How to skip the second part (JavaScript if statement) if the first part is false

A common idiom is to use an operator &&

like this

var errorMsg = window.content.document.getElementById("error-msg");

if (errorMsg && errorMsg.offsetParent) {
    ...
}

      

        
        
        
      

    

Here JavaScript will evaluate first errorMsg

, and if it's Truthy then it will evaluate the part errorMsg.offsetParent

. The condition will be met only if both expressions in &&

are Truthy.

Note. ... Truthy's evaluation will return false

if the expression being tested 0

, false

etc. (see list of Falsy values here ). So, if you want to check if they are not null

, just write it explicitly like this

if (errorMsg !== null && errorMsg.offsetParent !== null) {
    ...
}

      

        
        
        
      

    

On the other hand, the operator ||

will evaluate the second operator only if the first expression is Falsy. In your case, if

(window.content.document.getElementById("error-msg") != null) 

      

        
        
        
      

    

is true

, it means it getElementById("error-msg")

returns null

. Since the first expression evaluates to Truthy, it evaluates to another expression and effectively tries to check

null.offsetParent !== null

      

        
        
        
      

    

This is why it fails.