Displaying MySQL results with Drop Down Menus

[kjm7267]

Hello from Palo Alto, California.

Please, I'm stuck on something basic. I have a MySQL table with four records. Each record has two fields, gender and weight. Here's the data:

Mary: female, 150 pounds
Susan: female, 140 pounds
Jack: male, 140 pounds
Bob: male, 150 pounds

I want to code a web page that has two drop down menus -- one for gender, and one for weight -- and a "search" button. It produce a list of names that fit the entered criteria. For example, if "female" and "140" are entered, the result should be a one-name list of "Susan." The HTML part of the code would, I think, be something like the following, but I don't know what PHP code to wrap around it:

<FORM>
Find a
<select name="gender" size="1">
<option value> </option>
<option value="male">male</option>
<option value="female">female</option>
</select>
with a weight of
<select name="weight" size="1">
<option value> </option>
<option value="140">140</option>
<option value="150">150</option>
</select>
<INPUT TYPE="submit" value="Search"></FORM>

Please, any suggestions? Thank you!

[frozin]

This is horrible off-the-top-of-my-head code, but if youve done database requests before, you'll get the point.

if(isset($_POST['acres'])) { //if form was submitted
$gender = $_POST['acres']; //set variables for "acres" dropdown
$weight = $_POST['state']; //set variable for "state" dropdown

$result = "SELECT * FROM TABLENAME_HERE WHERE gender = $gender & weight = $weight"; //select only where those two fields match up

while (mysql_fetch_array(result)) { //will only loop through rows in which both of those feilds match.
name = row(name);
weight = row(weight);
gender = row(gender);
echo $name $gender $weight;
}