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五、条件构造器和常用接口
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
- AbstractWrapper : 用于查询条件封装,生成 sql 的 where 条件
2、QueryWrapper
1、例1:组装查询条件
java
@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);
}
2、例2:组装排序条件
java
@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);
}
3、例3:组装删除条件
java
@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);
}
4、例4:条件的优先级
java
@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("user@test.com");
int result = userMapper.update(user, queryWrapper);
System.out.println("受影响的行数:" + result);
}
java
@Test
public void test04() {
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
//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
//UPDATE t_user SET age=?, email=? WHERE (username LIKE ? AND (age > ? ORemail IS NULL))
//lambda表达式内的逻辑优先运算
queryWrapper
.like("username", "a")
.and(i -> i.gt("age", 20).or().isNull("email"));
User user = new User();
user.setAge(18); user.setEmail("user@test.com");
int result = userMapper.update(user, queryWrapper);
System.out.println("受影响的行数:" + result);
}
5、例5:组装select子句
java
@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);
}
6、例6:实现子查询
java
@Test
public void test06() {
//查询id小于等于3的用户信息
//SELECT id,username AS name,age,email,is_deleted FROM t_user WHERE (id IN(select id from t_user where id <= 3))
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.inSql("id", "select id from t_user where id <= 3");
List<User> list = userMapper.selectList(queryWrapper);
list.forEach(System.out::println);
}
3、UpdateWrapper
java
@Test
public void test07() {
//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
//组装set子句以及修改条件
UpdateWrapper<User> updateWrapper = new UpdateWrapper<>();
//lambda表达式内的逻辑优先运算
updateWrapper
.set("age", 18)
.set("email", "user@test.com")
.like("username", "a")
.and(i -> i.gt("age", 20).or().isNull("email"));
//这里必须要创建User对象,否则无法应用自动填充。如果没有自动填充,可以设置为null
//UPDATE t_user SET username=?, age=?,email=? WHERE (username LIKE ? AND (age > ? OR email IS NULL))
//User user = new User(); //user.setName("张三");
//int result = userMapper.update(user, updateWrapper);
//UPDATE t_user SET age=?,email=? WHERE (username LIKE ? AND (age > ? OR email IS NULL))
int result = userMapper.update(null, updateWrapper);
System.out.println(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:
java
@Test
public void test08() {
//定义查询条件,有可能为null(用户未输入或未选择)
String username = null;
Integer ageBegin = 10;
Integer ageEnd = 24;
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
//StringUtils.isNotBlank()判断某字符串是否不为空且长度不为0且不由空白符(whitespace)构成
if(StringUtils.isNotBlank(username)){
queryWrapper.like("username","a");
}
if(ageBegin != null){
queryWrapper.ge("age", ageBegin);
}
if(ageEnd != null){
queryWrapper.le("age", ageEnd);
}
//SELECT id,username AS name,age,email,is_deleted FROM t_user WHERE (age >= ? AND age <= ?)
List<User> users = userMapper.selectList(queryWrapper);
users.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
java
@Test
public void test08UseCondition() {
//定义查询条件,有可能为null(用户未输入或未选择)
String username = null;
Integer ageBegin = 10;
Integer ageEnd = 24;
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
//StringUtils.isNotBlank()判断某字符串是否不为空且长度不为0且不由空白符(whitespace)构成
queryWrapper
.like(StringUtils.isNotBlank(username), "username", "a")
.ge(ageBegin != null, "age", ageBegin)
.le(ageEnd != null, "age", ageEnd);
//SELECT id,username AS name,age,email,is_deleted FROM t_user WHERE (age >= ? AND age <= ?)
List<User> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}
5、LambdaQueryWrapper
java
@Test
public void test09() {
//定义查询条件,有可能为null(用户未输入)
String username = "a";
Integer ageBegin = 10;
Integer ageEnd = 24;
LambdaQueryWrapper<User> queryWrapper = new LambdaQueryWrapper<>();
//避免使用字符串表示字段,防止运行时错误
queryWrapper
.like(StringUtils.isNotBlank(username), User::getName, username)
.ge(ageBegin != null, User::getAge, ageBegin)
.le(ageEnd != null, User::getAge, ageEnd);
List<User> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}
6、LambdaUpdateWrapper
java
@Test
public void test10() {
//组装set子句
LambdaUpdateWrapper<User> updateWrapper = new LambdaUpdateWrapper<>();
updateWrapper
.set(User::getAge, 18)
.set(User::getEmail, "user@test.com")
.like(User::getName, "a")
.and(i -> i.lt(User::getAge, 24).or().isNull(User::getEmail)); //lambda表达式内的逻辑优先运算
User user = new User();
int result = userMapper.update(user, updateWrapper);
System.out.println("受影响的行数:" + result);
}