verify, if the string starts with given substring

Question

i have a string in $str variable.

how can i verify is it starts with some word?

---

example

$str = "http://somesite.com/somefolder/somefile.php";

when i wrote the following script

if(strpos($str, "http://") == '0') echo "yes"; //returns yes

BUT it returns yes even when i wrote

if(strpos($str, "other word here") == '0') echo "yes"; //returns yes too

i think it strpos returns zero if it can't find substring too(or empty value).

so, what can i do, if i want to verify the word, which in the start of string?(maybe i must use === in this case?)

Thanks

Answer 1

You should check with the identity operator (===), see the documentation.

Answer 2

You need to do:

if (strpos($str, "http://") === 0) echo "yes"

The === operator is a strict comparison that doesn't coerce types. If you use == then false, an empty string, null, 0, an empty array and a few other things will be equivalent.

See Type Juggling.

Answer 3

if(substr($str, 0, 7)=="http://") {
    echo("Statrs with http://");
}

Answer 4

There's a big red warning in the documentation about this:

This function may return Boolean FALSE, but may also return a non-Boolean value which evaluates to FALSE, such as 0 or "". Please read the section on Booleans for more information. Use the === operator for testing the return value of this function.

strpos may return 0 or false. 0 is equal to false (0 == false). It is not identical to false however, which you can test with 0 === false. So the correct test is if (strpos(...) === 0).

Be sure to read up on the difference, it's important: http://php.net/manual/en/language.operators.comparison.php

Answer 5

check with

if(strpos($str, "http://") === 0) echo "yes";

as == will turn positive for both false & 0 check the documentation