Reputation: 21781
My goal just debug
function dbg($var){
echo "you have passed $var";
}
call dbg($test)
output:
you have passed test
call dbg("var")
output:
you have passed "var"
In php .anyone could help me to do that?
Upvotes: 0
Views: 199
Reputation: 168853
Well, the second case is fairly straightforward - you're passing a string and you want to display the string. No worries.
But for the first case, I'm afraid the answer is: No you can't.
Once inside the function, PHP doesn't know anything about the variable that was passed into it other than the value.
I can't really see that it would be of much value though. It would be trivial to change your code to pass in a name and a value -- ie something like this:
function dbg($name,$value) {
print "You passed $name, and the value was $value";
}
dbg('test',$test);
That's not really all that great either though -- you may as well just use print_r()
and friends.
If you really want more powerful debugging tools, you should look into xDebug. It's a proper debugging tool for PHP, which allows you to step through the code line-by-line, and see the contents of variables at any point during the program run (among many other good features). It also integrates nicely with several popular IDEs.
Upvotes: 1
Reputation: 146360
Try this if $var
is a global variable:
function dbg($var){
echo "you have passed {$GLOBALS[$var]}";
}
Upvotes: 2