        echo "y";
    else
        echo "n";
?>

print_r($a) yields:

Array (
   [0] => stdClass Object ( [address] => 192.168.100.253 )
   [1] => stdClass Object ( [address] => 192.168.100.253 )
)

What's wrong with this code? The answer is always no!

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1  
Please improve the question title. It doesn't really say anything. –  Lightness Races in Orbit Jun 8 '11 at 11:12

4 Answers 4

up vote 0 down vote

The $a var is an object, not an array.

See http://www.php.net/manual/en/function.in-array.php#103983 for a function that also works on objects.

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No, it's an array of objects. –  DanielB Jun 8 '11 at 11:14
up vote 2 down vote

Your array is made of objects, not strings.

By doing in_array("192.168.100.253", $a) you are looking for the string "192.168.100.253" inside $a, and as you can see on the print_r - it's inside an object.

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up vote 1 down vote 
$flag = false;
foreach($a as $obj){
  if($obj->address == "192.168.100.253"){
     $flag = true;
     break;
  }
}

if($flag){
   echo 'Y';
}
else{
   echo 'N';
}

$a is array of stdObjects, and you are treating them as normal values.

You are required to use a foreach loop to iterate through each element of $a.

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up vote 0 down vote

its simple try dis

$a = $monitoring->getMonitoringServers();

if (!in_array("192.168.100.253", $a->address ) echo "y"; else echo "n"; ?> u l get output as n

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