How to populate a hidden form field with a value passed from another page

I'm using PHP/MySQL and DW CS4.
I am trying to obtain the external key for a table, and include it as a hidden field in a form for a second table.
The user selects a "need" from a list and is taken to a new page which displays the need selected in the prior page and a form the user can fill out with details of his offer, there should also be a hidden field in this form that contains the index to the needs table, this hidden field holds the external key. Most of the code is working, except for populating the hidden field with the external key. I have proven(by printing it to the screen) that I have obtained the external key and stored it in a variable ($saveNeedId) . What I'm unable to do is assign this variable to the hidden field in the form I'm about to store in a table. Sometimes I get zero and sometimes I get the index to the first need in the table. This ought to be simple but I can't get it to work, I must be missing something obvious - still very new to PHP.
Here's the code that sets up the variable and prints it to the screen for test purposes
      $saveNeedId = "-1";
          if (isset($_GET['needId'])) {
            $saveNeedId = $_GET['needId'];
            print $saveNeedId;
Here's the code that sets up the hidden fields in the form, the one I'm trying to set up is the first one, needId
     <input type="hidden" name="needId" value="<?php echo $row_rsNeedsUnmet['needId']; ?>" />
     <input type="hidden" name="offerId" value="" />
     <input type="hidden" name="MM_insert" value="form1" />
The page where the user sees the list of needs is here www.hollisterairshow.com/weneed.php
I'd really appreciate sone help with this, I've tried all combinations of double quotes, percent signs and nothing works...sigh.
Tony

Here's the code that sets up the variable and prints it to the screen for test purposes
      $saveNeedId = "-1";
          if (isset($_GET['needId'])) {
            $saveNeedId = $_GET['needId'];
            print $saveNeedId;
Here's the code that sets up the hidden fields in the form, the one I'm trying to set up is the first one, needId
     <input type="hidden" name="needId" value="<?php echo $row_rsNeedsUnmet['needId']; ?>" />
     <input type="hidden" name="offerId" value="" />
     <input type="hidden" name="MM_insert" value="form1" />
<input type="hidden" name="needId" value="<?php echo $_GET['needId']; ?>" />
I looked at your page. It looks like you figured it out.

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             if(source==scroller1)     textField1.setText(String.valueOf(fps));
             else if(source==scroller2)     textField2.setText(String.valueOf(fps));
// Property Change
    public void propertyChange(PropertyChangeEvent e) {
        if ("value".equals(e.getPropertyName())) {
            Number value = (Number)e.getNewValue();
            if (scroller1 != null && value != null) {
                scroller1.setValue(value.intValue());
             else if (scroller2 != null && value != null) {
                scroller2.setValue(value.intValue());
    // ACTION PERFORMED
    public void actionPerformed(ActionEvent event) {
    if (!textField1.isEditValid()) { //The text is invalid.
        Toolkit.getDefaultToolkit().beep();
        textField1.selectAll();
    } else try {                    //The text is valid,
        textField1.commitEdit();     //so use it.
    } catch (java.text.ParseException exc) { }
         if (!textField2.isEditValid()) { //The text is invalid.
        Toolkit.getDefaultToolkit().beep();
        textField2.selectAll();
    } else try {                    //The text is valid,
        textField2.commitEdit();     //so use it.
    } catch (java.text.ParseException exc) { }
...

if :p3_note_id is null
then
insert into notes (project_id, note, notes_month, notes_year) So, p3_note_id is NULL.
Another option is that you have a trigger on table NOTES that generates a new note_id even for an update.

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