如何在 Java 中查找和替换字符串中所有出现的子字符串？

Question

I looked at the existing topics, but my task is slightly different from those on the forum. I have the value:

int j = 37;

And I have one of the strings:

1) String s1 = "5+2+5+2 --j *2*7+3";
2) String s1 = "5+2+5+2 j-- *2*7+3";
3) String s1 = "5+2+5+2 ++j *2*7+3";
4) String s1 = "5+2+5+2 j++ *2*7+3";

I need with regular expression find --j , j-- , ++j , j++ and replace this occurrences of substring in string with number value j ;

Result string must be like this:

1) 5+2+5+2 36 *2*7+3
2) 5+2+5+2 36 *2*7+3
3) 5+2+5+2 38 *2*7+3
4) 5+2+5+2 38 *2*7+3

With pattern str.split(search).join(replacement) I can replace the char j with number 37 :

str.split("j").join(37);

Then I get:

1) String s1 = "5+2+5+2 --37 *2*7+3";
2) String s1 = "5+2+5+2 37-- *2*7+3";
3) String s1 = "5+2+5+2 ++37 *2*7+3";
4) String s1 = "5+2+5+2 37++ *2*7+3";

Question:
But how to perform the arithmetic operation ( increment , decrement ) at this time?

Answer 1

public class Test
{
    public static void main(String[] args)
    {
        int j = 37;
        String s1 = "5+2+5+2 ++j *2*7+3";
        
        // this line swap "j--" or "--j" on j--
        s1 = s1.replaceAll("j--|--j", j-1 + "");
        
        // this line swaps "j++" or "++j" on j++
        s1 = s1.replaceAll("(j\\+\\+)|(\\+\\+j)", j+1 + "");
        
        System.out.println(s1);
    }
}

Answer 2

String.replaceAll may be used as follows:

int j = 37;
String[] d = {
    "5+2+5+2 --j *2*7+3",
    "5+2+5+2 j-- *2*7+3",
    "5+2+5+2 ++j *2*7+3",
    "5+2+5+2 j++ *2*7+3"
};
        
Arrays.stream(d)
      .map(s -> s.replaceAll("--j|j--", Integer.valueOf(j - 1).toString()))
      .map(s -> s.replaceAll("\\+\\+j|j\\+\\+", Integer.valueOf(j + 1).toString()))
      .forEach(System.out::println);

to provide expected output:

5+2+5+2 36 *2*7+3
5+2+5+2 36 *2*7+3
5+2+5+2 38 *2*7+3
5+2+5+2 38 *2*7+3

Answer 3

Note that in the question, the operations do not follow the Java behaviour, ie pre- and postincrement both behave like preincrement.

To implement this behaviour, we need two regular expressions:

• One expression matching all --j and j-- to replace them with the value of j - 1 :

  [-]{2}j|j[-]{2} Regex101 demo

• One expression matching all ++j and j++ to replace them with the value of j + 1 :

  [+]{2}j|j[+]{2} Regex101 demo

Putting it together, we can write the following method:

public static String replaceJWith(String s, int valueForJ) {
    s = s.replaceAll("[-]{2}j|j[-]{2}", Integer.toString(valueForJ - 1));
    return s.replaceAll("[+]{2}j|j[+]{2}", Integer.toString(valueForJ + 1));
}

Ideone demo

This implementation assumes that each occurrence of j is pre- or suffixed by ++ or -- . Also, since we never really change the value of j , using multiple pre-, and postincrement operators in a single String will result in unexpected behaviour.

---

If we want to mimic the Java behaviour (ie distinguish between pre- and postincrement), we need three different regular expressions since we have to replace j with three different values:

• One regex to replace j(++|--) with the original value for j :

  j(?:[+]{2}|[-]{2}) Regex101 demo

• One regex to replace ++j with the value j + 1 :

  [+]{2}j Regex101 demo

• One regex to replace --j with the value j - 1 :

  [-]{2}j Regex101 demo

Setting it all together, we can write the following Java method:

public static String replaceJWith(String s, int valueForJ) {
    s = s.replaceAll("j(?:[+]{2}|[-]{2})", Integer.toString(valueForJ));
    s = s.replaceAll("[+]{2}j", Integer.toString(valueForJ + 1));
    return s.replaceAll("[-]{2}j", Integer.toString(valueForJ - 1));
}

Ideone demo

This solution has the same limitations as the first solution.

Answer 4

You can use a series of if statements that check the string for the presence of each of those patterns using contains() and then replace the pattern with the corresponding value:

if (s1.contains("--j") {
  s1.replace("--j", String.valueOf(j-1)); //assuming you have declared int j = 37; before 
} else if (s1.contains("j--")) { 
  //etc.
} [...]