asked on

PHP code to hide/show text box based on answer to previous drop down list

I am looking for php code that will enable me to hide/show a text box based on the users previous drop down list response.

(1) I want the "additional" text box to ONLY display if the user answers the drop down list with "yes"
(2) If the user answers "no" to the previous drop down list I don't want the "additional" text box displayed
(3) When the form is originally opened I dont want it to display the "additional" text box.

Thank you very much!

b0lsc0tt

Response on the previous page or the same page? If the same page and the page hasn't gone back to the server then PHP is useless here. You need to use clientside script, like Javascript.

PHP can be used to write the Javascript and send it to the browser. If you provide your PHP then I can help you do this in PHP but the html/javascript you need is like ...

<select onchange="document.getElementById('additionalBox').style.display = (this.value='yes')? '' : 'none';">

and the input would look like ...

<input type="text" name="additionalBox" id="additionalBox" style="display: none;">

That will change the box depending on what is selected and the box will not be shown by default (i.e. when page loads).

Let me know if you have a question or need more info.

bol

beatboxradio

Hello,

You can't do that in PHP unless you want to reload the page everytimes you select a value in the listbox. What I suggest is to use Javascript, which will allow you to do exactly what you need, but dynamically.

This code will do exactly what you need:
(Don't forget to put the javascript in the <head></head> )
#################################
<script language="JavaScript" type="text/javascript">
      function showTextBox(selectValue, objTextBox) {
            switch(selectValue) {
                  case 'yes':
                        objTextBox.style.visibility = 'visible';
                  break;

                  case 'no':
                        objTextBox.style.visibility = 'hidden';
                  break;
            }
      }
</script>

<form>
<select id="select_choice" name="select_choice" onchange="showTextBox(this.value, this.form.tb_choice)">
<option value="0">Choose</option>
<option value="yes">Yes</option>
<option value="no">No</option>
</select>
<input type="text" id="tb_choice" name="tb_choice" style="visibility:hidden" />
</form>

b0lsc0tt

I had a typo in part of my post. I left out an equal sign when I needed 2.

<select onchange="document.getElementById('additionalBox').style.display = (this.value=='yes')? '' : 'none';">

bol

cofcmgr

ASKER

I understand what you are saying about using javascript. The values that are captured in the form I need to pass forward to additional pages, and email. The PHP form that I would like to use in place of "additionalbox" is:

FormFieldLabel("Reseller (If Selected)", $_POST["Reseller"], "1", "1", $_POST["Target"]);
      echo "<br>  ";
            FormText("Reseller", $_POST["Reseller"], "50", "50");
      echo <<<END

Is that possible?

b0lsc0tt

What is FormFieldLabel and FormText? They aren't built-in PHP functions to my knowledge. If it is some function or routing that makes the fields then that function or routine would need to support this. I would need to know more about them and see code to know if that is possible. Most likely it is but the PHP you showed me isn't enough to know for sure or say how to do it.

bol

b0lsc0tt

By the way, if the show/hide is done without the page being "refreshed" then none of what we are doing will effect what is sent to other pages or emailed. It is all Javascript and would leave the info in the form intact and unchanged. The page is what is changed.

bol

cofcmgr

ASKER

Ok, I think I am getting closer to answer. I want to take the value that is entered into the input box that we created and becomes visible when the user selects "yes" and pass it to other forms.

ASKER CERTIFIED SOLUTION

b0lsc0tt

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basiclife

Just a slight correction:

if (isset($_POST['additionalBox']))

will always be true - it will always be set (assuming noone is messing with your form), it just may be empty

if (isset($_POST['additionalBox']) && trim($_POST['additionalBox']) != "")

is what you need :)

b0lsc0tt

I'm glad I could help. Thanks for the grade, the points and the fun question.

bol