使用方法在 Java 列表中搜索匹配项答案

【问题标题】：Using a method to search a match in a list in Java使用方法在 Java 列表中搜索匹配项
【发布时间】：2012-10-07 08:45:08
【问题描述】：

我正在尝试创建一个程序，该程序将使用名为“防弹”的包（完成，工作正常）从用户那里获取输入，但是当我找到匹配项时出现无限循环错误。我有点不知所措，我觉得我不太理解我的代码为什么不起作用。我不是在寻找直接的答案，因为我正在努力学习这一点，但任何帮助都将不胜感激。谢谢，我的代码如下。

--编辑-- 我得到了程序的工作。谢谢大家的帮助，你们太棒了。

import bulletproof.*;

public class A26_1 {
public static void main(String[] args) {
    BPScanner kb = new BPScanner();
    String reservedWordToCheck = kb.getStringFromUser("Enter a word to see if it's reserved or enter leave: ");

    System.out.println(ReservedWordChecker(reservedWordToCheck));
}

public static String ReservedWordChecker(String reservedWordToCheck) {
    String[] table = {
            "abstract", "assert", "boolean", "break", "byte", "case", "catch", "char", "class", "const",
            "continue", "default", "do", "double", "enum", "extends", "final", "finally", "float",
            "for", "goto", "if", "implements", "import", "instanceof", "int", "interface", "long", "native",
            "new", "package", "private", "protected", "public", "return", "short", "static", "strictfp", "super",
            "switch", "synchronized", "this", "throw", "throws", "transient", "try", "void", "volatile", "while"
        };

    while (true) {                      
        if (reservedWordToCheck.equalsIgnoreCase("leave"))
            break;                
        boolean found = false;

        for (int i=0; i < table.length; i++) {
            if (reservedWordToCheck.equalsIgnoreCase(table[i])) {
                found = true;
                break;
            }               
            if (found)
                System.out.println("Yeah, " + reservedWordToCheck + ", is reserved as a Java identifier.");
            else
                System.out.println("Nope, the word " + reservedWordToCheck + ", isn't reserved. Go hog wild.");                      
        }      
        System.out.println("OK BYE");
    } return reservedWordToCheck;
}

}

【问题讨论】：

一般输入法？就像我在该列表中输入一个单词，它应该匹配并说“是的”，然后单词“，是保留字。”。程序检查方法内部提供的列表。我希望能澄清一些事情。
您好，一个好的做法是使用如下所示的字符串相等函数：if ("leave".equalsIgnoreCase(reservedWordToCheck))。这样做可以防止发生 NullPointerException。如果 reservedWordToCheck 为空，则条件为假，正如我们所料，但避免了 RuntimeException。
你还应该考虑修改你的代码只有一个循环，你不需要外部的while。 if (table[i].equalsIgnoreCase(reservedWordToCheck)) { return reservedWordToCheck;} 之前的 if 和最后你可以简化你的代码。还要记住在您的方法之外将 String [] 表声明为常量，每次使用 ReservedWordChecker 时都创建一个新实例是不可取的。我还认为该方法应该被称为 existsReservedWord 并且可以返回一个布尔值，而不是您正在搜索的单词。

标签： java string list methods

【解决方案1】：

for 循环中的break 打破了for 循环而不是while。因此，while 循环将无限继续。

查看更改，我已经完成，它应该可以正常工作。

while (true) 
{
    .......
    .......

for (int i=0; i < table.length; i++) 
{                
    if (reservedWordToCheck.equalsIgnoreCase(table[i])) 
    {                    
        found = true; 
        break; // remove break from here          
    }

    if (found)                    
    {
        System.out.println("Yeah, " + reservedWordToCheck + ", is reserved as a Java identifier.");
        break;
    }                                  
} // end for
if(found)
{
    System.out.println("OK BYE");
}
else                    
{
    System.out.println("Nope, the word " + reservedWordToCheck + ", isn't reserved. Go hog wild.");
}
break;
} // end while

