Hi,

I've currently got this line:


to show how many records there are when you first log on. As my database is still fairly small, I was wondering if there would be a way to also greet them with the number of registered users (about 4 times the number as there are records -- looks much more impressive)?

Thanks,
Dan

You can use either

$user_count = $user_count--;

or

$user_count = $user_count-1;

or

$user_count = --$user_count;

or

--$user_count;

or (probably)

$user_count--;

Lots of options. Much of Perl is whatever you prefer.

I'm really grateful for Adam's input. I don't mind at all when he answers things for me. Gives me more time to do other stuff.

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JPD

You're getting the count by opening the .db file and putting the contents into an array called @lines, then using
$count = scalar(@lines);
right?

You can do the same thing with the password file. Just copy the code for counting the number of records and use $auth_pw_file instead of $db_file_name. Change
$count to something else -- $user_count would work. Then your printout would be


You may need to open your password file and take out the first line, though. There's a comment line at the beginning of the file. Then again, leaving it in will only inflate your user count by 1.


------------------
JPD

Or if you want to leave it in there, just after you get the number, go;

$user_count = $user_count--;

No reason to leave it in there though...



adam

[This message has been edited by dahamsta (edited May 29, 1999).]

Thank you so much for the quick help. That was sure easy! Carol, you were correct as to how I was getting the records count. I took Adam's advice and subtracted one, as I sort of like having the comment line in the password file to remind me what the fields signify.

Adam, I'm assuming you meant to write
$user_count = $user_count-1;
? That's what worked for me.

By the way, me thinks Adam is subtly pushing up on your territory, Carol. His response time is within single digit minutes of yours. Is there a moderator position up for grabs?

Dan

Hmm, that's odd. The first method ( $user_count--; ) didn't work for me. No errors, but it didn't subtract one. One of these days I need to get a book and try to get a grasp on this little Perl language...

Dan

[This message has been edited by Dan Kaplan (edited May 29, 1999).]

This is a little explanation from "Programming Perl."

"If you place one of the auto operators before the variable, it is known as a pre-incremented (pre-decremented) variable. Its value will be changed before it is referenced. If it is placed after the variable, it is known as a post-increment(post-decrement) variable and is changed after it is used."

Since you were referencing the same variable, I thought it would work to have the decrement operator after the variable. I guess not.

You should be able to use

--$user_count;

though, which is about as short a command as you can get.


------------------
JPD

Well, perlop says:


I'm a bit fuzzy headed this morning, so I'm not sure what that means...



adam

Hi folks,

I'm not going to go into it now, because I'm very (very) drunk, but I can tell you that:

$var--;

and:

--$var;

...are quite different. From what I understand, the way to decrement a variable is:

$var--;

I shall do my best to study this in the intervening moments. Now, if you'll pardon moi, we have a grand prix commencing in a few short hours, and I've been told (quite reliably) that I require (at least) two hours sleep beforehand. Mammy's huh?

adam

(why am i here?)

Maybe an example will help. Again, from "Programming Perl":


Which is still unclear in the context of
$user_count = $user_count--;

But you can use either

$user_count--;

or

--$user_count;

Same result.

------------------
JPD