为什么style.display没办法工作?
问题说明
为什么这行不通?应该检查display css样式的元素.
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 type="text" name="annonsera_price" id="annonsera_price">
我也没有用CSS设置它！
I haven't set it with css either!
正确答案
首先尝试仅警告style.display并查看其包含的内容:
First try alerting just the style.display and see what it contains:
var el = document.getElementById("annonsera_price");
alert(el.style.display);
style仅引用样式属性中的内容(或通过样式属性上的JS设置).显示可能是通过该类设置的，因此el.style.display将不会显示任何内容.
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.
[更新]:鉴于更新，没有警报发生的原因是style.display将是"，而不是内联"或阻止".除非您在style属性中或通过设置.style.display对其进行设置，否则它不会是内联"或阻止".
[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.
这篇好文章是转载于：学新通技术网
