die, exit doesnt die...sort of.

Anon

So...
'K, this really isn't a problem but just
something I'm curious about.

1) I'm new to this.
2) I'm writing a script and on line say, 20
I code this:

if ($this_var != $that_var){
     die ("HEY! this does NOT equal 
           that");

Now when I set the two vars to not equal it should die at line 23 or so.

BUT... While debugging (in an IDE program) I
got a parse error on line 87. I had neglected
to end a prior line with a semi-colon.

After fixing the semi-colon issue the script
ran fine and died appropriatly, but why did
it continue to run AFTER it should have died at line 23? I mean how did it find the parse
error on line 87 when it should have died on
line 23?

See what I mean?

Just curious-
Scott

scottbp

One easy way around this is to try:

if ($this_var != $that_var){
echo ("HEY! this does NOT equal that");
exit;
}

I hope this helps,

Scott

Anon

This is because PHP is a compiled language; what that means is that PHP reads in the entire program and parses it into a machine-usable form before it begins running it. The practical upshot is that it finds parse errors on line 87 before it even executes line 1 (let alone line 23).

A good thing - you don't want a script to be up and running for six months and then without warning a weird combination of circumstances throws up a parse error all because you forgot a semicolon!