文章目录
一、wrapper 介绍
1、Wrapper 家族
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
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2、创建测试类
@SpringBootTest
public class WrapperTests {
@Resource
private UserMapper userMapper;
}
二、QueryWrapper
例1:组装查询条件
@Test
public void test1() {
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper
.like("name","n")
.between("age", 10, 20)
.isNotNull("email");
List<User> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}
例2:组装排序条件
@Test
public void test2() {
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper
.orderByDesc("age")
.orderByAsc("id");
List<User> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}
例3:组装删除条件
@Test
public void test3() {
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.isNull("email");
int result = userMapper.delete(queryWrapper); //条件构造器也可以构建删除语句的条件
System.out.println("delete return count = " + result);
}
例4:条件的优先级
@Test
public void test4() {
//修改条件
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper
.like("name", "n")
.and(i -> i.lt("age", 18).or().isNull("email")); //lambda表达式内的逻辑优先运算
User user = new User();
user.setAge(18);
user.setEmail("user@qq.com");
int result = userMapper.update(user, queryWrapper);
System.out.println(result);
}
例5:组装select子句
@Test
public void test5() {
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.select("name", "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);//返回值是Map列表
maps.forEach(System.out::println);
}
例6:实现子查询
@Test
public void test6() {
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper.inSql("id", "select id from user where id <= 3");
//selectObjs的使用场景:只返回一列
List<Object> objects = userMapper.selectObjs(queryWrapper);//返回值是Object列表
objects.forEach(System.out::println);
}
但上面的方式容易引发sql注入
queryWrapper.inSql("id", "select id from user where id <= 3 or true"); // 或插叙出所有用户id
可以使用下面的查询方式替换
queryWrapper.in("id", 1, 2, 3 );
// 或
queryWrapper.le("id", 3 );
三、UpdateWrapper
例7:需求同例4
@Test
public void test7() {
//组装set子句
UpdateWrapper<User> updateWrapper = new UpdateWrapper<>();
updateWrapper
.set("age", 18)
.set("email", "user@qq.com")
.like("name", "n")
.and(i -> i.lt("age", 18).or().isNull("email")); //lambda表达式内的逻辑优先运算
//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
User user = new User();
int result = userMapper.update(user, updateWrapper);
System.out.println(result);
}
四、condition
例8:动态组装查询条件
@Test
public void test8() {
//定义查询条件,有可能为null(用户未输入)
String name = null;
Integer ageBegin = 10;
Integer ageEnd = 20;
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
if(StringUtils.isNotBlank(name)){
queryWrapper.like("name","n");
}
if(ageBegin != null){
queryWrapper.ge("age", ageBegin);
}
if(ageEnd != null){
queryWrapper.le("age", ageEnd);
}
List<User> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}
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 test8Condition() {
//定义查询条件,有可能为null(用户未输入)
String name = null;
Integer ageBegin = 10;
Integer ageEnd = 20;
QueryWrapper<User> queryWrapper = new QueryWrapper<>();
queryWrapper
.like(StringUtils.isNotBlank(name), "name", "n")
.ge(ageBegin != null, "age", ageBegin)
.le(ageEnd != null, "age", ageEnd);
List<User> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}
五、LambdaXxxWrapper
例9:Query - 需求同例8
@Test
public void test9() {
//定义查询条件,有可能为null(用户未输入)
String name = null;
Integer ageBegin = 10;
Integer ageEnd = 20;
LambdaQueryWrapper<User> queryWrapper = new LambdaQueryWrapper<>();
queryWrapper
//避免使用字符串表示字段,防止运行时错误
.like(StringUtils.isNotBlank(name), User::getName, "n")
.ge(ageBegin != null, User::getAge, ageBegin)
.le(ageEnd != null, User::getAge, ageEnd);
List<User> users = userMapper.selectList(queryWrapper);
users.forEach(System.out::println);
}
例10:Update - 需求同例4
@Test
public void test10() {
//组装set子句
LambdaUpdateWrapper<User> updateWrapper = new LambdaUpdateWrapper<>();
updateWrapper
.set(User::getAge, 18)
.set(User::getEmail, "user@qq.com")
.like(User::getName, "n")
.and(i -> i.lt(User::getAge, 18).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(result);
}