On Fri, 19 Jan 2001, Jason Jacobs wrote:
> I tried this, but it didn't work. I need the value inside the form still
> because the value of the select will be written to an array that gets passed
> to the next page (as hidden inputs), which will in turn be compiled into a
> value list for a db query. I tried making the name of the select be the
> array posistion (value[7] in this case), but when I changed values, IE
> errored and said that value.7 isn't an object. If I don't change the
> option, ie, I use the default select value, the code works. But as soon as
> I change the option, IE gives me an error and can't change the value. Would
> this work: make the onChange event trigger a page refresh to set the form
> values? Or will that simply draw the page again? Thanks.
>
> Here is the code I'm using...all of the php variables have values that work,
> the problem is in changing the value of the select
> -----BEGIN--------
> <script>
> <!--
> function changeValue(newValue) {
> document.addstuff.value[7].value = newValue;
Whoops. Try:
document.addstuff["value[7]"].value = newValue;
> }
> file://-->
> </script>
>
> echo "<td colspan=\"$numfields\"> Select which location you want this to
> apply to.<br><select name=\"pickloc\"
> onChange=\"changeValue(this.options[this.selectedIndex].value);\">";
>
> $i=0;
>
> while ($locidval = mysql_fetch_object($locinfo)){
> echo "<option value=\"$locidval->LocationID\">Location:
> $locidval->Area</option><br>\n";
> }//end while
>
> echo "</td></select><input type=\"hidden\"
> name=\"value[$locidnum]\"></tr><tr>\n";
>
> ------END-----------
>
--
Ignacio Vazquez-Abrams <[EMAIL PROTECTED]>
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