正則表達式匹配模式和匹配

Question

我正在從文件中讀取輸入，它包含一組行，如下所示：

BDI100                 172.20.1.5      YES TFTP   up                    up
    BDI500                 172.20.1.50     YES TFTP   up                    up
    BDI600                 172.20.1.58     YES TFTP   up                    up

我必須提取僅包含172.20.1.5的完整行

以下是我的代碼：

while ((line = lineNumberReader.readLine()) != null) {
        Pattern p = Pattern.compile(expr.trim()); /*expr is filter contains 172.20.1.5 */
        Matcher m = p.matcher(line);
        if(m.find()){
            System.out.println(line);
        }
    }

我將輸出視為：

BDI100                 172.20.1.5      YES TFTP   up                    up

但是打印所有的線條。

Answer 1

這將有助於您：

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Test {

    public static void main(String[] args) {
        String[] lines = new String[]{"BDI100                 172.20.1.5      YES TFTP   up                    up",
            "BDI500                 172.20.1.50     YES TFTP   up                    up",
            "BDI600                 172.20.1.58     YES TFTP   up                    up"};
        Pattern p = Pattern.compile("172\\.20\\.1\\.5\\s");
        for (String line : lines) {
            Matcher m = p.matcher(line);
            if (m.find()) {
                System.out.println(line);
            }
        }
    }

}

Answer 2

你在做什么似乎有點過分。 也許嘗試這個：

String need = "172.20.1.5";
String line = "";
while ((line = lineNumberReader.readLine()) != null) {
    if (line.trim().equals("")) {
        continue:
    }
    if (line.contains(need) {
        System.out.println(line);
        break;
    }
}

Answer 3

/*Intialize expr within braces */
String expr = "(172.20.1.5 )";
while ((line = lineNumberReader.readLine()) != null) {
        Pattern p = Pattern.compile(expr.trim()); /*expr is filter contains 172.20.1.5 */
        Matcher m = p.matcher(line);
        if(m.find()){
            System.out.println(line);
        }
    }