PHP If Statement not working correctly, not null

Question

I'm struggling to understand why my if statement below always results in false. I am creating a function which will test incoming connections to a script which will reject connections made by certain bots.

In my test below, on applying the if logic, I'm expecting a TRUE as both the array $value and $test value should match... resulting in a NOT NULL?

$bots = array(0 => "PaperLiBot", 1 => "TweetmemeBot", 2 => "Appsfirebot", 3 => "PycURL", 4 => "JS-Kit", 5 => "Python-urllib");

$test = strtolower("PaperLiBot");


foreach($bots as $value)
{   
    $i = strtolower(strpos($value, $test));

    if ($i != NULL)
    {
        echo "Bot is found";
        exit;
    }else
    {
        echo "not found";
    }

}

strotolower(strpos()) makes no sense. strpos() returns the position as an integer, or false. Did you mean it to say strtolower($value)? — Pekka
– Pekka, Commented Jun 28, 2011 at 21:52
BTW, if you want contiguous numeric keys starting at 0 (0, 1, 2, ...) in your array you can omit them: array("PaperLiBot", "TweetmemeBot", "Appsfirebot", "PycURL", "JS-Kit", "Python-urllib") — David Harkness
– David Harkness, Commented Jun 28, 2011 at 22:09

GWW · Accepted Answer · 2011-06-28 21:54:33Z

3

I think you are trying to accomplish this

foreach($bots as $value)
{   
    $i = strpos(strtolower($value), $test);

    if ($i !== false){
        echo "Bot is found";
        exit;
    }else{
        echo "not found";
    }

}

answered Jun 28, 2011 at 21:54

GWW

44.3k11 gold badges119 silver badges110 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Rufinus · Accepted Answer · 2011-06-28 22:06:58Z

1

you want to write:

if(stristr($value, $test)){
 // found
}else{
 // not found

}

edited Jun 28, 2011 at 22:06

answered Jun 28, 2011 at 21:54

Rufinus

30.9k7 gold badges71 silver badges85 bronze badges

1 Comment

GWW Over a year ago

That is incorrect if the position is 0 the if statement will fail.

Marc B · Accepted Answer · 2011-06-28 21:54:32Z

0

null in PHP is mutually type-castable to '0', '', ' ', etc... You need to use the strict comparisons to check for a 0-based array index:

if ($i !== NULL) { 
 ...
}

note the extra = in the inequality operator. It forces PHP to compare type AND value, not just value. When comparing value, PHP will typecast however it wants to in order to make the test work, which means null == 0 is true, but null === 0 is false.

answered Jun 28, 2011 at 21:54

Marc B

362k44 gold badges433 silver badges508 bronze badges

3 Comments

NikiC Over a year ago

strpos returns false, not null ;)

Marc B Over a year ago

true, but OP is explicitly checking for null for some reason.

Damo Over a year ago

I was looking for null by mistake. Changed to False which now works. Thanks

Fabrizio · Accepted Answer · 2011-06-28 21:56:03Z

0

the correct way to check if a variable is set to NULL is:

if(!is_null($var)){
    [my code]
}

the strpos returns a boolean, not a NULL value.

if you are unsure of the content of a variable, you can always debug it with a simple

var_dump($var);

answered Jun 28, 2011 at 21:56

Fabrizio

3,8062 gold badges31 silver badges34 bronze badges

Collectives™ on Stack Overflow

PHP If Statement not working correctly, not null

4 Answers 4

Comments

1 Comment

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

Comments

1 Comment

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related