java：检索以特定字符串开头和结尾的字符串的一部分

Question

This is the program and i have a string strLineText from which i need to extract the words that contain target in them.

Ex. In the string "random string with IWANTTHISABC-123 and more" i need to extract IWANTTHISABC-123 . Similarly if the string is "random string with IWANTTHISBBC-001" i need to extract `IWANTTHISBBC-001. The prefix is fixed

I have tried it with substring() (Method1) but the logic doesn't work for Strings that end with this target word ie, nothing is outputted

I tried the split() (Method2) and it works for all four combinations.

Can you help me with achieving using the substring() (Method1) for all four combinations

public static void main(String[] args) throws IOException {

    String target = "IWANTTHIS";

    //Four possible inputs
    String strLineText = "random string with IWANTTHISABC-123 and more";   //works
    String strLineText = "IWANTTHISCBC-45601 and more";                    //works
    String strLineText = "IWANTTHISEBC-1";                                 //doesn't work
    String strLineText = "random string with IWANTTHISKBC-55545";          //doesn't work

    //Method1
    System.out.println("O/P 1:" + strLineText.substring(strLineText.indexOf(target), 
            strLineText.indexOf(target) + strLineText.substring(strLineText.indexOf(target)).indexOf(" ") + 1).trim());

    //Method2
    for (String s : strLineText.split(" "))
        if (s.contains(target))
            System.out.println("O/P 2:" + s.trim());
}

Answer 1

strLineText.substring(strLineText.indexOf(target)).indexOf(" ") will be -1 if strLineText contains no spaces after your target string. You could check if strLineText.substring(strLineText.indexOf(target)) contains spaces, and if not, take the substring until the end of strLineText :

//Method1
int beginIndex = strLineText.indexOf(target);
String substring = strLineText.substring(beginIndex);
int endIndex = substring.contains(" ") ? beginIndex + substring.indexOf(" ") : strLineText.length();
System.out.println("O/P 1:" + strLineText.substring(beginIndex, endIndex));

Answer 2

I think it is pretty straightforward, you just need to compute the end index starting from begin index. Here is the snippet that works for all cases.

int begin = strLineText.indexOf(target);
int end = strLineText.indexOf(" ", begin);
if(end == -1) end = strLineText.length();

System.out.println(strLineText.substring(begin, end));

Answer 3

Assumes that your definition of "word" is a sequence of alphas, excluding numbers, symbols, etc. For other definitions of "word," the regular expression can be adjusted accordingly. If you want to include parts of the word previous to the target string, you can add a loop that counts backwards from startIndex, examining characters to see if they are alpha.

public class Foo
{
  public static void main(String[] args)
  {
    String target = "IWANTTHIS";

//    String candidate = "random string with IWANTTHISABC-123 and more";
      String candidate = "IWANTTHISCBC-45601 and more";
//    String candidate = "IWANTTHISEBC-1";
//    String candidate = "random string with IWANTTHISKBC-55545";

    int startIndex = -1;
    int endIndex = -1;

    if(candidate.contains(target))
    {
      System.out.println("Target located.");

      startIndex = candidate.indexOf(target);

      System.out.println("target starts at " + startIndex);

      // keep adding characters until first non-alpha char

      endIndex = startIndex + target.length();

      boolean wordEnded = false;

      while(!wordEnded && (endIndex >= candidate.length()))
      {
        String foo = Character.toString(candidate.charAt(endIndex + 1));

        if(foo.matches("[a-zA-Z]"))
        {
          endIndex++;
        }
        else
        {
          wordEnded = true;
        }
      }

      String full = candidate.substring(startIndex, endIndex + 1);

      System.out.println("Full string = " + full);
    }
    else
    {
      System.out.println("No target located. Exiting.");
    }
  }
}