使用正则表达式解析非常简单的JSON文件 [英] Using regex to parse a very simple JSON file

问题描述

我不确定是否可以完成，但是我想将一个非常简单的JSON文件解析为字符串数组.

I'm not sure if it can be done, but I'd like to parse a very simple JSON file to an array of Strings.

示例文件:

["String1", "String2", "oneMoreString"]

到目前为止，我还以为我会使用带有模式的Scanner来获取输出，但未能做到这一点.

So far I thought I'd use Scanner with a pattern to get my output, but failed to do this.

    ArrayList<String> strings = new ArrayList<String>();
    File f = new File("src/sample.txt");
    String pattern = "\\s*[\"\"]\\s*";
    try {
        InputStream is = new FileInputStream(f);
        Scanner s = new Scanner(is);
        s.useDelimiter(pattern);
        while (s.hasNext()){
            strings.add(s.next());
        }
        s.close();
        is.close();
    } catch (Exception e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

因为该模式显然是错误的，因为它会适当地考虑，"，但是我希望它不会包含在内...:S

because the pattern is clearly wrong, since it considers ", " as it fits, but I'd like it wouldn't be included... :S

我也接受一些建议，这些建议可能会以任何其他可以解析的方式起作用.也许是JSON解析器?但是由于该文件是如此简单，所以我认为没有必要.

I also accept suggestions that may work to any other way this can be parsed. Maybe a JSON parser? but because the file is so simple I didn't consider it necessary.

推荐答案

最好使用Jackson映射器之类的JSON解析器来解析JSON字符串.

It is better to use a JSON parser like Jackson Mapper to parse a JSON String.

但是，如果您有一个简单的String，则可以快速使用示例正则表达式.

But to if you have a simple String you can use a sample Regular expression to it quickly.

尝试一下:

    String str = "[\"String1\", \"String2\", \"oneMoreString\"]";

    Pattern pattern = Pattern.compile("\"(.+?)\"");
    Matcher matcher = pattern.matcher(str);

    List<String> list = new ArrayList<String>();
    while (matcher.find()) {
        // System.out.println(matcher.group(1));.
        list.add(matcher.group(1));
    }

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