Question : php drop down boxes

What I need is the following:

I have a php drop down list box with numeric items in it from 1 to 40, name "sel1"
When another text box (name: "intPce") is populated with a value which is > than 1, I need the drop down list box to show a value of 1 and the label 1 - the value is used in a calculation and the label is just a visual text in the box

I assume we can use php for this, but java will also be fine

Bill

Answer : php drop down boxes

>> Error: Expected ','
>> Error: Object Expected
is there any line numbers mentioned in the error console?
Take a look at your code carefully. Something is missing...
Probably you have some variables with one and the same name...?!
Change
function onChngIntPceAll() {
for (i=0; i<20; i++) {
onChngIntPce(i);
}
}
to
function onChngIntPceAll() {
  for (indx=1; indx < 22; indx++) onChngIntPce(indx);
}
Also place this debug msg:
alert('onChngIntPce invoked for : ' + idx); just after this line function onChngIntPce(idx) {
so your onChngIntPce(idx) function will look like:

function onChngIntPce(idx) {
alert('onChngIntPce invoked for : ' + idx); // remove this msg later
if (document.getElementById("intPce"+idx).value > 1) document.getElementById("sel"+idx).selectedIndex = 1;
calcVAT();
SumExVAT();
Totals();
proc();
}