[英]CSS selector select some part of div class name
Select one part of text in div class: 在div类中选择文本的一部分:
<div class="enabled_disabled disabled">
<div class="enabled_disabled">
<div class="enabled_disabled">
<div class="enabled_disabled">
<div class="enabled_disabled disabled">
I have those div tags, is there any xpath syntax or fizzler CSS selectors syntax to select just those div tags which have enabled_disabled only (the 3 in the middle)? 我有那些div标签,是否有任何xpath语法或fizzler CSS选择器语法只选择那些只有enabled_disabled的 div标签(中间的3)? not those with enabled_disabled disabled
而不是那些禁用enabled_disabled的人
var html = new HtmlDocument();
html.LoadHtml(getitems);
var doc = html.DocumentNode;
var items = (from r in doc.QuerySelectorAll("div.enabled_disabled:not(.disabled)")
let Name = r.InnerHtml//QuerySelector("div.enabled_disabled div.title_bar div.rate_title span.name").InnerText.CleanInnerText()
select new {
CName = Name
}).ToArray();
To select an element with only the class enabled_disabled
, you would use: 要选择仅具有
enabled_disabled
类的enabled_disabled
,您将使用:
[class='enabled_disabled']
Using a :not
selector is not available in vanilla Fizzler , but can be used if you grab FizzlerEx 使用
:not
选择器在vanilla Fizzler中不可用 ,但是如果你抓住FizzlerEx就可以使用它
To select an element with only the class enabled_disabled
, you would use: 要选择仅具有
enabled_disabled
类的enabled_disabled
,您将使用:
div[@class='enabled_disabled']
If the div's assigned classes starts with enabled_disabled
: 如果div的指定类以
enabled_disabled
:
div[class^=enabled_disabled ]
If the div's assigned classes contains enabled_disabled
如果div的已分配类包含
enabled_disabled
div[class*=enabled_disabled ]
If the div only has the class enabled_disabled
如果div 只有类
enabled_disabled
div[class=enabled_disabled ]
If the div has the class enabled_disabled
and not the class disabled
如果div 具有类
enabled_disabled
而不 disabled
该类
div.enabled_disabled:not(.disabled)
Given the HTML you list in your question, either of the last two will work for you. 鉴于您在问题中列出的HTML,最后两个中的任何一个都适合您。
more on attribute selectors from MDN , and, more on :not() 更多关于来自MDN的属性选择器 ,以及更多关于:not()
You could use this selector that will match the class and will avoid the other. 您可以使用与该类匹配的此选择器,并避免使用另一个。
$(".enabled_disabled:not('.disabled')");
and you can take out contents out of $()
你可以从
$()
取出内容
and it is valid css selector 它是有效的CSS选择器
.enabled_disabled:not(.disabled)
Use not()
selector in css.The :not pseudo-class
represents an element that is not represented by its argument. 在css中使用
not()
选择器。 :not pseudo-class
表示不由其参数表示的元素。
.enabled_disabled:not(.disabled){
}
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