在最后一次出现角色之前删除所有内容

I'm trying to perform the following action on a string :

我正在尝试对字符串执行以下操作：

find the last occurrence of the character "/";
找到字符“/”的最后一个匹配项;
remove everything before that character;
删除该角色之前的一切;
return the remains of the string;
返回字符串的遗骸;

To be more explicit, let's say I have the following string :

更明确一点，假设我有以下字符串：

var string = "/Roland/index.php"; // Which is a result of window.location.pathname

Now what I need to extract out of it is everything but the actual page, something like this :

现在我需要提取的是除了实际页面之外的所有内容，如下所示：

var result = "index.php" // Which is what I need to be returned

Of course, that is just an example, because obviously I will have different pages, but the same principles apply.

当然，这只是一个例子，因为显然我会有不同的页面，但同样的原则适用。

I was wondering if someone could help me out with a solution for it. I tried the next actions but with no success :

我想知道是否有人可以帮我解决问题。我尝试了下一步行动但没有成功：

var location = window.location.pathname;
var result = location.substring(location.lastIndexOf["/"]);

5 个解决方案

#1

You have the right idea just replace the brackets with parentheses.

您有正确的想法只需用括号替换括号。

var string = "/Roland/index.php";
var result = string.substring(string.lastIndexOf("/") + 1);

Here is an example in jsfiddle and here is an explanation of the .lastIndexOf() method on the Mozilla Developer Network.

这是jsfiddle中的一个示例，这里是对Mozilla Developer Network上的.lastIndexOf（）方法的解释。

#2

Personally I'd use a regular expression:

就个人而言，我会使用正则表达式：

var result = string.replace(/^.*\/(.*)$/, "$1");

If you're familiar with regular expressions (and you should be if not :-) then it's not as alien-looking as it is when they're unfamiliar.

如果你熟悉正则表达式（如果不是，那么你应该是这样的:-)那么当它们不熟悉时，它并不像外星人一样。

The leading ^ forces this regular expression to "anchor" the match at the start of the string. The \/ matches a single / character (the \ is to keep the / from confusing the regular expression parser). Then (.*)$ matches everything else from the / to the end of the string. The initial .* will swallow up as much as it can, including / characters before the last one. The replacement text, "$1", is a special form that means "the contents of the first matched group". This regex has a group, formed by the parentheses around the last .* (in (.*)$). That's going to be all the stuff after the last /, so the overall result is that the whole string is replaced by just that stuff. (If the pattern doesn't match because there aren't any / characters, nothing will happen.)

前导^强制此正则表达式在字符串的开头“锚定”匹配。 \ /匹配单个/字符（\是为了防止混淆正则表达式解析器）。然后（。*）$匹配从/到字符串末尾的所有其他内容。最初的。*将尽可能多地吞噬，包括最后一个之前的/字符。替换文本“$ 1”是一种特殊形式，表示“第一个匹配组的内容”。这个正则表达式有一个组，由最后一个括号括起来。*（in（。*）$）。这将是最后一个/之后的所有内容，因此整体结果是整个字符串被这些东西取代。（如果模式不匹配，因为没有任何/字符，则不会发生任何事情。）

#3

Split the string into an array on / and .pop() off the last element. Note, that you will first need to strip off a trailing slash if there is one.

将字符串拆分为最后一个元素上的/和.pop（）上的数组。请注意，如果有斜杠，首先需要去掉尾部斜杠。

var locationstring = window.location.pathname;
// replace() the trailing / with nothing, split on the remaining /, and pop off the last one
console.log(locationstring.replace(/\/$/, "").split('/').pop());

If in the case of a URL like /path/stuff/here/ where you have the trailing /, if that case should return an empty string rather than here, modify the above to remove the .replace() from the call chain. I assumed you would want the last component regardless of a trailing slash, but may have incorrectly assumed.

如果在像/ path / stuff / here /这样的URL的情况下你有尾随/，如果那种情况应该返回一个空字符串而不是在这里，修改上面的内容以从调用链中删除.replace（）。我假设你想要最后一个组件而不管尾部斜杠，但可能错误地假设了。

console.log(locationstring.split('/').pop());

#4

    var result = /\/([^\/]*)$/.exec(location)[1];

//"remove-everything-before-the-last-occurrence-of-a-character#10767835"

Note: location here is the window.location, not your var location.

注意：这里的位置是window.location，而不是你的var位置。

#5

var string = "/Roland/index.php";
var result = string.substring(0, string.lastIndexOf("/") + 0);

#1