一、基于QueryWrapper 组装条件

组装查询条件：

@Test
public void test01(){
    //查询用户名包含a，年龄在20到30之间，并且邮箱不为null的用户信息
    //SELECT id,username AS name,age,email,is_deleted FROM t_user WHERE is_deleted=0 AND (username LIKE ? AND age BETWEEN ? AND ? AND email IS NOT NULL)
    QueryWrapper<User> queryWrapper = new QueryWrapper<>();
    queryWrapper.like("username", "a")
            .between("age", 20, 30)
            .isNotNull("email");
    List<User> list = userMapper.selectList(queryWrapper);
    list.forEach(System.out::println);

组装排序条件:

@Test
public void test02(){
    //按年龄降序查询用户，如果年龄相同则按id升序排列
    //SELECT id,username AS name,age,email,is_deleted FROM t_user WHERE is_deleted=0 ORDER BY age DESC,id ASC
    QueryWrapper<User> queryWrapper = new QueryWrapper<>();
    queryWrapper
            .orderByDesc("age")
            .orderByAsc("id");
    List<User> users = userMapper.selectList(queryWrapper);
    users.forEach(System.out::println);
}

组装删除条件:

@Test
public void test03(){
    //删除email为空的用户
    //DELETE FROM t_user WHERE (email IS NULL)
    QueryWrapper<User> queryWrapper = new QueryWrapper<>();
    queryWrapper.isNull("email");
    //条件构造器也可以构建删除语句的条件
    int result = userMapper.delete(queryWrapper);
    System.out.println("受影响的行数：" + result);
}

and和or关键字使用(修改)：

@Test
public void test04() {
    QueryWrapper<User> queryWrapper = new QueryWrapper<>();
    //将年龄大于20并且用户名中包含有a或邮箱为null的用户信息修改
    //UPDATE t_user SET age=?, email=? WHERE username LIKE ? AND age > ? OR email IS NULL)
    queryWrapper
            .like("username", "a")
            .gt("age", 20)
            .or()
            .isNull("email");
    User user = new User();
    user.setAge(18);
    user.setEmail("[email protected]");
    int result = userMapper.update(user, queryWrapper);
    System.out.println("受影响的行数：" + result);
}

指定列映射查询：

@Test
public void test05() {
    //查询用户信息的username和age字段
    //SELECT username,age FROM t_user
    QueryWrapper<User> queryWrapper = new QueryWrapper<>();
    queryWrapper.select("username", "age");
    //selectMaps()返回Map集合列表，通常配合select()使用，避免User对象中没有被查询到的列值为null
    List<Map<String, Object>> maps = userMapper.selectMaps(queryWrapper);
    maps.forEach(System.out::println);
}

condition判断组织条件:

 @Test
public void testQuick3(){
    
    String name = "root";
    int    age = 18;

    QueryWrapper<User> queryWrapper = new QueryWrapper<>();
    //判断条件拼接
    //当name不为null拼接等于, age > 1 拼接等于判断
    //方案1: 手动判断
    if (!StringUtils.isEmpty(name)){
        queryWrapper.eq("name",name);
    }
    if (age > 1){
        queryWrapper.eq("age",age);
    }
    
    //方案2: 拼接condition判断
    //每个条件拼接方法都condition参数,这是一个比较运算,为true追加当前条件!
    //eq(condition,列名,值)
    queryWrapper.eq(!StringUtils.isEmpty(name),"name",name)
            .eq(age>1,"age",age);   
}

二、 基于 UpdateWrapper组装条件

使用queryWrapper:

@Test
public void test04() {
    QueryWrapper<User> queryWrapper = new QueryWrapper<>();
    //将年龄大于20并且用户名中包含有a或邮箱为null的用户信息修改
    //UPDATE t_user SET age=?, email=? WHERE username LIKE ? AND age > ? OR email IS NULL)
    queryWrapper
            .like("username", "a")
            .gt("age", 20)
            .or()
            .isNull("email");
    User user = new User();
    user.setAge(18);
    user.setEmail("[email protected]");
    int result = userMapper.update(user, queryWrapper);
    System.out.println("受影响的行数：" + result);
}

注意：使用queryWrapper + 实体类形式可以实现修改，但是无法将列值修改为null值！

使用updateWrapper:

@Test
public void testQuick2(){

    UpdateWrapper<User> updateWrapper = new UpdateWrapper<>();
    //将id = 3 的email设置为null, age = 18
    updateWrapper.eq("id",3)
            .set("email",null)  // set 指定列和结果
            .set("age",18);
    //如果使用updateWrapper 实体对象写null即可!
    int result = userMapper.update(null, updateWrapper);
    System.out.println("result = " + result);

}

使用updateWrapper可以随意设置列的值！！