如何做将改变元素样式显示的Javascript函数？

Question

function burgerMenu(){
            //alert(document.getElementById("hiddenMenuUL").style.display);
            if(document.getElementById("hiddenMenuUL").style.display="none"){
                document.getElementById("hiddenMenuUL").style.display="block";
            }
            else if(document.getElementById("hiddenMenuUL").style.display="block"){
                document.getElementById("hiddenMenuUL").style.display="none";
            }
    }

I want to toggle the burger menu icon. So if I press the icon the burgerMenu() function will be toggle and verify if its display: none then change to display:block and vice versa. hiddenMenuUL is the for list of links. This line below is the button line in HTML:

<button class="btn" onclick="burgerMenu()">&#9776;</button>

My CSS for hiddenMenuUL

#hiddenMenuUL{
        display: none;
        list-style: none;
        width: 100%;
        justify-content: center;
    }

Answer 1

When you are writing your if, you are not actually checking if the display === "none", you are trying to set it to none.

What you should be doing looks like this:

function burgerMenu(){
            let menuStyle = document.getElementById("hiddenMenuUL").style.display;
            if(menuStyle === 'none'){
                document.getElementById("hiddenMenuUL").style.display="block";
            }
            else {
                document.getElementById("hiddenMenuUL").style.display="none";
            }
    }

Edit:

It seems display rule is defined in CSS, and isn't inline.

.style can only reach inline styles, for example:

<div id="hiddenMenuUL" style="display: none;"></div>

Adding 'display: none' inline like in the example above should be the fastest and easiest way to 'fix' it in your particular scenario, although it probably wouldn't be considered the best practice.