Java 基于第 n 个分隔符的字符串修剪 position

Question

I have a java method in which trying to parse a string where fields are delimited by char ^A. Sample string like below.

HDR^A1^A20220106^ATYPE^AXXX^AJAPAN^AUNIFORM^AHELP^AEXAMPLE^A

I have attempted to use apache ordinalIndexOf but so far not yet successful with the same.

Could you please help to suggest if any other alternative approach available for this scenario.

public class HelloWorld{

public static int ordinalIndexOf(String str, String substr, int n) {

    int pos = str.indexOf(substr);

    while (--n > 0 && pos != -1)

        pos = str.indexOf(substr, pos+1);

    return pos;

}
     public static void main(String []args){
//Just kept it here as I need to use standards.. not using in below code
         String CTRL_A = Character.valueOf((char) 0x01).toString();    
     String str = "HDR^ABYE^A20220103065014^Agoogle.com_29958^ABUDDY^A1.0^A123456789012^AHAI^ABYE";

              int position = ordinalIndexOf(str,"^A",6);

        System.out.println(str.substring(0,position));

     }

}

Expected Output String:

EVENT_HDR^ABYE^A20220103065014^Agoogle.com_29958^ABUDDY^A1.0^A123456789012

Referred Link

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Answer 1

You could try to use the.indexOf() method. So, in this case try:

String str = "HDR^A1^A20220106^ATYPE^AXXX^AJAPAN^AUNIFORM^AHELP^AEXAMPLE^A";
System.out.println(str.indexOf("^A"));

This will print the index of the first appearance of the given substring. Alternatively, you could do:

String str = "HDR^A1^A20220106^ATYPE^AXXX^AJAPAN^AUNIFORM^AHELP^AEXAMPLE^A";
System.out.println(str.indexOf("^A", 10)); // Start at index 10.

Please refer to the documention here for more: https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#indexOf(java.lang.String)

Answer 2

This will remove the 7th ^A and everything following it:

str = str.replaceAll("((.*?^A){6})^A.*", "$1");