如何将字符串中的数字加1? [英] How to increment the number in a String by 1?

问题描述

我在字符串下面有一个-

I have a below String as -

String current_version = "v2";

这里的版本也可以是这些字符串之一- v3 或 v4 或 v10 或 v100 或 v500 或 v1000

And here version can be either of these string as well - v3 or v4 or v10 or v100 or v500 or v1000

现在，我希望将current_version从 v2 增加到 v3 .这意味着我总是需要将current_version增加1.

Now I am looking to increment the current_version from v2 to v3. Meaning I always need to increment the current_version by 1.

如果current_version是 v5 ，那么我将其递增到 v6 .

If current_version is v5, then I will increment it to v6.

我该如何使用字符串?

注意:-

我正在从某种方法获取 current_version 字符串值，该方法将始终仅向我返回此v2，v3，v4.

I am getting current_version string value from some method which will always return me like this v2, v3, v4 only..

推荐答案

如果您想将v后面的数字解析为int并将其递增1，只需通过对字符串进行子串化，将数字加1并将其串联即可.所得的递增数字返回为带"v"前缀的字符串:

If you want to parse the number after the v as an int and increment it by 1, just do so by substringing, adding 1 to the number and concatenating the resulting incremented number back into a string with "v" prepended:

    String version = "v1";
    String newVersion = "v" + (Integer.parseInt(version.substring(1,version.length()))+1);
    System.out.println(newVersion);

或者，您可以在 int 中跟踪当前版本，对其进行递增，并通过将int附加到"v"来构造新的 String ，而无需调用 substring().

Alternatively, you could keep track of the current version in an int, increment that and construct your new String by appending that int to a "v", without having to call substring().

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