但是，您不需要 while(true) 构造来检查这一点。只有for 就足够了。

【讨论】：

【解决方案2】：

那是因为您的未标记 break 仅完成 for 语句，但 while(true) 将继续迭代。

解决方案可以使用如下标记的中断：

  search: 
  while (true) {                      
        if (reservedWordToCheck.equalsIgnoreCase("leave"))           
           break;                
        boolean found = false;

        for (int i=0; i < table.length; i++) {
            if (reservedWordToCheck.equalsIgnoreCase(table[i])) {
                found = true;
                break search;
            }               
            if (found)
                System.out.println("Yeah, " + reservedWordToCheck + ", is reserved as a Java identifier.");
            else
                System.out.println("Nope, the word " + reservedWordToCheck + ", isn't reserved. Go hog wild.");                      
        }      
        System.out.println("OK BYE");
    } return reservedWordToCheck;

【讨论】：

【解决方案3】：

你没有在正确的位置打破：

for (int i=0; i < table.length; i++) {
    if (reservedWordToCheck.equalsIgnoreCase(table[i])) {
        found = true;
        break; <--
    }               
    if (found)
        System.out.println("Yeah, " + reservedWordToCheck + ", is reserved as a Java identifier.");
    else
        System.out.println("Nope, the word " + reservedWordToCheck + ", isn't reserved. Go hog wild.");                      
}

找到该单词后，您会跳出 for 循环，除了 OK BYE 之外什么都不显示。

我不明白你为什么在那里放一个无限循环？

类似的东西会起作用：

public static String ReservedWordChecker(String reservedWordToCheck) {
    String[] table = { ... };

    if (reservedWordToCheck.equalsIgnoreCase("leave")) {
        break;
    }

    boolean found = false;
    for (String entry : table) {
        if (entry.equalsIgnoreCase(reservedWordToCheck)) {
            found = true;
            break;
        }
    }

    if (found) {
        System.out.println("Yeah, " + reservedWordToCheck + ", is reserved as a Java identifier.");
    } else {
        System.out.println("Nope, the word " + reservedWordToCheck + ", isn't reserved. Go hog wild.");                      
    }
    System.out.println("OK BYE");
    return reservedWordToCheck;
}

}

【讨论】：

【解决方案4】：

找到匹配项后，您并没有退出 while 循环：-

您应该将您的if-else 移出您的for-loop 并在您的if 中添加中断，这将中断您的while，无论是否找到匹配项：-

for (int i=0; i < table.length; i++) {
    if (reservedWordToCheck.equalsIgnoreCase(table[i])) {
        found = true;
        break;
    }
} 
if (found) {
     System.out.println("Yeah, " + reservedWordToCheck + 
                        ", is reserved as a Java identifier.");
     break;
} else {
     System.out.println("Nope, the word " + reservedWordToCheck + 
                        ", isn't reserved);
     break;
}

但我不明白，为什么你需要一个while 循环。它有点毫无意义。你的for-loop 足以做你想做的事。

【讨论】：

【解决方案5】：

我认为最简单的方法是 -

public static boolean ReservedWordChecker(String reservedWordToCheck) {
    String[] table = {
            "abstract", "assert", "boolean", "break", "byte", "case", "catch", "char", "class", "const",
            "continue", "default", "do", "double", "enum", "extends", "final", "finally", "float",
            "for", "goto", "if", "implements", "import", "instanceof", "int", "interface", "long", "native",
            "new", "package", "private", "protected", "public", "return", "short", "static", "strictfp", "super",
            "switch", "synchronized", "this", "throw", "throws", "transient", "try", "void", "volatile", "while"
        };

       for (String data : table) {
            if (reservedWordToCheck.equalsIgnoreCase(data)) {
                return true;
            }
        }
      return false;
}

【讨论】：

哇，我听说我可以将它转换为布尔值，但我没想到会这么简单。这对我帮助很大！谢谢！
就我个人而言，我非常相信一个入口和一个出口点，但这只是我

【解决方案6】：

您需要在 for 循环之外编写以下代码，因为您的 for 在您获得匹配时中断，但您的 while 循环继续运行

 if (found){
            System.out.println("Yeah, " + reservedWordToCheck + ", is reserved as a Java identifier.");
            break;
 }
 else{
            System.out.println("Nope, the word " + reservedWordToCheck + ", isn't reserved. Go hog wild.");
            break;
 }

【讨论】：