PHP: name of variable being passed?

[justravis]

In this simple example, I would like the function to output, the name of the var, the letter 'x'

 

<?php

function out($var)
{
    echo "$var<br />";
    #echo $$var;
    #echo func_get_args($var);
    #$varlist=get_defined_vars();
    #echo $varlist[0];
}

$x=3;

out($x);

?>

 

Any ideas?

 

 

Thanks for yr time!

[Jessica]

Uhm, it will always be "var".  It is not x, because out takes the argument $var, not $x.

[gerkintrigg]

you could possibly do something with a FOR loop to get the name of the variable AND the name of the value.

[ToonMariner]

think you may have to use variable variables here.

 

<?php

function out($varname, $varavl)
{
    echo "$varname = $varval<br />";
}

$x = 'x';
$$x=3;

out($x,$$x);

?>

[justravis]

since the function cant even tell the name of the var being passed to it, how would a for loop inside the function make a difference?

 

 

THANK YOU.