如何修复 RuntimeException?

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英文:

How to fix RuntimeException?

问题

import java.sql.Connection;

public static void main(String[] args) throws SQLException, RuntimeException {

    try {
        DriverManager.registerDriver(new com.mysql.cj.jdbc.Driver());

        String connectionURL = "jdbc:mysql://localhost:3306/aaa?useUnicode=true&useJDBCCompliantTimezoneShift=true&useLegacyDatetimeCode=false&serverTimezone=UTC";
        String login = "root";
        String password = "aaa";

        try (Connection conn = DriverManager.getConnection(connectionURL, login, password)) {
            Statement stmt = conn.createStatement();
            
            ResultSet rs = stmt.executeQuery("SELECT * FROM Lista1");
            
            while(rs.next()) {
                System.out.println(rs.getString("nazwa"));
            }
            rs.close();
            
            stmt.close();
        }
    } catch (SQLException e) {
        System.out.println("Blad polaczenia:");
        e.printStackTrace();
    } catch (RuntimeException e) {
        System.out.println("Blad polaczenia:");
        e.printStackTrace();
    }
}

The console log throws:

Exception in thread "main" java.lang.RuntimeException:
    at cccc.web.Main.main(Main.java:1)
    
I tried to delete return, but it still threw bad issue.
I tried many things but I don't know how to fix it.
I don't know how to solve it. And I spent many many many times to fight with it!!!!
英文:
import java.sql.Connection;
public static void main(String[] args) throws SQLException, RuntimeException {
try{
DriverManager.registerDriver(new com.mysql.cj.jdbc.Driver());
String connectionURL="jdbc:mysql://localhost:3306/aaa?useUnicode=true&useJDBCCompliantTimezoneShift=true&useLegacyDatetimeCode=false&serverTimezone=UTC";
String login="root";
String password="aaa";
try (Connection conn = DriverManager.getConnection(connectionURL, login, password)) {
Statement stmt = conn.createStatement();
ResultSet rs = stmt.executeQuery("SELECT * FROM Lista1");
while(rs.next()) {
System.out.println(rs.getString("nazwa"));
}
rs.close();
stmt.close();
}
}
catch(SQLException e) {
System.out.println("Blad polaczenia:");
e.printStackTrace();
}
catch(RuntimeException e) {
System.out.println("Blad polaczenia:");
e.printStackTrace();
}

The console log throws:

    Exception in thread "main" java.lang.RuntimeException: 
at cccc.web.Main.main(Main.java:1)

I tried to delete return, but it still threw bad issue.
I tried many things but I don't know how to fix it.
I don't know how to solve it. And I spent many many many times to fight with it!!!!

答案1

得分: 1

异常对象包含两个信息,告诉您发生了什么:异常的具体类型(RuntimeException是一个基类),如果幸运的话,会有一个文本消息告诉您问题所在。

在您的异常处理程序中打印它们。

System.out.println("Exception: " + e);

当然,堆栈跟踪会告诉您导致问题的源文件中的确切行。

这实际上不是“修复”异常的问题,而是理解发生了什么,并编写代码来处理这种情况。

英文:

The exception object contains two things that tell you what happened: the specific type of the exception (RuntimeException is a base), and if you're lucky, a text message telling you what the problem is.

Print them in your exception handler.

    System.out.println("Exception: " + e)

And, of course, the stacktrace tells you the exact line of your source file that caused the problem.

This is not really a matter of 'fixing' an exception, it's a matter of understanding what is going on and writing code to handle that situation.

答案2

得分: 0

以下是翻译好的内容:

当错误告诉您异常发生在第一行时,这非常像是一个解析问题。请看这里:

try (Connection conn = DriverManager.getConnection(connectionURL, login, password)) {
    Statement stmt = conn.createStatement();

    ResultSet rs = stmt.executeQuery("SELECT * FROM Lista1");

    while(rs.next()) {
        System.out.println(rs.getString("nazwa"));
    }
    rs.close();

    stmt.close();
}

上述代码是一个没有catchtry块。它位于外部的另一个try中。但是一旦您使用了try,就需要一个catch。修复此解析错误,让我们看看是否还有其他问题。

英文:

When an error tells you that the exception happened on the first line, that very much looks like a parsing issue. Take a look here:

    try (Connection conn = DriverManager.getConnection(connectionURL, login, password)) {
Statement stmt = conn.createStatement();
ResultSet rs = stmt.executeQuery("SELECT * FROM Lista1");
while(rs.next()) {
System.out.println(rs.getString("nazwa"));
}
rs.close();
stmt.close();
}

The code above is a try without a catch. It's inside an outer try. But once you are using a try, you need a catch. Fix this parsing error and let's see whether there are other problems as well.

huangapple
  • 本文由 发表于 2020年9月23日 07:10:34
  • 转载请务必保留本文链接:https://go.coder-hub.com/64018852.html
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