Java 字符串用“."分割(点) [英] Java string split with "." (dot)

问题描述

为什么这段代码的第二行会抛出ArrayIndexOutOfBoundsException?

Why does the second line of this code throw ArrayIndexOutOfBoundsException?

String filename = "D:/some folder/001.docx";
String extensionRemoved = filename.split(".")[0];

虽然这有效:

String driveLetter = filename.split("/")[0];

我使用 Java 7.

I use Java 7.

推荐答案

如果要在文字点上拆分，则需要对点进行转义:

You need to escape the dot if you want to split on a literal dot:

String extensionRemoved = filename.split("\.")[0];

否则，您将在正则表达式 . 上拆分，这意味着任何字符".
请注意在正则表达式中创建单个反斜杠所需的双反斜杠.

Otherwise you are splitting on the regex ., which means "any character".
Note the double backslash needed to create a single backslash in the regex.

您得到一个 ArrayIndexOutOfBoundsException 因为您的输入字符串只是一个点，即 "."，这是一个边缘情况，在拆分时会产生一个空数组点;split(regex) 从结果中删除所有尾随空白，但由于在一个点上拆分一个点只留下两个空白，因此在删除尾随空白后，您会得到一个空数组.

You're getting an ArrayIndexOutOfBoundsException because your input string is just a dot, ie ".", which is an edge case that produces an empty array when split on dot; split(regex) removes all trailing blanks from the result, but since splitting a dot on a dot leaves only two blanks, after trailing blanks are removed you're left with an empty array.

为了避免在这种边缘情况下出现 ArrayIndexOutOfBoundsException，请使用 split(regex, limit)，它有第二个参数是结果数组的大小限制.当 limit 为 negative 时，禁用从结果数组中删除尾随空白的行为:

To avoid getting an ArrayIndexOutOfBoundsException for this edge case, use the overloaded version of split(regex, limit), which has a second parameter that is the size limit for the resulting array. When limit is negative, the behaviour of removing trailing blanks from the resulting array is disabled:

".".split("\.", -1) // returns an array of two blanks, ie ["", ""]

即，当 filename 只是一个点 "." 时，调用 filename.split("\.", -1)[0] 将返回一个空白，但调用 filename.split("\.")[0] 将抛出一个 ArrayIndexOutOfBoundsException.

ie, when filename is just a dot ".", calling filename.split("\.", -1)[0] will return a blank, but calling filename.split("\.")[0] will throw an ArrayIndexOutOfBoundsException.

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