以下代码返回错误,但我不确定原因.需要更改什么才能进行编译？
switch (DAO.class) {
case BookDAO.class:
return bookDAO;
}
如果我写这样的话:
public static void main(String[] args) {
try {
} catch (MalformedURLException e) {
e.printStackTrace()
};
}
Java编译器在catch子句中显示编译错误MalformedURLException.如果我URL url = new URL("HI");在try块中插入行
,抱怨停止.我认为java必须使用包绑定这些已检查的异常.然后我通过CookieManager manager = new CookieManager();在try块中插入" "来尝试另一类java.net包 .再次开始编译错误.
那么JVM如何将这些经过检查的异常与java类绑定在一起来编译这样的编译时错误？
在运行代码时我遇到了这个错误:
Exception in thread "Thread-5" java.lang.NullPointerException at
com.example.example1.service.RecordData.run(RecordData.java:87)
这是我的计划:
public class SearchParser {
public static void main(String[] args) {
String url = "jdbc:mysql://localhost:3306/";
String dbName = "share";
String driver = "com.mysql.jdbc.Driver";
String userName = "root";
String password = "";
try {
List<RecordData> list = new ArrayList<RecordData>();
Connection conn = DriverManager.getConnection(url + dbName,
userName, password);
Statement stmt = conn.createStatement();
ResultSet mysqldata = stmt
.executeQuery("select * from product_links");
while (mysqldata.next()) {
RecordData data = new RecordData();
data.setId(mysqldata.getInt("id"));
data.setDomain_name(mysqldata.getString("domain_name"));
data.setCommon_link_name(mysqldata
.getString("common_link_name"));
data.setHead_element_pattern(mysqldata
.getString("head_element_pattern")); …我是android的新手.这是JSONString我已经存储在我的项目中.
public class MainActivity extends Activity {
MenuFields menuFieldInstance[];
Gson gson = new Gson();
String jsonObjectString = "{\n\n\n [\n \"id\": 22,\n \"menu_id\": 1,\n \"field_type_id\": 1,\n \"c4w_code\": \"1234\",\n \"field_label\": \"Customer No\",\n \"field_values\": \"\",\n \"date_Created\": \"2012-09-16 05:11:23\",\n \"date_modified\": \"2013-11-20 10:33:23\",\n \"is_required\": 0,\n \"is_static\": 1,\n \"field_order\": 1\n ],\n [\n \"id\": 23,\n \"menu_id\": 1,\n \"field_type_id\": 1,\n \"c4w_code\": \"1234\",\n \"field_label\": \"Company Name\",\n \"field_values\": \"\",\n \"date_Created\": \"2012-09-16 05:11:56\",\n \"date_modified\": \"2013-11-20 10:33:23\",\n \"is_required\": 1,\n \"is_static\": 1,\n \"field_order\": 3\n ]\n\n}";
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main); …数学表达
任何人都可以知道这个的原因.因为我只知道++ - 操作,但在这种情况下,运算符是'+ +'并且仍然没有给出错误.
我需要使用Java在字符串中的空格后删除文本。
输入：
0 21343434
2 2323
6 232312
输出：
0
2
6
有人知道怎么做吗？只需编辑字符串即可删除空格后的所有内容？谢谢！
我正在做一个24小时.我有下面的代码.我的问题是int[]时间等于[0, 0, 0]并在每次迭代时打印出来.
一些可能对您有帮助的信息.我有Array TimeCounteras [hours, minutes, seconds].reset方法将选择的int值设置为零,保持不变或增加1.嵌套的while循环用于通过时钟.
public static int hours;
public static int minutes;
public static int seconds;
public static int [] TimeCounter = {hours, minutes, seconds};
//End of day
public static void reset(int [] midNight){
hours = 0;
minutes = 0;
seconds = 0;
}
//End of a minute
public static void resetSeconds(int [] endOfMin){
minutes++;
seconds = 0;
}
//End of a Hour
public static void resetMinutes(int [] endOfHour){ …假设我有一组Employee对象.
单个员工对象具有以下参数:名称,年龄和部门.
我需要按年龄对Employee对象进行排序.
除了一部分,我知道如何做到这一点.
当我使用compareTo方法时,我需要指定如何对整数进行排序.
如何使它们成为整数或列表数组？
编辑
这是我的代码,以进一步阐明我需要做什么.
public class Company {
public static void main(String [] args){
Employee[] e = new Employee[13];
PrimeAgeChecker p = new PrimeAgeChecker();
Department d = new Department();
e[0] = new Employee("Counting Guru",55,"Accounting");
e[1] = new Employee("Counting Pro",45,"Accounting");
e[2] = new Employee("Counting Savvy",40,"Accounting");
e[3] = new Employee("Counting Novice",25,"Accounting");
e[4] = new Employee("Sales Guru",50,"Marketing");
e[5] = new Employee("Sales Pro",48,"Marketing");
e[6] = new Employee("Sales Savvy",38,"Marketing");
e[7] = new Employee("Hiring Guru",58,"Human Resrouces");
e[8] = new Employee("Hiring Pro",47,"Human …