I have a mysql table t1 and t2
In t1 I have columns id,name,and t2_id
In t2 I have columns t2_id and job
Then I make a html/php form that is used to populate t1.
In the form there is field name where user type a name
There is also a dropdownlist populated from job in t2 (relation is t2_id).
I have a page where you can wiev all posts. After every posts I have a link called update. When submit update I send a $_GET to the next page wich presents a sticky form with the post related to the $_GET
Again the dropdownmeny for job shall be in the update form. I want it to be sticky and I can´t solve it myself. My code is:
echo "<select>";
while($rows=mysql_fetch_array($result)){
if ($x == $y)
{
echo "<option selected value=\"$x\";
echo "> $x</option>";
}
else
echo "<option value="$rows['t2_id']">$rows['t2_id']</option>";
}
echo "</select>";
Hi i've seen read the problem and i can provide you the solution for that as the expression is itself is not proper, Kindly revert for proper solution.
Regards