如何在字符串中找到第n次出现的字符? [英] How to find nth occurrence of character in a string?

问题描述

类似于发布的问题 here，正在查找寻求 Java 中的解决方案.

即如何从一个字符串中找到一个字符/字符串第n次出现的索引?

示例:/folder1/folder2/folder3/".在这种情况下，如果我要求第三次出现斜杠 (/)，它会出现在 folder3 之前，并且我希望返回这个索引位置.我的实际意图是从第 n 次出现的字符中将其子串起来.

Java API 中是否有任何方便/随时可用的方法，或者我们是否需要自己编写一个小逻辑来解决这个问题?

还有，

1. 我在 Apache Commons Lang 的 StringUtils，但我没有找到.
2. 正则表达式在这方面有帮助吗?

解决方案

如果您的项目已经依赖于 Apache Commons，您可以使用 StringUtils.ordinalIndexOf，否则，这是一个实现:

public static int ordinalIndexOf(String str, String substr, int n) {int pos = str.indexOf(substr);而 (--n > 0 && pos != -1)pos = str.indexOf(substr, pos + 1);返回位置；}

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这篇文章已被改写为文章这里.>

Similar to a question posted here, am looking for a solution in Java.

That is, how to find the index of nth occurrence of a character/string from a string?

Example: "/folder1/folder2/folder3/". In this case, if I ask for 3rd occurrence of slash (/), it appears before folder3, and I expect to return this index position. My actual intention is to substring it from nth occurrence of a character.

Is there any convenient/ready-to-use method available in Java API or do we need to write a small logic on our own to solve this?

Also,

1. I quickly searched whether any method is supported for this purpose at Apache Commons Lang's StringUtils, but I don't find any.
2. Can regular expressions help in this regard?

解决方案

If your project already depends on Apache Commons you can use StringUtils.ordinalIndexOf, otherwise, here's an implementation:

public static int ordinalIndexOf(String str, String substr, int n) {
    int pos = str.indexOf(substr);
    while (--n > 0 && pos != -1)
        pos = str.indexOf(substr, pos + 1);
    return pos;
}

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