document.getelementbyId will return null if element is not defined?

问题

In my code, I see this:

if (document.getElementById('xx') !=null) {
    //do stuff
}

if xx element is not defined, will this evaluate to true or false?

Should I write:

if (document.getElementById('xx'))

to be safe?

回答1:

console.log(document.getElementById('xx') ) evaluates to null.

document.getElementById('xx') !=null evaluates to false

You should use document.getElementById('xx') !== null as it is a stronger equality check.

回答2:

getElementById is defined by DOM Level 1 HTML to return null in the case no element is matched.

!==null is the most explicit form of the check, and probably the best, but there is no non-null falsy value that getElementById can return - you can only get null or an always-truthy Element object. So there's no practical difference here between !==null, !=null or the looser if (document.getElementById('xx')).

回答3:

Yes it will return null if it's not present you can try this below in the demo. Both will return true. The first elements exists the second doesn't.

Demo

Html

<div id="xx"></div>

Javascript:

   if (document.getElementById('xx') !=null) 
     console.log('it exists!');

   if (document.getElementById('xxThisisNotAnElementOnThePage') ==null) 
     console.log('does not exist!');

来源：https://stackoverflow.com/questions/15666163/document-getelementbyid-will-return-null-if-element-is-not-defined