I should have been more exact in my original reply. The second query
isn't necessary. Try this:
File: login.php
# if $employee_1, query db workgroups table to check if $emp_login_id
belongs to any groups
$sql_2 = "SELECT * FROM workgroups WHERE emp_id='$emp_login_id'";
$result_2 = @mysql_query
ark) for all of your help.
O From Now 'Till Then,
\->Reginald Alex Mullin
/\ 212-894-1690
> -Original Message-
> From: Mullin, Reginald
> Sent: Monday, March 18, 2002 2:45 PM
> To: '[EMAIL PROTECTED]'
> Subject: FW: [PHP] Creating arrays using
On Tuesday 19 March 2002 03:44, Mullin, Reginald wrote:
> I've just added another record to the table. Now they're a total of 3
> records matching the "WHERE emp_id='$emp_login_id" criteria. When I
> "print_r(array_values($emp_login_grp_id));" I get the following values:
> Array ( [0] => 222 [1]
reason, it seems to be skipping
the first record.
O From Now 'Till Then,
\->Reginald Alex Mullin
/\ 212-894-1690
> -Original Message-
> From: Mullin, Reginald
> Sent: Monday, March 18, 2002 2:29 PM
> To: 'Mark Heintz PHP Mailing Lists'; [EMAIL
>Reginald Alex Mullin
/\ 212-894-1690
> -Original Message-
> From: Mark Heintz PHP Mailing Lists [SMTP:[EMAIL PROTECTED]]
> Sent: Monday, March 18, 2002 1:26 PM
> To: Mullin, Reginald
> Cc: [EMAIL PROTECTED]
> Subject: Re: [PHP] Cre
You have to call mysql_fetch_array for each record in your result set...
$emp_login_wkgrp_id = array ();
$emp_login_grp_id = array ();
$emp_login_role_id = array ();
$i = 0;
while($employee_2 = mysql_fetch_array($result_2)){
$emp_login_wkgrp_id[$i] = $employee_2["wkgrp_id"];
$emp_login_grp_i
Hi Guys,
I've been experiencing some problems when trying to build 3 arrays with the
ID values of all of the groups a user belongs to. (I then want to register
these arrays into the current session). The arrays only appear to be
getting the first value (group ID) instead of all of the values th
[posted and emailed]
On 12 Apr 2001 12:48:52 -0700, [EMAIL PROTECTED] (Jeffrey
Greer) wrote:
>
>// putting the values in an array is trivial, but you should do it
>like this
>// or you will confuse your numeric ids with the ids that php
>puts in an array automatically
Whoops. I was wrong a
[posted and emailed]
If you're looking for a simple way to work with databases I would use
phplib. Phplib is a db abstraction layer which supports over a dozen
databases including mysql. You can get the libraries at
http://phplib.netuse.de/download/phplib-7.2c.tar.gz You will need to
use "db_
Do you understand the options that the other people have explained?
-- Rodney
"Ashley M. Kirchner" wrote:
> "Rodney J. Woodruff" wrote:
>
> > http://www.php.net/manual/en/function.msql-fetch-array.php
>
> Okay, call me dense. I can't figure this out. This is what I'm trying to
> do:
>
>
On Wed, Apr 11, 2001 at 03:55:38PM -0600, Ashley M. Kirchner wrote:
> "Rodney J. Woodruff" wrote:
>
> > http://www.php.net/manual/en/function.msql-fetch-array.php
(snip)
> I need that result into the following:
> $items = array(0 => "Undefined", 1 => "Work", 2 => "Personal");
Here's a
"Rodney J. Woodruff" wrote:
> http://www.php.net/manual/en/function.msql-fetch-array.php
Okay, call me dense. I can't figure this out. This is what I'm trying to
do:
$sql = "select p_id, project from proj where uid=$uid";
$result = mysql_db_query($database,$sql);
(the resulti
try
";
print_r( $the_array_I_want );
echo "";
?>
Be aware that adding an element to $the_array_I_want will give it the next
sequential number which would look like an ID but have nothing to do with
the contents of your database.
Unless it is absolutely necessary I would reco
http://www.php.net/manual/en/function.msql-fetch-array.php
Hope this helps.
"Ashley M. Kirchner" wrote:
> I need to convert an MySQL result into an Array...somehow. The
> query returns the following:
>
> ++---+
> | ID | Project |
> ++---+
> | 1 |
I need to convert an MySQL result into an Array...somehow. The
query returns the following:
++---+
| ID | Project |
++---+
| 1 | Home |
| 2 | Work |
| 3 | Family |
| 4 |Misc. |
| . | ... |
| . |
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