resource id help!!!

[myraleen]

hello.
Help please!!!!

here is my code ( or the beginning of it )

$user_agency_id = mysql_query("SELECT agency_id FROM users WHERE username = 'usrname'");

Now, what it is suposed to do, is take the user agency id and input it into a table, like this:
$agency_id = "$user_agency_id";

Instead i get resource id # 4.

i know this has something to do with mysql_num_rows, but everything i try just returns the resource id number.

It is suposed to return the user that is logged on, return their agency id, please help, i've been working on this for days and days.

[NYSiRacer]

It's doing what it is supposed to do (The $user_agency_id you have is supposed to return a resource id #)
Look up [man]mysql_query/man & [man]mysql_fetch_array/man
I think this is what you want:

$user_agency_id = mysql_query("SELECT agency_id FROM users WHERE username = 'usrname'");
while ($row = mysql_fetch_array($user_agency_id)) {
$usrname = $row['username'];
}

[myraleen]

No, that is just giving me parse error's and doesn't like the {} i take those out and i get a variable error.

This is SOOOOOOO fustrating.

[myraleen]

ok
using this code:

$user_agency_id = mysql_query ("SELECT agency_id FROM users WHERE username = '$usrname')

$result = mysql_fetch_row($user_agency_id)
$agency_id = $result['agency_id']

i get this error when trying to get this to print to the screen
$agency_id = "$user_agency_id";

Parse error parse error, unexpected TVARIABLE in blah blah blah on line 199 which is the line for the
$agency_id = $result['agency_id']

that might explain more

[NYSiRacer]

Put this portion of code in a separate application:

$connection = mysql_connect('localhost', 'ur_username', 'ur_password');
mysql_select_db('ur_database',$connection);
$usrname = 'value here'; // enter a name that exists in your db
$result = mysql_query("select * from users where username = '$usrname'",$connection);
while ($row = mysql_fetch_array($result)) {
$myname = $row['username'];
}

// now echo $myname
echo $myname;

# no errors, right?
# if there're no errors here, then you've done something wrong in your page, not my codes =)

[myraleen]

of course i can get the username.
that's not the problem.
i can't get the agency id of the user that is logged in trying to add another user.

this is what i'm doin:

the page is for adding a user, guy would be the admin, going in, to add a user. But guy is only allowed to add users for his own division, or agency... therefore i want his division to appear in the table, where normally the user would have a selection box that is set to their personal agency id.
That's why i was using user_agency_id... so that it would be for guy alone, (or whoever was logged on) and not simply the user that is being created or edited.
~sigh~

[myraleen]

i got it! thanks. 🙂

[NYSiRacer]

$connection = mysql_connect('localhost', 'ur_username', 'ur_password');

mysql_select_db('ur_database',$connection);

$usrname = 'value here'; // enter a name that exists in your db

$result = mysql_query("select * from users where username = '$usrname'",$connection);

while ($row = mysql_fetch_array($result)) {
$agency_id = $row['agency_id'];
}

echo $agency_id;
// this will do