我可以从方法中捕获抛出异常吗？

Question

当我尝试从方法声明中抛出异常时，出现错误 "Unreachable catch block for ClassNotFoundException. This exception is never thrown from the try statement body"。

代码是这样的：

public class MenuSQL {
    private static String sentence = "";
    private static int option;
    Statement sentenceSQL = ConnectSQL.getConexion().createStatement();

public MenuSQL(int option) throws ClassNotFoundException, SQLException {
    super();
    this.option = option;
    try {
        System.out.print("Introduce the sentence: ");
        System.out.print(sentence);
        sentence += new Scanner(System.in).nextLine();
        System.out.println(MenuSentence.rightNow("LOG") + "Sentence: " + sentence);

        if (opcion == 4) {
            MenuSentence.list(sentence);
        } else {
            sentenceSQL.executeQuery(sentence);
        }
    } catch (SQLException e) {
        System.out.println(MenuSentence.rightNow("SQL") + "Sentence: " + sentence);
    } catch (ClassNotFoundException e) {
        System.out.println(MenuSentence.rightNow("ERROR") + "Sentence: " + sentence);
    }
}
}

我怎样才能赶上ClassNotFoundException？提前致谢。

Answer 1

try{...} catch(){...}语句的catch块只能捕获try{...}块抛出的异常。（或该异常的超类）

try {
    Integer.parseInt("1");
    //Integer.parseInt throws NumberFormatException
} catch (NumberFormatException e) {
    //Handle this error
}

但是，您要做的基本上是这样的：

try {
    Integer.parseInt("1");
    //Integer.parseInt throws NumberFormatException
} catch (OtherException e) {
    //Handle this error
}

因为 none 你的 try{...} 块中的语句抛出 OtherException，编译器会给你一个错误，因为它知道 nothing 在你的 try{...} 块中将 ever 抛出那个异常，所以你不应该尝试 catch 永远不会 thrown.[=22 的东西=]

在你的例子中，你的 try{...} 块中没有任何东西会抛出 ClassNotFoundException，所以你不需要捕捉它。您可以从代码中删除 catch (ClassNotFoundException e) {...} 以修复错误。

我可以从方法中捕获抛出异常吗？

Can I catch a throw exception from method?

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