A
Anonymous
Guest
Hi all,
this time the problem is abt transferring a var to a drop down list......it goes on something like this....
i wan to edit a project list by means of proj_id..its primary key...ok fine i did it...i am also transferring it....in a drop down list "in the other file" i have a list of projectleaders names in it........now their user_id and proj_leadid are equal.......
what i want is when i a passing the proj_id....to the next file the drop down list shud show the CORRESPONDING projectleader and after that the others in the drop down....
this below code just gives the user_id and the usernames this is in the "otherfile"......help me to show the CORRESPONDING projectleader
$sql="select user_name,user_id from mast_user";
$result=mysql_query($sql,$conn);
$user = "<select name='user3'\n";
while($row = mysql_fetch_array($result))
{
$user.="<option value='".$row["user_id"]."'>".$row["user_name"]."</option>";
}
$user.="</select>";
Sansoft
this time the problem is abt transferring a var to a drop down list......it goes on something like this....
i wan to edit a project list by means of proj_id..its primary key...ok fine i did it...i am also transferring it....in a drop down list "in the other file" i have a list of projectleaders names in it........now their user_id and proj_leadid are equal.......
what i want is when i a passing the proj_id....to the next file the drop down list shud show the CORRESPONDING projectleader and after that the others in the drop down....
this below code just gives the user_id and the usernames this is in the "otherfile"......help me to show the CORRESPONDING projectleader
$sql="select user_name,user_id from mast_user";
$result=mysql_query($sql,$conn);
$user = "<select name='user3'\n";
while($row = mysql_fetch_array($result))
{
$user.="<option value='".$row["user_id"]."'>".$row["user_name"]."</option>";
}
$user.="</select>";
Sansoft