Brandon Metcalf wrote:
> d == dev@archonet.com writes:
>
> d> Brandon Metcalf wrote:
> d> > Yep, it seems that's the problem. If I pass in $table and use a
> d> > lexical variable defined inside do_delete(), the problem goes away.
> d> > So, this is where my understanding of how triggers work lacks. For a
> d> > given session, each execution of a trigger isn't completely
> d> > independent?
>
> d> Nothing to do with triggers - it's all to do with your Perl code.
>
>
> I respectfully disagree because if I don't execute a DELETE on foo2 as
> shown in my original email, the problem doesn't occur.
Of course not.
> Somewhere in
> the trigger execution it's remembering the first table on which the
> trigger fired.
Yes. in your "sub do_delete" with it's local variable.
> So, the information about foo2 is coming from
> somewhere and it's in the Perl code.
Yes, your local copy of $table in do_delete.
> In other words, I performing two
> different DELETEs which cause two different invocations of the same
> trigger.
You've written your code such that do_delete has a local copy of $table.
In fact, the way it actually works iirc is that when you exit the
trigger function "my $table" goes out of scope and vanishes, but the
"$table" in do_delete doesn't vanish and persists from call to call. You
might call this a static variable in C terms.
> d> #!/usr/bin/perl
>
> d> sub foo {
> d> my $x = shift;
> d> print "foo x = $x\n";
> d> bar();
> d> return;
>
> d> sub bar {
> d> print "bar x = $x\n";
> d> }
> d> }
>
> d> foo(1);
> d> foo(2);
> d> exit;
This code mirrors _exactly_ what is happening with your trigger. On the
first call to foo $x is set to 1, on the second it's set to 2. That
doesn't affect "sub bar" though because its copy of $x is still the one
from the first call.
Maybe the following makes it clearer:
#!/usr/bin/perl
sub foo {
my $x = shift;
print "foo x = $x, ";
bar();
return;
sub bar {
print "bar x = $x\n";
$x--;
}
}
for my $i (1..5) { foo($i); }
exit;
$ ./perl_example.pl
foo x = 1, bar x = 1
foo x = 2, bar x = 0
foo x = 3, bar x = -1
foo x = 4, bar x = -2
foo x = 5, bar x = -3
The two $x variables go their separate ways and the one in "bar" doesn't
go out of scope at the end of the function.
--
Richard Huxton
Archonet Ltd