Re: questions regarding PHP and Forms

At 05:39 PM 1/28/2006, Paul Goepfert wrote:

Thanks for the help everyone. I have decided to use javascript to deal with my second problem. I understand that this is not a JavaScript fourm but I am having some trouble getting my JS function to work correctly. Here is my JS code. function getEquipment () {

        var mylist = document.getElementById("equip");
        if (mylist.options[mylist.selectedIndex].value=="other")
        {
                document.writeln("<tr>");
                document.writeln("      <td align="left">Other</td>");
                document.writeln("      <td><input type="text" name="other" size="20"></td>");
                document.writeln("</tr>");
        }

} This is my code for where the function is called <select name="equipment" size="1" id="equip" onChange="getEquipment()">

                            <option value="treadmill">Treadmil</option>
                            <option value="exercise bike">Exercise Bike</option>
                            <option value="other">Other</option>
                            <option value="rowing machine"

selected>Rowing Machine</option>

                          </select>

My problem is when the page loads in a web browser the Option "Other" is selected every time. From what you can see from my code "Rowing machine is the one that I want to be shown when the page loads. Anyway can anyonne tell me why my code is not executing correctly. I am not too familiar with java script. What I am trying to accomplish is if Other is selected then add a new table row below the drop down menu with the content that I have in the function. If no one can help with this would someone please direct me to a Java Script Forum. Thanks Paul

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