while() meeting 2 expectations
Posted: Fri Apr 14, 2006 12:00 am
To start, I'm going to post the code of my issue.
Basically, my main focus is on this:
What I'm wanting it to do is only run through the while() loop while $count < $employees_left and as long as the $state of the employee is "1". For example, I have the following in my table:
In this case, I would only want it to echo out and and not echo out since Ben's state is "0".
I'm sure you guys can think of something to make this simple and make it work. I've been up too long. Thanks in advance.
Code: Select all
//Used for the employee menu and evaluations left to be completed
$employee_list="SELECT * FROM employees";
$total_employees=mysql_query($employee_list);
$employees_left=mysql_numrows($total_employees);
$eval_remaining=($employees_left - $eval_complete);
//Evaluates a variable for 'state' == 1
$state = mysql_query('SELECT state FROM employees');
$state_menu=mysql_result($state,"state");
if ($employees_left==0)
{
echo "<option>--No employees in database--</option>";
}
else
{
$count=0;
while ($count < $employees_left && $state_menu == '1') {
$name_menu=mysql_result($total_employees,$count,"name");
echo "<option>$name_menu</option>";
$count++;
}
}Code: Select all
while ($count < $employees_left && $state_menu == '1')Code: Select all
Name | State
------------------
Tristan | 1
Ben | 0
Aaron | 1Code: Select all
<option>Tristan</option>Code: Select all
<option>Aaron</option>Code: Select all
<option>Ben</option>I'm sure you guys can think of something to make this simple and make it work. I've been up too long. Thanks in advance.