jQuery 可见性属性问题
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Visibility attribute question
提问by Misha Moroshko
What is the difference between
之间有什么区别
$("#myid").attr("style", "visibility: hidden")
$("#myid").attr("style", "visibility: hidden")
and
和
$("#myid").css("visibility", "hidden")
$("#myid").css("visibility", "hidden")
?
?
回答by Nick Craver
Doing this:
这样做：
$("#myid").attr("style", "visibility: hidden")
Will leave onlythis style attribute, while doing this:
将仅保留此样式属性，同时执行以下操作：
$("#myid").css("visibility", "hidden")
Will add(or set) this style attribute.
将添加（或设置）此样式属性。
Here's an example, the first will alwaysresult in this:
这是一个例子，第一个总是会导致这个：
style="visibility: hidden;"
The second just adds visibilityso your style may now be:
第二个只是添加visibility所以你的风格现在可能是：
style="width: 50px; color: red; visibility: hidden;"
回答by Justin Niessner
Nothing. Just two ways to accomplish the same goal.
没有。只有两种方法可以实现相同的目标。
The first will overwrite any existing style settings. If you had:
第一个将覆盖任何现有的样式设置。如果你有：
<div style="font-weight: bold;" />
It would become:
它会变成：
<div sytle="visibility: hidden;" />
The second will add the visibility setting to the existing styles. So:
第二个将向现有样式添加可见性设置。所以：
<div style="font-weight: bold;" />
Woudl become:
会变成：
<div style="font-weight: bold; visibility: hidden;" />
If there's no style attribute already set, then the two will produce the same end result.
如果尚未设置样式属性，则两者将产生相同的最终结果。
回答by Josh Wright
There isn't really any difference.$.css() is just a shortcut method for accessing the css style attribute of a dom element.
真的没有什么区别。$.css() 只是访问 dom 元素的 css 样式属性的快捷方法。
EDIT: As justin pointed out, there isa difference in that the .attr() method will overwrite any existing style attributes.
编辑：贾斯汀指出的那样，是在该.attr（）方法将覆盖所有现有的样式属性的差异。
