Question

使用包含方法，我们为所有这些日志获取true：

console.log("https://example.com/test/media/instructions/listen_again_long/1.mp3".includes('listen_again_long')); // true
console.log("https://example.com/test/media/instructions/listen_again_long/2.mp3".includes('listen_again')); // true
console.log("https://example.com/test/media/instructions/listen_again_long/3.mp3".includes('listen')); // true

但是我们知道只有第一个日志应该返回true，因为在较长的字符串中恰好有listen_again_long。

如果我们认为此部分固定：https://example.com/test/media/instructions/

我们如何只返回true的第一个日志和false的其余日志？

Answer 1

您实际上正在寻找用/括起来的某个字符串，因此一种选择是简单地在传递给String.prototype.includes()的参数中同时包含两个/：

console.log("https://example.com/test/media/instructions/listen_again_long/1.mp3".includes('/listen_again_long/'));
console.log("https://example.com/test/media/instructions/listen_again_long/2.mp3".includes('/listen_again/'));
console.log("https://example.com/test/media/instructions/listen_again_long/3.mp3".includes('/listen/'));

您还可以使用RegExps和RegExp.prototype.test()做同样的事情：

console.log(/\/listen_again_long\//.test("https://example.com/test/media/instructions/listen_again_long/1.mp3"));
console.log(/\/listen_again\//.test("https://example.com/test/media/instructions/listen_again_long/2.mp3"));
console.log(/\/listen\//.test("https://example.com/test/media/instructions/listen_again_long/3.mp3"));

在两种情况下，如果您要确保匹配不在其他地方发生，您可以将/listen_again_long/替换为整个内容：

"...".includes("https://example.com/test/media/instructions/listen_again_long/");

或者，使用RegExp：

/https:\/\/example.com\/test\/media\/instructions\/listen_again_long\//.test("...");

Answer 2

您必须提取要与参数进行比较的子字符串，然后进行直接的===比较。

var url = <your passed in mp3 file>;
var s = "https://example.com/test/media/instructions/"
var substring = url.substring(url.indexOf(s) + url.length);
substring = substring.substring(0, url.indexOf("/");

substring === "listen_again_long"

如果只有确切的包含，则返回true

2 个答案: