javascript - 有没有办法验证 Sizzle 选择器？

Question

有没有办法在不运行 Sizzle 选择器的情况下验证（验证其构造是否正确）？

Answer 1

好吧，正如 Russ 所说，由于 Sizzle 解释了选择器，它无法在不评估它的情况下验证它。

但是，您可以捕获 Sizzle 抛出的异常以确定选择器是否有效：

function isSelectorValid(selector)
{
    try {
        $(selector);
    } catch (x) {
        return false;
    }
    return true;
}

您可以在此处测试此解决方案。

---

编辑：为了历史，我原来的（和过度设计的）答案是：

However, it's possible to temporarily override Sizzle's error management in order to extract a boolean value from the error status of its last parse operation. The following solution takes advantage of the fact that jQuery exposes Sizzle through $.find (so far):

function isSelectorValid(selector)
{
    var oldErrorMethod = $.find.error;
    try {
        $.find.error = function(msg) {
            valid = false;
            oldErrorMethod(msg);
        };
        $(selector);
        return true;
    } catch (x) {
        return false;
    } finally {
        $.find.error = oldErrorMethod;
    }
}

That can arguably be considered as a horrible hack, but it works: you can test it here.

Answer 2

不完全是，Sizzle 引擎没有被编译，所以检查选择器有效性的唯一方法是选择它。

但是，您可以执行以下操作：

var selector = ...construct your selector ...
if ($(selector).length > 0) {
 // it worked.
}