[英]jQuery selector, contains to equals
I have the folowing selector 我有下面的选择器
var likeComperssionOption = $('select[id*=ComparisionType]').eq(0)
.find("option:contains('LIKE')");
this checks for an option which contains the word 'like' right? 这会检查包含 “喜欢”这个词的选项吗?
How do i find an option which is exactly with the word 'like'? 我如何找到一个与“喜欢”这个词完全相同的选项?
Somthing like this: 像这样的东西:
var likeComperssionOption = $('select[id*=ComparisionType]').eq(0)
.find("option:equals('LIKE')");
只需选择从所有选项select
,并通过它们进行过滤text()
值:
var likeComperssionOption = $('select[id*=ComparisionType]:first option').filter(function() { return $(this).text() == "LIKE" });
If you want to select based on the value
attribute of the options, you need to use the attribute equals selector to select elements with a specific value for their value
attribute: 如果要根据选项的value
属性进行选择,则需要使用属性equals selector选择具有其value
属性的特定值的元素:
var likeComperssionOption = $('select[id*=ComparisionType]').eq(0)
.find("option[value='LIKE']")
Otherwise, to select based on the display text of the options, use: 否则,要根据选项的显示文本进行选择,请使用:
var likeComperssionOption = $('select[id*=ComparisionType]').eq(0)
.find("option").filter(function() {
return $(this).text() == 'LIKE';
});
Update: You might be having some problems with your initial selector, it looks very strange. 更新:你的初始选择器可能有些问题,看起来很奇怪。 You should probably change 你可能应该改变
$('select[id*=ComparisionType]').eq(0)
to something simple like 简单的事情
$('#ComparisionType')
The concept of this answer works fine, you can see it in action here . 这个答案的概念很好,你可以在这里看到它 。
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