使用模式匹配从文件中排序，Java

Question

所以我已经让我的程序正确地分隔文本文件的行，甚至可以匹配第一行文本的模式，但我还需要能够检测和分隔文本文件的地址行并根据它们的方向或 street/broadway 对它们进行排序，但我什至无法为地址设置检测到初始模式。我是否使用了错误的正则表达式，这就是为什么无法正确检测到地址部分的原因？

代码

package csi311;

// Import some standard Java libraries.
import java.io.BufferedReader;
import java.io.FileReader;
import java.util.regex.Pattern;
import java.util.regex.Matcher;
import java.util.ArrayList;
/**
 * Hello world example.  Shows passing in command line arguments, in this case a filename. 
 * If the filename is given, read in the file and echo it to stdout.
 */
public class HelloCsi311 {

    /**
     * Class construtor.
     */
    public HelloCsi311() {
    }


    /**
     * @param filename the name of a file to read in 
     * @throws Exception on anything bad happening 
     */
    public void run(String filename) throws Exception {
        if (filename != null) {
            readFile(filename); 
        }
    }


    /**
     * @param filename the name of a file to read in 
     * @throws Exception on anything bad happening 
     */
    private void readFile(String filename) throws Exception {
        System.out.println("Dumping file " + filename); 
        // Open the file and connect it to a buffered reader.
        BufferedReader br = new BufferedReader(new FileReader(filename));  
        ArrayList<String> foundaddr = new ArrayList<String>();
        String line = null;  
        String pattern = "^\d\d\d-[A-Za-z][A-Za-z][A-Za-z]-\d\d\d\d";
        String address[] = new String[4];
        address[0] = "\d{1,3}\s\[A-Za-z]{1,20}";
        address[1] = "\d{1,3}\s\[A-Za-z]{1,20}\s\d{1,3}\[A-Za-z]{1,20}\s\[A-Za-z]{1,20}";
        address[2] = "\d{1,3}\s\d{1,3}\[A-Za-z]{1,20}\s\[A-Za-z]{1,20}";
        address[3] = "\d\d\s\[A-Za-z]{1,20}";
        Pattern r = Pattern.compile(pattern);
        // Get lines from the file one at a time until there are no more.
        while ((line = br.readLine()) != null) {
            if(line.trim().isEmpty()) {
                continue;
            }
            String sample = line.replaceAll("\s+,", ",").replaceAll(",+\s",",");
            String[] result = sample.split(",");
            String pkgId = result[0].trim().toUpperCase();
            String pkgAddr = result[1].trim();


            Float f = Float.valueOf(result[2]);
            for(String str : result){
                // Trying to match for different types
                for(String pat : address){
                    if(str.matches(pat)){
                        System.out.println(pat);
                    }
                }



                if(f < 50 && !pkgId.matches(pattern)) {
                    Matcher m = r.matcher(str);
                    if(m.find()) {
                        foundaddr.add(str);
                    }
                }
            }
        }

        if(foundaddr != null) {
            System.out.println(foundaddr.size());
        }   

        // Close the buffer and the underlying file.
        br.close();
    }



    /**
     * @param args filename
     */
    public static void main(String[] args) {
        // Make an instance of the class.
        HelloCsi311 theApp = new HelloCsi311();
        String filename = null; 
        // If a command line argument was given, use it as the filename.
        if (args.length > 0) {
            filename = args[0]; 
        }
        try { 
            // Run the run(), passing in the filename, null if not specified.
            theApp.run(filename);
        }
        catch (Exception e) {
            // If anything bad happens, report it.
            System.out.println("Something bad happened!");
            e.printStackTrace();
        }    
    }
}

文本文件

123-ABC-4567, 15 W. 15th St., 50.1
456-BGT-9876,22 Broadway,24
QAZ-456-QWER, 100 East 20th Street,50
Q2Z-457-QWER, 200 East 20th Street, 49
6785-FGH-9845 ,45 5th Ave, 12.2,
678-FGH-9846 ,45 5th Ave, 12.2

123-ABC-9999, 46 Foo Bar, 220.0
347-poy-3465, 101 B'way,24

下面是应该能够处理地址行的代码行，但由于某种原因，它不匹配正确分隔地址行的模式和输出，可以在 for 循环上方的打印语句中看到处理地址，但出于某种原因，地址行甚至没有被检测为匹配项，我对为什么会这样感到困惑。

代码行问题在于

  for(String str : result){
      //System.out.println(str);
      // Trying to match for different types
      for(String pat : address){
          if(str.matches(pat)){
              System.out.println(pat);
          }
      }

所需输出 - 按要求编辑 -

22 Broadway
45 5th Ave
101 B'way

Answer 1

我认为问题出在您的正则表达式上。例如\d\d\s\[A-Za-z]{1,20}，全部转义后变成\d\d\s\[A-Za-z]{1,20}。细分如下：

\d：匹配任意数字
\d：匹配任意数字
\s：匹配任意空白字符
\[：匹配[字符
A-Za-z：匹配文字A-Za-z
]：匹配文字字符]
- {1,20}：匹配前面的字符（]）1-20次。

您可能需要的正则表达式是 \d\d\s[A-Za-z]{1,20}，作为转义字符串，它是 \d\d\s[A-Za-z]{1,20}。请注意 [.

之前没有 \

还有一点要记住，正则表达式可以匹配字符串中的任何位置。例如，正则表达式 a 会匹配字符串 a，但也会匹配 abc、bac、abracadabra 等。要避免这种情况，您必须使用锚定符号 ^ 和 $ 分别匹配开始和结束。然后你的正则表达式变成 ^\d\d\s[A-Za-z]{1,20}$.

我还注意到您使用 for 循环 for(String str : result){ 将每一列与正则表达式进行匹配。在我看来，您应该只匹配 result[1] 或 pkgAddr.

最后一点，看看Regex 101。它将允许您针对一堆输入测试您的正则表达式，看看它们是否匹配。

使用模式匹配从文件中排序，Java

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