一、insert

User user = new User();
user.setId(1);
user.setAge(18);
int rows = userMapper.insert(user);

二、select

1、普通查询

User user = userMapper.selectById(11);

2、批量查询

List<Long> ids = Arrays.asList(123L,124L,125L);
List<User> userList = userMapper.selectBatchIds(ids);

3、名字包含小雨并且年龄小于30

QueryWrapper<User> queryWrapper = new QueryWrapper<User>();
queryWrapper.like("username","小雨").lt("age",30);
List<User> userList = userMapper.selectList(queryWrapper);

4、名字包含小雨并且年龄大于等于20且小于等于40并且email不为空

QueryWrapper<User> queryWrapper = new QueryWrapper<User>();
queryWrapper.like("username","小雨").between("age",20,30).isNotNull("email");
List<User> userList = userMapper.selectList(queryWrapper);

5、名字姓肖或者年龄大雨等于20，按照年龄降序排列，年龄相同按照id生序排列

QueryWrapper<User> queryWrapper = new QueryWrapper<User>();
queryWrapper.likeRight("username","肖")
        .or().ge("age",20).orderByDesc("age").orderByAsc("id");
List<User> userInfoList = userMapper.selectList(queryWrapper);

6、创建日期为2019年10月2日并且直属上级名字为王姓

QueryWrapper<User> queryWrapper = new QueryWrapper<User>();
queryWrapper.apply("date_format(create_time,'%Y-%m-%d')={0}","2019-10-07")
        .inSql("parent_id","select id from user where username like '王%'");
List<User> userList = userMapper.selectList(queryWrapper);

7、只取指定的字段

QueryWrapper<User> queryWrapper = new QueryWrapper<User>();
queryWrapper.select("id","username");
List<User> userList = userMapper.selectList(queryWrapper);

8、只获取一条、返回数量

QueryWrapper<User> queryWrapper = new QueryWrapper<User>();
queryWrapper.like("name","小雨").lt("age",30);
User user = userMapper.selectOne(queryWrapper);
int count = userMapper.selectCount(queryWrapper);

三、update

1、根据id进行更新

//把id为1的用户年龄更新为18，其他属性不变
User user = new User();
user.setId(1);
user.setAge(18);        
int rows = userMapper.updateById(user);

2、条件构造器作为参数进行更新

//把名字为rhb的用户年龄更新为18，其他属性不变
UpdateWrapper<User> updateWrapper = new UpdateWrapper<>();
updateWrapper.eq("name","rhb");
User user = new User();
user.setAge(18);
int rows = userMapper.update(user, updateWrapper);

3、lambda构造器

//把名字为rhb的用户年龄更新为18，其他属性不变
LambdaUpdateWrapper<User> lambdaUpdateWrapper = new LambdaUpdateWrapper<>();
lambdaUpdateWrapper.eq(User::getName, "rhb").set(User::getAge, 18);
int rows = userMapper.update(null, lambdaUpdateWrapper);

四、delete

1、根据id进行删除

int rows = userMapper.deleteById(1351456313578713090L);

2、根据map删除

Map<String, Object> map = new HashMap<>();
map.put("name", "张三");
map.put("age", 25);
int rows = userMapper.deleteByMap(map);

3、以条件构造器为参数删除

QueryWrapper wrapper = new QueryWrapper();
wrapper.eq("name","张三");
wrapper.eq("age",25);
int rows = userMapper.delete(wrapper);

LambdaQueryWrapper<User> wrapper = new LambdaQueryWrapper();
wrapper.eq(User::getName,"张三").eq(User::getAge,25);
int rows = userMapper.delete(wrapper);

五、LambdaQueryWrapper条件构造器

setSqlSelect    SELECT 查询字段    举例
where    WHERE 语句，拼接 + WHERE 条件    
and    AND 语句，拼接 + AND 字段=值    
andNew    AND 语句，拼接 + AND (字段=值)    
or    OR 语句，拼接 + OR 字段=值    .eq(" name “,” 木 子 “). or ( ). eq (” id ", 1 ) => name=“木子” or id=1 注意事项：主动调用or表示下一个方法不是and连接，不调用默认为使用and连接
orNew    OR 语句，拼接 + OR (字段=值)    
eq    等于=    .eq(“name”,“木子”) => name=“木子”
allEq    基于 map 内容等于=    
ne    不等于<>    
gt    大于>    
ge    大于等于>=    
lt    小于<    
le    小于等于<=    
like    模糊查询 LIKE    .like(“name”,“木子”) => name like ‘%木子%’
notLike    NOT LIKE模糊查询    
in    IN 查询    .in(“age”,{1,2,3}) => age in (1,2,3)
notIn    NOT IN 查询    
isNull    NULL 值查询    
isNotNull    IS NOT NULL    
groupBy    分组 GROUP BY    
having    HAVING 关键词    
orderBy    排序 ORDER BY    
orderAsc    Asc 排序 ORDER BY    
orderDesc    DESC 排序 ORDER BY    
exists    EXISTS 条件语句    
notExists    NOT EXISTS 条件语句    
between    BETWEEN 条件语句    .between (“age”,1,2) => age between 1 and 2
notBetween    NOT BETWEEN 条件语句    
addFilter    自由拼接 SQL    
last    拼接在最后，例如：last(“LIMIT 1”)