如果以给定字符开始，则返回变量的剩余部分

Question

If the value of a variable (i.e. $var="^hello";) starts with a given character (i.e. ^), how could I return the remainder of that variable (i.e. hello), and returnFALSE (or NULL or 0 if easier) if it doesn't start with the given character? Note that there is no guarantee that the variable will be a string. I tried the following, but it results in errors if the value is NULL.

$var='^hello'; //Works
$var=123; //Works
$var=NULL; //Doesn't work
echo ($var[0]=='^')?substr($var,1):false;
echo($new);

Answer 1

What you've done actually works, but when you echo false, nothing will appear on the screen, so just wrap false with double quotes to display "false" on the screen

$var='^hello'; //Works
$var=123; //Works
echo ($var[0]=='^')?substr($var,1):"false"; // Add double quotes to false

Live demo

EDIT. Note that this will not work if $var is set to NULL. To prevent an error, just change ($var[0]=='^') to ($var && $var[0]=='^')