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  1. #1
    SitePoint Enthusiast
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    May 2006
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    Displaying MySQL results with Drop Down Menus

    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!

  2. #2
    SitePoint Enthusiast
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    Apr 2005
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    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;
    }


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