删除开头部分字符串，最后在java中删除部分字符串

Question

I want a dynamic code which will trim of some part of the String at the beginning and some part at last. I am able to trim the last part but not able to trim the initial part of the String to a specific point completely. Only the first character is deleted in the output.

public static String removeTextAndLastBracketFromString(String string) {
  StringBuilder str = new StringBuilder(string);
  int i=0;
  do {
    str.deleteCharAt(i);
    i++;
  } while(string.equals("("));
  str.deleteCharAt(string.length() - 2);
  return str.toString();
}

This is my code. When I pass Awaiting Research(5056) as an argument, the output given is waiting Research(5056 . I want to trim the initial part of such string till ( and I want only the digits as my output. My expected output here is - 5056 . Please help.

Answer 1

You don't need loops (in your code), you can use String.substring(int, int) in combination with String.indexOf(char) :

public static void main(String[] args) {
    // example input
    String input = "Awaiting Research(5056)";
    // find the braces and use their indexes to get the content
    String output = input.substring(
                        input.indexOf('(') + 1, // index is exclusive, so add 1
                        input.indexOf(')')
                    );
    // print the result
    System.out.println(output);
}

Output:

5056

Hint:

Only use this if you are sure the input will always contain a ( and a ) with indexOf('(') < indexOf(')') or handle IndexOutOfBoundsException s, which will occur on most String s not matching the braces constraint.

Answer 2

If your goal is just to look one numeric value of the string, try split the string with regex for the respective numeric value and then you'll have the number separated from the string eg:

 Pattern pattern = Pattern.compile("\\d+");
 Matcher matcher = pattern.matcher("somestringwithnumberlike123");
 
 if(matcher.find()) {
     System.out.println(matcher.group());
 }

Answer 3

Using a regexp to extract what you need is a better option :

String test = "Awaiting Research(5056)";
Pattern p = Pattern.compile("([0-9]+)");
Matcher m = p.matcher(test);
if (m.find()) {
    System.out.println(m.group());
}

Answer 4

For your case, battery use regular expression to extract your interested part.

Pattern pattern = Pattern.compile("(?<=\\().*(?=\\))");
Matcher matcher = pattern.matcher("Awaiting Research(5056)");
if(matcher.find())
{
    return matcher.group();
}

Answer 5

It is much easier to solve the problem eg using the String.indexOf(..) and String.substring(from,to) . But if, for some reason you want to stick to your approach, here are some hints:

Your code does what is does because:

• string.equals("(") is only true if the given string is exacly "("
• the do {code} while (condition) -loop executes code once if condition is not true -> think about using the while (condition) {code} loop instead
• if you change the condition to check for the character at i, your code would remove the first, third, fifth and so on: After first execution i is 1 and char at i is now the third char of the original string (because the first has been removed already) -> think about always checking and removing charAt 0.