Session IDs (capturing the uid)

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Session IDs (capturing the uid)

Wednesday, November 14 2001 @ 11:26 am EST
Contributed by: Anonymous
Views: 4,954

I'm trying to add some new features to my geeklog, but I can't seem to find out where the session starts or how to capture the uid. Can you point me in the right direction?

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Well...

Authored by: Tony on Thursday, November 15 2001 @ 08:49 pm EST

First, what version of Geeklog are you using? I\'m assuming 1.2.5-1. If so, there is a global array that gets populated with each page called $USER...that\'s assuming the user is logged in. If they aren\'t logged in the $USER is empty.

If you want to access the user ID from a function then here you do the following:

function foo() {
global $USER;

echo \"The user id is \" . $USER[\'uid\'];
}

That would print the user id for the current user. If you want it to be a bit more elegant and handle anonymous users then:

function foo() {
$global $USER;

if (!empty($USER[\'uid\']) {
$uid = $USER[\'uid\'];
} else {
$uid = 1; // 1 is the uid for Anonymous
}
echo \"The user id is \" . $uid;
}

Does that help?

---
The reason people blame things on previous generations is that there's only one other choice.

Cheers..

Authored by: Anonymous (Anonymous User) on Thursday, November 15 2001 @ 09:32 pm EST

Tony thanks. That did the trick.

Matt

Little help..

Authored by: Anonymous (Anonymous User) on Thursday, November 15 2001 @ 09:54 pm EST

Tony I captured the uid as you pointed out, but the insert into the table isn't doing as it's told.
I realize this isn't php builder, but the insert looks like the following. $sql = "INSERT INTO imagesubmission (uid,image1_name,) VALUES('$uid','$image1_name)";
now I know this should insert the uid into the table field, but maybe there's something more I'm missing relative to this.
Matt

Little help..

Authored by: Tony on Sunday, November 18 2001 @ 02:28 pm EST

Did you make sure that you declared $USER (or $_USER in 1.3) to be global? If so, I\'d recommend using errorlog (or COM_errorLog in 1.3) to print out debug messages.

Actually, it looks as though you are making a custom hack. A good idea is to declare a global debug variable and check that to see if you should print out debug messages:

$_<libname>_DEBUG = false;

then in your functions:

function foo() {
global $_<libname>_DEBUG;

if ($_<libname>_DEBUG) {
COM_errorLog(\"<some message or variable\", 1);
}

This will print your message to error.log. That way when your or someone else wants to debug the code they just set $_<libname>_DEBUG = true and wala....you get debug messages.

Replace <libname> with a logical name for your code library. If you are working on an image library, for example, use IMG for your lib name. Then your debug variable would be $_IMG_DEBUG.

Make sense?

---
The reason people blame things on previous generations is that there's only one other choice.

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