05、条件构造器和常用接口
5.1、wapper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Wrapper
LambdaUpdateWrapper
:Lambda
更新封装Wrapper
5.2、QueryWrapper
5.2.1、例1:组装查询条件
@Test
public void test01() {
//查询用户名包括e,年龄在20到30之间,邮箱信息不为null的用户信息
//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 BETWEEN ? AND ? AND email IS NOT NULL)
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.like("user_name","e")
.between("age",20,30)
.isNotNull("email");
List<User> list = userMapper.selectList(queryWrapper);
list.forEach(System.out::println);
}
5.2.2、例2:组装排序条件
@Test
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
//SELECT uid AS id,user_name AS name,age,email,is_deleted FROM t_user WHERE is_deleted=0 ORDER BY age DESC,uid ASC
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.orderByDesc("age")
.orderByAsc("uid");
List<User> list = userMapper.selectList(queryWrapper);
list.forEach(System.out::println);
}
5.2.3、例3:组装删除条件
@Test
public void test03() {
//删除邮箱地址为null的用户信息
//UPDATE t_user SET is_deleted=1 WHERE is_deleted=0 AND (email IS NULL)
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.isNull("email");
int result = userMapper.delete(queryWrapper);
System.out.println("result:" + result);
}
5.2.4、例4:条件的优先级
@Test
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
//UPDATE t_user SET user_name=?, email=? WHERE is_deleted=0 AND (age > ? AND user_name LIKE ? OR email IS NULL)
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.gt("age",20)
.like("user_name","a")
.or()
.isNull("email");
User user = new User();
user.setName("田七");
user.setEmail("tianqi@123.com");
int result = userMapper.update(user, queryWrapper);
System.out.println("result:" + result);
}
@Test
public void test05() {
//将(年龄大于20或邮箱为null)并且用户名中包含有a的用户信息修改
//lambda中的条件优先执行
//UPDATE t_user SET user_name=?, email=? WHERE is_deleted=0 AND (user_name LIKE ? AND (age > ? OR email IS NULL))
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.like("user_name", "a")
.and(i->i.gt("age",20).or().isNull("email"));
User user = new User();
user.setName("小明");
user.setEmail("ming@123.com");
int result = userMapper.update(user, queryWrapper);
System.out.println("result:" + result);
}
5.2.5、例5:组装select子句
@Test
public void test06() {
//查询用户的用户名、年龄、邮箱信息
//SELECT user_name,age,email FROM t_user WHERE is_deleted=0
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.select("user_name","age","email");
List<Map<String, Object>> maps = userMapper.selectMaps(queryWrapper);
maps.forEach(System.out::println);
}
5.2.6、例6:实现子查询
@Test
public void test07() {
//查询id小于等于100的用户信息
//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))
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);
}
5.3、UpdateWrapper
@Test
public void test08() {
//将(年龄大于20或邮箱为null)并且用户名中包含有a的用户信息修改
//UPDATE t_user SET user_name=?,email=? WHERE is_deleted=0 AND (user_name LIKE ? AND (age > ? OR email IS NULL))
UpdateWrapper<User> updateWrapper = new UpdateWrapper<>();
updateWrapper.like("user_name","a")
.and(i -> i.gt("age",20).or().isNull("email"));
updateWrapper.set("user_name", "小红").set("email","hong@123.com");
int result = userMapper.update(null, updateWrapper);
System.out.println("result:"+ result);
}
5.4、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:
@Test
public void test09() {
//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 <= ?)
String username = "a";
Integer ageBegin = null;
Integer ageEnd = 30;
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
if (StringUtils.isNotBlank(username)) {
//isNotBlank判断某个字符串是否不为空字符串,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
@Test
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);
}
5.5、LambdaQueryWrapper
@Test
public void test11() {
//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 <= ?)
String username = "a";
Integer ageBegin = null;
Integer ageEnd = 30;
LambdaQueryWrapper<User> lambdaQueryWrapper = new LambdaQueryWrapper<>();
lambdaQueryWrapper.like(StringUtils.isNotBlank(username), User::getName, username)
.ge(ageBegin != null, User::getAge, ageBegin)
.le(ageEnd != null, User::getAge, ageEnd);
List<User> list = userMapper.selectList(lambdaQueryWrapper);
list.forEach(System.out::println);
}
5.6、LambdaUpdateWrapper
@Test
public void test12() {
//将(年龄大于20或邮箱为null)并且用户名中包含有a的用户信息修改
//UPDATE t_user SET user_name=?,email=? WHERE is_deleted=0 AND (user_name LIKE ? AND (age > ? OR email IS NULL))
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, "小红").set(User::getEmail,"hong@123.com");
int result = userMapper.update(null, updateWrapper);
System.out.println("result:"+ result);
}