Question

我有一个字符串：String test = "A1=CA2=BOA2=RA4=OA11=O";
现在我想要输出：A1,A2,A2,A4,A11

如果我的字符串如下：

String test = "[HEADER_1] [DATA] \n A10=T,(\"Re-N RO-25 M-N P-N (B)\"),1.0:0.8,R=25.0 F-7.829215,-4.032765 A20=B,R2,M=XY,R=29.999999999999996 F564.997550,669.454680";


public static void main(String[] args) {
    String test = "A1=CA2=BOA2=RA4=O";
    String data[] = test.split("[A-a]\\d{1,100}=");

    for (String str : data) {
      System.out.println("Split data:"+str);
    }      
}

那是什么样的正则表达式？

Answer 1

我不会使用拆分。您想要的部件的正则表达式为/a\d+/i。如果您使用此解决方案和Create array of regex matches中的解决方案来收集列表中的所有匹配项，则可以轻松创建所需的输出。

在Java-Parlour中转换正则表达式：

public static void main(String[] args) {
    String test = "A1=CA2=BOA2=RA4=OA11=O";
    List<String> allMatches = new ArrayList<String>();
    Matcher m = Pattern.compile("(a\\d+)", Pattern.CASE_INSENSITIVE).matcher(test);

    while (m.find()) {
       allMatches.add(m.group());
    }
}

allMatches现在包含A1 A2 A2 A4 A11

Answer 2

您可能正在寻找的正则表达式是[A-a][0-9]{1,2}。

所以在Java中，

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


String test = "A1=CA2=BOA2=RA4=OA11=O";
Matcher m = Pattern.compile("[A-a][0-9]{1,2}").matcher(test);
while (m.find()) {
    System.out.print(m.group(0)+",");
}

Answer 3

String test = "A1=CA2=BOA2=RA4=OA11=O";
String[] parts = test.split("=");
String[] output = new String[parts.length];
for (int i=0; i < parts.length; ++i) {
    output[i] = parts[i].replaceAll("^.*(A\\d*)$", "$1");
    System.out.println(parts[i] + " becomes " + output[i]);
}

<强>输出：

A1 becomes A1
CA2 becomes A2
BOA2 becomes A2
RA4 becomes A4
OA11 becomes A11
O becomes O

从java中的字符串中提取特定字符串

3 个答案: