如何覆盖由 jquery/javascript 设置的 css 高度？
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how to overwrite css height set by jquery/javascript?
提问by frequent
I have this:
我有这个：
var setHeight = $(this).outerHeight();
// returns e.g. 687
$("#someElement").css({'height': $setHeight+"px !important" });
// I want to override this jquery-set height
I'm not sure this is the right way... probably not, since it's not working.
我不确定这是正确的方法……可能不是，因为它不起作用。
Thanks for helping out!
感谢您的帮助！
采纳答案by Charlotte
setHeight, not $setHeight. and the !importantis unneeded.
setHeight，不是$setHeight。并且!important是不需要的。
回答by Rory McCrossan
Your variable name doesn't have a leading $. Also, the !importantflag will cause this not to work in Firefox, however as you're applying this style directly to the element, you shouldn't need it.
您的变量名称没有前导$. 此外，该!important标志将导致这在 Firefox 中不起作用，但是当您将此样式直接应用于元素时，您不应该需要它。
$("#someElement").css('height', setHeight + "px");
Also note that if you're only setting the element's height you can also just shorten this to a height()call:
另请注意，如果您只设置元素的高度，您也可以将其缩短为height()调用：
$("#someElement").height(setHeight);
回答by faino
Your variables don't match; remember that punctuation and proper spelling are important to calling the variable properly; try changing the second line to:
您的变量不匹配；请记住，标点符号和正确拼写对于正确调用变量很重要；尝试将第二行更改为：
$("#someElement").css('height',setHeight+'px !important');
回答by John Shepard
Take out dollar sign ;) 'setHeight'
取出美元符号 ;) 'setHeight'
回答by brenjt
That should totally work except that your variable is setHeightand you are trying to use $setHeight. Make sure they are the same. Make sure your selector is correct and obtaining the element. I believe that you don't even need the !important. The style tag should overwrite any .css definition unless you have !importantin it.
除了您的变量是setHeight并且您正在尝试使用$setHeight. 确保它们相同。确保您的选择器正确并获取元素。我相信你甚至不需要!important. style 标签应该覆盖任何 .css 定义，除非你有!important它。
