[Solved] Why the prevoius value still exists after reinitialaize variable – PHP

Question

$carname; does not in fact do anything. It’s neither using the variable for any operation, nor is it assigning anything to it. In PHP variables are “initialised” by assigning something to them. If anything, $carname; is the same as echo $carname; without the echo, it’s just a NOOP. In fact, the first line doesn’t do anything either, the first var_dump triggers an Undefined variable carname notice.

Accepted Answer

$carname; does not in fact do anything. It’s neither using the variable for any operation, nor is it assigning anything to it. In PHP variables are “initialised” by assigning something to them. If anything, $carname; is the same as echo $carname; without the echo, it’s just a NOOP. In fact, the first line doesn’t do anything either, the first var_dump triggers an Undefined variable carname notice.