Question

我试图将确切的AdvanceJava关键字与给定的inputText字符串匹配，但它同时执行if和else条件，而不是我只想匹配AdvanceJava关键字。

String inputText = ("iwanttoknowrelatedtoAdvancejava").toLowerCase().replaceAll("\\s", "");

String match = "java";
List keywordsList = new ArrayList<>();//where keywordsList{advance,core,programming} -> keywordlist fetch 
// from database 

Enumeration e = Collections.enumeration(keywordsList);
int size = keywordsList.size();
while (e.hasMoreElements()) {
    for (int i = 0; i < size; i++) {
        String s1 = (String) keywordsList.get(i);
        if (inputText.contains(s1) && inputText.contains(match)) {
            System.out.println("Yes we providing " + s1);
        } else if (!inputText.contains(s1) && inputText.contains(match)) {
            System.out.println("Yes we are working on java");
        }
    }
    break;
}

由于

Answer 1

您可以通过使用模式和匹配器类来完成此操作

    Pattern p = Pattern.compile("java");
    Matcher m = p.matcher("Print this");
    m.find();

如果你想在一行中找到多个匹配项，你可以反复调用find（）和group（）来全部提取它们。

Answer 2

以下是使用模式匹配实现目标的方法。

在第一个例子中，我按原样输入了您的输入文本。这只会改进您的算法，该算法具有O（n ^ 2）性能。

String inputText = ("iwanttoknowrelatedtoAdvancejava").toLowerCase().replaceAll("\\s", "");

        String match = "java";
        List<String> keywordsList = Arrays.asList("advance", "core", "programming");

        for (String keyword : keywordsList) {
            Pattern p = Pattern.compile(keyword.concat(match));

            Matcher m = p.matcher(inputText);
            //System.out.println(m.find());
            if (m.find()) {
                System.out.println("Yes we are providing " + keyword.concat(match));
            }
        }

但我们可以通过更好的实施来改进这一点。这是上述实现的更通用版本。此代码在匹配之前不会操作输入文本，而是提供更通用的正则表达式，它忽略空格并匹配不区分大小写的方式。

String inputText = "i want to know related to Advance java";

        String match = "java";
        List<String> keywordsList = Arrays.asList("advance", "core", "programming");

        for (String keyword : keywordsList) {
            Pattern p = Pattern.compile(MessageFormat.format("(?i)({0}\\s*{1})", keyword, match));
            Pattern p1 = Pattern.compile(MessageFormat.format("(?i)({0})", match));

            Matcher m = p.matcher(inputText);
            Matcher m1 = p1.matcher(inputText);
            //System.out.println(m.find());
            if(m.find()) {
                System.out.println("Yes we are providing " + keyword.concat(match));
            } else if(m1.find()) {
                System.out.println("Yes we are working with " + match);
            }
        }

Answer 3

@sithum - 谢谢，但它在输出中执行if else的条件。请参阅我附加的屏幕截图。

我应用了以下逻辑，它运行正常。请参考，谢谢。

    String inputText = ("iwanttoknowrelatedtoAdvancejava").toLowerCase().replaceAll("\\s", "");
    String match = "java"; 
     List<String> keywordsList = session.createSQLQuery("SELECT QUESTIONARIES_RAISED FROM QUERIES").list();  // Fetch values from database (advance,core,programming)
    String uniqueKeyword=null;
    String commonKeyword= null;
    int size =keywordsList.size();
    for(int i=0;i<size;i++){
               String s1 = (String) keywordsList.get(i);//get values one by one from list
    if(inputText.contains(match)){
    if(inputText.contains(s1) && inputText.contains(match)){
    Queries q1 = new Queries();
    q1.setQuestionariesRaised(s1); //set matched keyword to getter setter method
    keywordsList1=session.createQuery("from Queries sentence where questionariesRaised='"+q1.getQuestionariesRaised()+"'").list(); // based on matched keyword fetch according to matched keyword sentence which stored in database
    for(Queries ob  : keywordsList1){

    uniqueKeyword=  ob.getSentence().toString();// Store fetched sentence to on string variable
      }
    break;
    }else {
      commonKeyword= "java only"; 
         }
       }
     }}
  if(uniqueKeyword!= null){
   System.out.println("Yes we providing......................" + uniqueKeyword);        
  }else if(commonKeyword!= null){
   System.out.println("Yes we providing " + commonKeyword);
    }else{      
}

如何使用java从给定的字符串中获取完全匹配关键字？

3 个答案: