I have a PHP function, like this:
function($foo = 12345, $bar = false){}
What I want to do, is call this function with the default argument of $foo passed, but $bar set to true, like this (more or less)
function(DEFAULT_VALUE, true);
How do I do it? How do I pass an argument as a function's default value without knowing that value?
Thanks in advance!
This is not natively possible in PHP. There are workarounds like using arrays to pass all parameters instead of a row of arguments, but they have massive downsides.
The best manual workaround that I can think of is defining a constant with an arbitrary value that can't collide with a real value. For example, for a parameter that can never be -1:
define("DEFAULT_ARGUMENT", -1);
and test for that:
function($foo = DEFAULT_ARGUMENT, $bar = false){}
put them the other way round:
function($bar = false, $foo = 12345){}
function(true);
The usual approach to this is that if (is_null($foo)) the function replaces it with the default. Use null, empty string, etc. to "skip" arguments. This is how most built-in PHP functions that need to skip arguments do it.
<?php
function($foo = null, $bar = false)
{
if (is_null($foo))
{
$foo = 12345;
}
}
?>
NULL is actually NULL and not an empty string (a la ===)PHP can't do exactly that, so you'll have to work around it. Since you already need the function in its current form, just define another:
function blah_foo($bar)
{
blah(12345, $bar);
}
function blah($foo = 12345, $bar = false) { }
i think, it's better sorted based on the argument that the most frequently changed, $bar for example, we put it first,
function foo( $bar = false, $foo = 12345){
// blah blah...
}
so when you want to pass $bar to true, and $foo to 12345, you do this,
foo(true);
""andnullare identical