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单击按钮后,我试图向我的 mysql 数据库发送一些信息,但我的屏幕冻结然后退出。这是代码:

    b1.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View v) 
        {
            try
            {
                new SendInfo(email.getText().toString(),password.getText().toString());
            }catch(Exception e){
                error.setText((CharSequence) e);
            }

        }
    });

这是 SendInfo 类:

public SendInfo(String email, String pass) throws Exception
{
    Connection conn = null;
    String url = "jdbc:mysql://ip...:3306/"; // info changed
    String dbName = "db"; //info changed
    String driver = "com.mysql.jdbc.Driver";
    String userName = "user"; //info changed
    String password = "pass"; //info changed
    String table = "users";

    Class.forName(driver).newInstance();
    DriverManager.setLoginTimeout(100);
    conn = DriverManager.getConnection(url+dbName,userName,password);

    Statement st = conn.createStatement();
    java.util.Calendar cal = java.util.Calendar.getInstance();
    java.util.Date utilDate = cal.getTime();
    java.sql.Date sqlDate = new Date(utilDate.getTime());

    st.executeUpdate("INSERT INTO users VALUES('1','Komal','asd',"+sqlDate+")");
    conn.close();

}

我什至没有使用我得到的值作为参数来使它更容易,但它仍然不起作用。

4

1 回答 1

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看起来下面的行产生了错误:

error.setText((CharSequence) e);

将其更改为

error.setText(e.toString());

解决它。

于 2012-12-20T02:10:02.203 回答