Problem with if-condition

[BennySHS]

Hi there,

every few month I need a little javascript, and every time I get into problems :/
I hope you can explain this to me:

I want to write a small functions which set some elements visible / invisible, dependent if the related checkbox is checked or not.
Should be very simple, but sadly not for me.

The ideea:

function test()
{
if (document.getElementById('zuteiler').checked = true)
{
// set the elements visible
}
// hide the elements
}


But this doesnt work. With this function I event can't uncheck my checkbox even more...

It seems to me that javascript makes out of the comparison "if (document.getElementById('zuteiler').checked = true)" a command and set allways the chechbox to checked ??!

I hope you can help me once more,
thx a lot,
greets ben

 

[trollacious]

Your problem is that Javascript uses a DOUBLE equal sign in an if condition. Your statement SHOULD be:

if (document.getElementById('zuteiler').checked =[red]=[/red] true)

Lee

 

[BennySHS]

thx for your annotation, but that's not the reason :/

 

[theniteowl]

BennySHS
trollacious is correct that if you are using a single equal sign it will treat it as an assignment rather than a comparison. With two equal signs it will evaluate properly unless there are other problems with your code.

You do not even need the equal sign though.
(document.getElementById('zuteiler').checked) is a statement in itself that returns true or false based on the status of the checkbox.
if (document.getElementById('zuteiler').checked)
{
alert("It was checked");
}
else
{
alert("It was not checked");
}
The above code evaluates the field and responds correctly. It does nothing to SET the checked property, only to test it and respond.


Paranoid? ME?? WHO WANTS TO KNOW????