call a function by omitting function argument - php

Question

function givemeposts($user, $pos){
$sql = "select * from posts where user ='" . $user . "' and pos = '" . $pos . "' order by inde";
...
givemeposts('public', 'home01'); // this works as usual
Now I nedd to call the above function, but without $pos param, i.e. whatever $pos value is.
Something like:
givemeposts('public', whatever);
Any way to do this?

Answer 1

You can use default parameter :
function givemepost( $user , $pos=NULL ){
if($pos==NULL){
//some code
}lese{
$sql = "select * from posts where user ='" . $user . "' and pos = '" . $pos . "' order by inde";
}
}
This way you can call the function with one paramter, and set what the function have to do, inside the if branch.

Answer 2

In PHP you can set default values for arguments
The default value must be a constant expression, not (for example) a variable, a class member or a function call.
Note that when using default arguments, any defaults should be on the right side of any non-default arguments; otherwise, things will not work as expected.
If you do this, the default value for $pos will be '' (an empty string):
function givemeposts($user, $pos = ''){
Or you can something else than an empty string, whatever value you want:
function givemeposts($user, $pos = 'SomeValue'){
Then you can simply call your function like this:
givemeposts($user);

Answer 3

You can try this:
function givemeposts($user, $pos = '')
{
if(!empty($pos)) {
$pos = "AND pos = '" . $pos . "'";
}
$sql = "SELECT * FROM posts WHERE user ='" . $user . "' $pos ORDER BY inde";
}
givemeposts('public');