Wrapper介绍
Wrapper
: 条件构造抽象类,最顶端父类
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AbstractWrapper
: 用于查询条件封装,生成 sql 的 where 条件
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QueryWrapper
: 查询条件封装 -
UpdateWrapper
: Update 条件封装 -
AbstractLambdaWrapper
: 使用Lambda 语法
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LambdaQueryWrapper
:用于Lambda语法使用的查询Wrapper -
LambdaUpdateWrapper
: Lambda 更新封装Wrapper
QueryWrapper
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)
@Test
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);
}
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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
@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
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.orderByDesc("age").orderByAsc("uid");
List<User> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}
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组装删除条件
执行SQL: UPDATE t_user SET is_deleted=1 WHERE is_deleted=0 AND (email IS NULL)
@Test
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);
}
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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)
@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
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.gt("age", 20).like("username", "a").or().isNull("email");
User user = new User();
user.setName("test04");
user.setEmail("[email protected]");
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);
}
控制台打印查询结果
执行SQL: UPDATE t_user SET username=?, email=? WHERE is_deleted=0 AND (username LIKE ? AND (age > ? OR email IS NULL))
@Test
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("test05");
user.setEmail("[email protected]");
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);
}
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组装select子句
执行SQL: SELECT username,age FROM t_user WHERE is_deleted=0
@Test
public void test06() {
//查询用户信息的username和age字段
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.select("username", "age");
List<Map<String, Object>> maps = userMapper.selectMaps(queryWrapper);
maps.forEach(System.out::println);
}
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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))
@Test
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> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}
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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
执行SQL:
@Test
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", "[email protected]");
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);
}
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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 <= ?)
@Test
public void test09() {
//定义查询条件,有可能为null(用户未输入或未选择)
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("username", 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);
}
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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), "username", username)
.ge(ageBegin != null, "age", ageBegin)
.le(ageEnd != null, "age", ageEnd);
List<User> list = userMapper.selectList(queryWrapper);
list.forEach(System.out::println);
}
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。
@Test
public void test11() {
//定义查询条件,有可能为null(用户未输入)
String username = "a";
Integer ageBegin = null;
Integer ageEnd = 30;
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, "a")
.ge(ageBegin != null, User::getAge, ageBegin)
.le(ageEnd != null, User::getAge, ageEnd);
List<User> list = userMapper.selectList(queryWrapper);
list.forEach(System.out::println);
}
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。
@Test
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")
//lambda表达式内的逻辑优先运算
.and( i -> i.gt(User::getAge, 20).or().isNull(User::getEmail))
//组装set子句
.set(User::getEmail, "[email protected]");
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);
}