使用正则表达式解析数组语法 [英] Parsing array syntax using regex

问题描述

我认为我要问的是非常琐碎的或已经问过的，但是我很难找到答案。

I think what I am asking is either very trivial or already asked, but I have had a hard time finding answers.

我们需要捕获内部数字字符在给定字符串中的括号之间。

We need to capture the inner number characters between brackets within a given string.

给定字符串

StringWithMultiArrayAccess[0][9][4][45][1]

和正则表达式

^\w*?(\[(\d+)\])+?

我希望有6个捕获组并可以访问内部数据。
但是，我最终只能捕获捕获组2中的最后一个 1字符。

I would expect 6 capture groups and access to the inner data. However, I end up only capturing the last "1" character in capture group 2.

如果重要的是我的java junit测试，

If it is important heres my java junit test:

@Test
public void ensureThatJsonHandlerCanHandleNestedArrays(){
    String stringWithArr = "StringWithMultiArray[0][0][4][45][1]";
    Pattern pattern = Pattern.compile("^\\w*?(\\[(\\d+)\\])+?");


    Matcher matcher = pattern.matcher(stringWithArr);
    matcher.find();

    assertTrue(matcher.matches()); //passes

    System.out.println(matcher.group(2));  //prints 1 (matched from last array symbols)

    assertEquals("0", matcher.group(2)); //expected but its 1 not zero
    assertEquals("45", matcher.group(5));  //only 2 capture groups exist, the whole string and the 1 from the last array brackets

}

推荐答案

为了捕获每个数字，您需要更改正则表达式，以便它（a）捕获单个数字，并且（b）不锚定于-并且因此受字符串的任何其他部分限制（ ^ \w *？将其锚定到字符串的开头）。然后，您可以遍历它们：

In order to capture each number, you need to change your regex so it (a) captures a single number and (b) is not anchored to--and therefore limited by--any other part of the string ("^\w*?" anchors it to the start of the string). Then you can loop through them:

Matcher mtchr = Pattern.compile("\\[(\\d+)\\]").matcher(arrayAsStr);
while(mtchr.find())  {
   System.out.print(mtchr.group(1) + " ");
}

输出：

0 9 4 45 1

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