Question

假设我的文件结构如下：

第0行：

354858 Some String That Is Important AA 其他STUFF SOMESTUFF   应该被忽视

第1行：

543788 Another String That Is Important AA 其他东西   应该被忽视的SOMESTUFF

依旧......

现在我想获取我的示例中标记的信息（参见灰色背景）。序列 AA 始终存在（并且可以用作中断并跳到下一行），而信息字符串的长度不同。

解析信息的最佳方法是什么？一个带有if, then, else的缓冲读卡器，或者你可以告诉某种解析器，读取一些lenth XYZ 然后将所有内容读入一个字符串，直到找到 AA 然后跳过。

Answer 1

我会逐行读取文件，并将每一行与正则表达式相匹配。我希望我在下面的代码中的评论足够详细。

// The pattern to use
Pattern p = Pattern.compile("^([0-9]+)\\s+(([^A]|A[^A])+)AA");

// Read file line by line
BufferedReader br = new BufferedReader(new FileReader(myFile));
String line;
while((line = br.readLine()) != null) {
  // Match line against our pattern
  Matcher m = p.matcher(line);
  if(m.find()) {
    // Line is valid, process it however you want
    // m.group(1) contains the number
    // m.group(2) contains the text between number and AA
  } else {
    // Line has invalid format (pattern does not match)
  }
}

我使用的正则表达式（Pattern）的说明：

^([0-9]+)\s+(([^A]|A[^A])+)AA

^               matches the start of the line
([0-9]+)        matches any integral number
\s+             matches one or more whitespace characters
(([^A]|A[^A])+) matches any characters which are either not A or not followed by another A
AA              matches the terminating AA

更新为评论回复：

如果每一行都有前面的|字符，则表达式如下所示：

^\|([0-9]+)\s+(([^A]|A[^A])+)AA

在JAVA中，你需要像这样逃避它：

"^\\|([0-9]+)\\s+(([^A]|A[^A])+)AA"

角色|在正则表达式中具有特殊含义，必须进行转义。

Answer 2

如果没有更多信息，就无法告诉您哪个问题最适合您。

一种解决方案可能是

String s = "354858 Some String That Is Important AA OTHER STUFF SOMESTUFF THAT SHOULD BE IGNORED";
String[] split = s.substring(0, s.indexOf(" AA")).split(" ", 2);
System.out.println("split = " + Arrays.toString(split));

<强>输出

split = [354858, Some String That Is Important]

Answer 3

您可以逐行阅读文件，并排除包含 AA charSequence的部分：

final String charSequence = "AA";
String line;
BufferedReader r = new BufferedReader(new InputStreamReader(new FileInputStream("yourfilename")));
try {
    while ((line = r.readLine()) != null) {
       int pos = line.indexOf(charSequence);
       if (pos > 0) {
            String myImportantStuff = line.substring(0, pos);
            //do something with your useful string
       }
    }
} finally {
    r.close();
}

Answer 4

使用正则表达式：.+?(?=AA) 。

检查 Here is the Demo

Answer 5

以下是您的解决方案：

public static void main(String[] args) {
    InputStream source; //select a text source (should be a FileInputStream)
    {
        String fileContent = "354858 Some String That Is Important AA OTHER STUFF SOMESTUFF THAT SHOULD BE IGNORED\n" +
                "543788 Another String That Is Important AA OTHER STUFF SOMESTUFF THAT SHOULD BE IGNORED";
        source = new ByteArrayInputStream(fileContent.getBytes(StandardCharsets.UTF_8));
    }

    try(BufferedReader stream = new BufferedReader(new InputStreamReader(source))) {
        Pattern pattern = Pattern.compile("^([0-9]+) (.*?) AA .*$");
        while(true) {
            String line = stream.readLine();
            if(line == null) {
                break;
            }
            Matcher matcher = pattern.matcher(line);
            if(matcher.matches()) {
                String someNumber = matcher.group(1);
                String someText = matcher.group(2);
                //do something with someNumber and someText
            } else {
                throw new ParseException(line, 0);
            }
        }
    } catch (IOException | ParseException e) {
        e.printStackTrace(); // TODO ...
    }
}

Answer 6

您可以使用正则表达式，但如果您知道每一行都包含AA并且您希望内容最多为AA，那么您可以简单地执行substring(int,int)来获取排队到AA

public List read(Path path) throws IOException {
    return Files.lines(path)
          .map(this::parseLine)
          .collect(Collectors.toList());
}

public String parseLine(String line){
    int index = line.indexOf("AA");
    return line.substring(0,index);
}

这是{Java}版read

public List read(Path path) throws IOException {
    List<String> content = new ArrayList<>();

    try(BufferedReader reader = new BufferedReader(new FileReader(path.toFile()))){
        String line;
        while((line = reader.readLine()) != null){
            content.add(parseLine(line));
        }
    }

    return content;
}

Java顺序解析文件中的信息

6 个答案: