为什么这个style.display不起作用？

Question

Why won't this work? It should check the element for display css style.

if (byId("annonsera_price").style.display=='inline' || byId("annonsera_price").style.display=='block'){
alert ("Hello");
}

function byId(x){
        return document.getElementById(x);
}

update2:

<input style="width:100px;" type="text" name="annonsera_price" id="annonsera_price">

I haven't set it with css either!

Answer 1

If the style is set via CSS or is a default style, you need to get the computed style of the element:

var el    = document.getElementById("annonsera_price");

// currentStyle for IE, getComputedStyle for standards-compliant browsers
var style = el.currentStyle || window.getComputedStyle(el);

if (style.display == "inline" || style.display == "block")
    alert("Hello");

Answer 2

First try alerting just the style.display and see what it contains:

var el = document.getElementById("annonsera_price");
alert(el.style.display);

style only references what is in the style attribute (or set through JS on the style property). It's possible that the display is set through the class, and therefore el.style.display won't show anything.

[ Update ]:Given the update, the reason no alert happens is because style.display will be "", not "inline" or "block". It won't be "inline" or "block" until you set it to something, either in the style attribute or by setting .style.display .