1 Wrapper 介绍

Wrapper: 条件构造抽象类,最顶端父类AbstractWrapper: 用于查询条件封装,生成 sql 的 where 条件QueryWrapper: 查询条件封装UpdateWrapper: Update 条件封装AbstractLambdaWrapper: 使用 Lambda 语法LambdaQueryWrapper:用于 Lambda 语法使用的查询 WrapperLambdaUpdateWrapper: Lambda 更新封装 Wrapper
2 QueryWrapper
2.1 组装查询条件
执行 SQL:
SELECT uid AS 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)
public void test01(){
//查询用户名包含 a,年龄在 20 到 30 之间,邮箱信息不为 null 的用户信息
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.like("username","a").between("age",20,30).isNotNull("email");
List<User> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}2.2 组装排序条件
执行 SQL:
SELECT uid AS id,username AS name,age,email,is_deleted FROM t_user WHERE is_deleted=0 ORDER BY age DESC,id ASC
public void test02(){
//查询用户信息,按照年龄的降序排序,若年龄相同,则按照 id 升序排序
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.orderByDesc("age").orderByAsc("id");
List<User> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}2.3 组装删除条件
执行 SQL:
UPDATE t_user SET is_deleted=1 WHERE is_deleted=0 AND (email IS NULL)
public void test03(){
//删除邮箱地址为 null 的用户信息
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.isNull("email");
int result = userMapper.delete(queryWrapper);
System.out.println(result > 0 ? "删除成功!" : "删除失败!");
System.out.println("受影响的行数为:" + result);
}2.4 条件的优先级
执行 SQL:
UPDATE t_user SET user_name=?, email=? WHERE is_deleted=0 AND (age > ? AND user_name LIKE ? OR email IS NULL)
public void test04(){
//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
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.gt("age",20).like("username","a").or().isNull("email");
User user = new User();
user.setName("Oz");
user.setEmail(" test@oz6.com ");
int result = userMapper.update(user, queryWrapper);
System.out.println(result > 0 ? "修改成功!" : "修改失败!");
System.out.println("受影响的行数为:" + result);
}执行 SQL:
UPDATE t_user SET username=?, email=? WHERE is_deleted=0 AND (username LIKE ? AND (age > ? OR email IS NULL))
public void test05(){
//将用户名中包含有 a 并且(年龄大于 20 或邮箱为 null)的用户信息修改
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.like("username","a").and(i->i.gt("age",20).or().isNull("email"));
User user = new User();
user.setName("Vz7797");
user.setEmail(" test@ss8o.com ");
int result = userMapper.update(user, queryWrapper);
System.out.println(result > 0 ? "修改成功!" : "修改失败!");
System.out.println("受影响的行数为:" + result);
}2.5 组装 select 子句
执行 SQL:
SELECT username,age,email FROM t_user WHERE is_deleted=0
public void test06(){
//查询用户的用户名、年龄、邮箱信息
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.select("username","age","email");
List<Map<String, Object>> maps = userMapper.selectMaps(queryWrapper);
maps.forEach(System.out::println);
}2.6 实现子查询
执行 SQL:
SELECT uid AS id,user_name AS name,age,email,is_deleted FROM t_user WHERE is_deleted=0 AND (uid IN (select uid from t_user where uid ⇐ 100))
public void test07(){
//查询 id 小于等于 100 的用户信息
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.inSql("uid", "select uid from t_user where uid <= 100");
List<User> list = userMapper.selectList(queryWrapper);
list.forEach(System.out::println);
}3 UpdateWrapper
UpdateWrapper 不仅拥有 QueryWrapper 的组装条件功能,还提供了 set 方法进行修改对应条件的数据库信息
另一种写法参考上面的优先级部分。
public void test08(){
//将用户名中包含有 a 并且(年龄大于 20 或邮箱为 null)的用户信息修改
UpdateWrapper<User> updateWrapper = new UpdateWrapper<>();
updateWrapper.like("username","a").and( i -> i.gt("age",20).or().isNull("email")).set("email"," svip@qq.com ");
int result = userMapper.update(null, updateWrapper);
System.out.println(result > 0 ? "修改成功!" : "修改失败!");
System.out.println("受影响的行数为:" + result);
}4 condition
在真正开发的过程中,组装条件是常见的功能,而这些条件数据来源于用户输入,是可选的,因此我们在组装这些条件时,必须先判断用户是否选择了这些条件,若选择则需要组装该条件,若没有选择则一定不能组装,以免影响 SQL 执行的结果
-
思路一
执行 SQL:
SELECT uid AS id,user_name AS name,age,email,is_deleted FROM t_user WHERE is_deleted=0 AND (user_name LIKE ? AND age ⇐ ?)
public void test09(){ String username = "a"; Integer ageBegin = null; Integer ageEnd = 30; QueryWrapper<User> queryWrapper = new QueryWrapper<>(); if(StringUtils.isNotBlank(username)){ //isNotBlank 判断某个字符创是否不为空字符串、不为 null、不为空白符 queryWrapper.like("user_name", username); } if(ageBegin != null){ queryWrapper.ge("age", ageBegin); } if(ageEnd != null){ queryWrapper.le("age", ageEnd); } List<User> list = userMapper.selectList(queryWrapper); list.forEach(System.out::println); } -
思路二
上面的实现方案没有问题,但是代码比较复杂,我们可以使用带 condition 参数的重载方法构建查询条件,简化代码的编写
public void test10(){ String username = "a"; Integer ageBegin = null; Integer ageEnd = 30; QueryWrapper<User> queryWrapper = new QueryWrapper<>(); queryWrapper.like(StringUtils.isNotBlank(username), "user_name", username) .ge(ageBegin != null, "age", ageBegin) .le(ageEnd != null, "age", ageEnd); List<User> list = userMapper.selectList(queryWrapper); list.forEach(System.out::println); }
5 LambdaQueryWrapper
功能等同于 QueryWrapper,提供了 Lambda 表达式的语法可以避免填错列名。
public void test11(){
String username = "a";
Integer ageBegin = null;
Integer ageEnd = 30;
LambdaQueryWrapper<User> queryWrapper = new LambdaQueryWrapper<>();
queryWrapper.like(StringUtils.isNotBlank(username), User::getName, username)
.ge(ageBegin != null, User::getAge, ageBegin)
.le(ageEnd != null, User::getAge, ageEnd);
List<User> list = userMapper.selectList(queryWrapper);
list.forEach(System.out::println);
}6 LambdaUpdateWrapper
功能等同于 UpdateWrapper,提供了 Lambda 表达式的语法可以避免填错列名。
public void test12(){
//将用户名中包含有 a 并且(年龄大于 20 或邮箱为 null)的用户信息修改
LambdaUpdateWrapper<User> updateWrapper = new LambdaUpdateWrapper<>();
updateWrapper.like(User::getName, "a")
.and(i -> i.gt(User::getAge, 20).or().isNull(User::getEmail));
updateWrapper.set(User::getName, "小黑").set(User::getEmail," abc@atguigu.com");
int result = userMapper.update(null, updateWrapper);
System.out.println("result:"+result);
}