change a checkbox if another checkbox is checked

[Mindphaser]

I'm new to coding, but I'm getting through every problem I've encountered thanks to the people here and on other forums.

Now I'm wondering:

- I have to 2 different arrays of check boxes (150 boxes each)
- when a user clicks a card-shuffle button, it builds a deck based on the status of array #2
- checkbox #1 in array #1 corrosponds to checkbox #1 in array #2.

I need a statement that says " if checkbox #1 in array #1 ISN'T checked, then uncheck checkbox #1 in array #2."

And so on and so forth for the next checkboxes in the arrays.

Is this possible?

Here's what I've been trying: (maybe I'm accessing the form elements incorrectly? I don't get any errors from this code, but it doesn't work)

function unflip_cards(){

if (form.item_1_array_1.checked == true)
	form.item_1_array_2.checked == true;

if (form.item_2_array_1.checked == true)
	form.item_2_array_2.checked == true;

if (form.item_3_array_1.checked == true)
	form.item_3_array_2.checked == true;

if (form.item_4_array_1.checked == true)
	form.item_4_array_2.checked == true;

if (form.item_5_array_1.checked == true)
	form.item_5_array_2.checked == true;

if (form.item_6_array_1.checked == true)
	form.item_6_array_2.checked == true;

if (form.item_7_array_1.checked == true)
	form.item_7_array_2.checked == true;

if (form.item_8_array_1.checked == true)
	form.item_8_array_2.checked == true;

}

[Logic Ali]

I need a statement that says " if checkbox #1 in array #1 ISN'T checked, then uncheck checkbox #1 in array #2."

The syntax would be:

if ( !form.item_1_array_1.checked )
  form.item_1_array_2.checked = false;

To do it for all elements:

for( var i = 1, elem; ( elem = form[ "item_" + i +"_array_1" ] ); i++ )
  if( !elem.checked )
    form[ "item_" + i +"_array_2" ].checked = false;

That relies on all elements being numbered contiguously.

[Mindphaser]

Thanks Logic Ali, I was actually able to figure it all out before I came back on the forum here and saw your solution.

Exactly what I did and needed.

Sorry a newbie such as myself had to post a question with such easy and obvious solutions. My coding skills leave lots to be desired. I'm getting better, but all my strengths are with the artistic side of web design.

Thanks again LA, if I never figureded it out, this would have helped me