how to declare

elmo25

hello,
i would like to ask how to declare a value that read from a .txt file to become an int so that i can use it in my next function.
ex:
this is how i read from a file.

$fp = fopen( "file.txt" , 'r' );
$str = fread( $fp , filesize( "file.txt" ) );
$fp = fclose( $fp );

how to delare the item i read from it become an INT?
INT $ABC = $str ????

thank you.

LarsLai

hi!
in php you do not have to declare anything.
that is the good / bad (as you like) about php.... it is "typeless" (I do not know whether this is the right term in english))

Roger_Ramjet

PHP uses dynamic typing. If you want to force a var to a specific type you can eg. add it to zero 0, it will automatically then be a number but with the same value; concatenate with an empty string to force a string.

Shrike

Originally posted by Roger Ramjet
PHP uses dynamic typing. If you want to force a var to a specific type you can eg. add it to zero 0, it will automatically then be a number but with the same value; concatenate with an empty string to force a string.

Or do it the correct way :[

$var = "0";
print gettype((int)$var) . "\n";
$var = 0;
print gettype((string)$var) . "\n";

rehfeld

you can also use settype()


$foo = '5';

settype($foo, 'integer');

// $foo is now an int



// but i think this way of casting is easier, like shown above

$foo = '5';
$foo = (int) $foo;


// you can even do cool stuff like this, although this example is pretty useless

foreach ( (array) 'string' as $foo) {

}

// it will turn 'string' into a numerically indexed array, w/ 1 element


but usually you wont even need to cast, because this works


$a = '5';
$b = '10';
$c = $a + $b;

echo $c; // 15

elmo25

Thank you very much . It help me a lot.