在java中編寫正則表達式代碼

Question

請幫助我們，我想知道如何編寫正則表達式代碼。

可以說，一個文件包含3句

[hi Tom how are you.hey Andy its nice to see you.where is your wife Tom.] 

所以，當我搜索Tom我希望程序打印的第一和最後一個句子，如果我搜索Andy程序應該只打印第二句。

我瘋了，因爲我所做的只是打印Tom或Andy。 這是我的代碼：

Pattern p =Pattern.compile("Tom\\w+") 

Answer 1

如果我理解你的問題吧，你想匹配2句爲Tom和1 Andy。你想這樣做：

String line = "hi Tom how are you.hey Andy its nice to see you.where is your wife Tom."; 

String pattern = "[\\w\\s]*Tom[\\w\\s]*[\\.]?"; 

Pattern r = Pattern.compile(pattern); 
Matcher m = r.matcher(line); 

while (m.find()) { 
    System.out.println("Find: " + m.group(0)); 
} 

輸出：

Find: hi Tom how are you. 
Find: where is your wife Tom. 

Answer 2

如果你是一個初學者，那麼試試下面的代碼。您可以通過在一次迭代中檢查多個鍵/人名來提高效率。

public class Tommy { 

    public static void main(String[] args) { 

     String junk = "hi Tom how are you.hey Andy its nice to see you.where is your wife Tom."; 
     System.out.println(junk); 
     String [] rez = extractor(junk, "Tom"); 
     printArray(rez); 

    } 

    public static String[] extractor(String text, String key) { 
     String[] parts = text.split("\\."); 
     String[] rez = new String[parts.length];// Use an arraylist ? 

     for (int i = 0; i < parts.length; i++) { 
      if (parts[i].contains(key)) { 
       rez[i] = parts[i]; 
      } 
     } 

     return rez; 
    } 

    public static void printArray(String[] ar) { 

     for (int i = 0; i < ar.length; i++) { 
      if (ar[i] != null) { 
       System.out.println(ar[i]); 
      } 
     } 

    } 

} 

Answer 3

假設每個句子"."結束，則：

public List<String> findMatches(String string, String name) { 
    List<String> result = new ArrayList<String>(); 
    Pattern p = Pattern.compile("[\\w\\s]*" + name + "[\\w\\s]*");  
    for (String s : string.split("\\.")) { 
     Matcher m = p.matcher(s); 
     if(p.matcher(s).matches()) 
      result.add(s); 
    } 
    return result; 
} 


String string = "hi Tom how are you. hey Andy its nice to see you. where is your wife Tom."; 
System.out.println(findMatches(string, "Tom"); 
System.out.println(findMatches(string, "Andy"); 

Answer 4

(?=[^.]*?\bTom\b[^.]*\.)([\w\s]+(?=\.)) 

嘗試此Tom。對於Andy與Andy.See演示代替湯姆。

http://regex101.com/r/vR4fY4/10