具有近似值的搜索属性[重复]

This question already has an answer here:

这个问题在这里已有答案:

• Can I use an attribute selector for CONTAINS in queryselector()? 1 answer

我可以在queryselector()中使用CONTAINS的属性选择器吗? 1个答案

I'm trying to find a JavaScript element ID and I only know the title.

我正在尝试查找JavaScript元素ID,我只知道标题。

var x = document.querySelectorAll('[title="My Workforce"]');

var id = x.getAttribute('id');

I don't know how to write the title so that it is any approximation of "My Workforce". I have tried using "*" (as in "*My Workforce*") but it didn't work. querySelectorAll() looks for an exact match.

我不知道如何写标题,以便它是“我的劳动力”的任何近似值。我尝试过使用“*”(如“* My Workforce *”),但它没有用。 querySelectorAll()查找完全匹配。

1 个解决方案

#1

You can use querySelector() instead. This allows the use of wildcards, like so:

您可以使用querySelector()代替。这允许使用通配符,如下所示:

https://jsfiddle.net/xtukey8o/

document.querySelector('[title*="My Workforce"]')