[RESOLVED] session variable as selected in a listbox

[boyangaZ]

hello im a new member here at PHPBuilder...i have read many topics because it helped me alot...just want to need help or advices for my problem...

here is the scenario :

1. admin
2. user
3. if admin logs in, items in a list box shows all
4. if user logs in and if the user is one in the list, his/her name WILL ONLY SHOW and selected

here's the working listbox/dropdown i've used for admin...

$content.="<td><select name='aDMIN' onChange='this.form.submit();'>";
$userRs = mysql_query("SELECT * FROM table');
while ($user_row = mysql_fetch_object($userRs)){
$evalid = $user_row->id;
if ($user_row->lastName){
if ($user_row->middleName){$middleName=$user_row->middleName;}
$fullName = ucfirst($user_row->lastName).", ".ucfirst($user_row->firstName)." ".ucfirst($middleName);
if ($evaluator == $user_row->id){$content.="<option value='$user_row->id' selected> $fullName </option>";}
else $content.="<option value='$user_row->id'> $fullName </option>";

i want for the user....if someone can help....

[bradgrafelman]

Welcome to PHPBuilder! When posting PHP code, please use the board's [noparse]

..

[/noparse] bbcode tags as they make your code much easier to read and analyze.

As for your issue... how do you keep track of whether a "user" or "admin" is logged in? How do you know which "user" is logged in? If you can answer those questions, then it sounds like all you need to do is add a WHERE clause to your SELECT statement such that only the current user's row is selected.

[boyangaZ]

oops sorry for not using the boards bbcode tag....my apologies

first, thanks for replying.....by means of their roleId...i've already done that, but still my dropdown/list box does not change even i use the account of a 'user' because it always displays the full list...i've tried reading about 'sessions' but i dont know or i cant run...

maybe can you provide me a sample code for session using the above scenario...

[boyangaZ]

please, anyone....

[boyangaZ]

FROM this post -- > http://board.phpbuilder.com/showthread.php?10390331-session-variable-as-selected-in-a-listbox&p=11031597#post11031597

hello im a new member here at PHPBuilder...i have read many topics because it helped me alot...just want to need help or advices for my problem...

here is the scenario :

1. admin
2. user
3. if admin logs in, items in a list box shows all
4. if user logs in and if the user is one in the list, his/her name WILL ONLY SHOW and selected

here's the working listbox/dropdown i've used for admin...

$content.="<td><select name='aDMIN' onChange='this.form.submit();'>";
$userRs = mysql_query("SELECT * FROM table');
while ($user_row = mysql_fetch_object($userRs)){
$evalid = $user_row->id;
if ($user_row->lastName){
if ($user_row->middleName){$middleName=$user_row->middleName;}
$fullName = ucfirst($user_row->lastName).", ".ucfirst($user_row->firstName)." ".ucfirst($middleName);
if ($evaluator == $user_row->id){$content.="<option value='$user_row->id' selected> $fullName </option>";}
else $content.="<option value='$user_row->id'> $fullName </option>";

i want for the user....if someone can help....

:help:

[boyangaZ]

boyangaZ;11031597 wrote:

please, anyone....

woohoo, SOLVED BY ME...thanks