Sql 使用返回多个值的函数更新多个列

我试图用函数的多个返回值更新一个表

我已经创建了一个类型

CREATE OR REPLACE TYPE city_state AS OBJECT
(
  city VARCHAR2(30),
  state VARCHAR2(2)
);
/

我有一个函数，返回这种类型的变量

CREATE OR REPLACE FUNCTION closestcity(lat IN NUMBER, lon IN NUMBER) RETURN city_state IS
...

我需要更新一个包含城市、州、纬度和经度列的表。对于lat/lon，我应该调用函数并使用结果更新城市/州的值。我只想为每一行调用一次函数，并且只有一些行需要更新（假设city为NULL）

这就是我目前得到的

UPDATE (SELECT * FROM t t1 WHERE city IS NULL)
SET (city, state) = (
                     SELECT newcity.city, newcity.state FROM
                       ( SELECT closestcity(latitude, longitude) newcity
                         FROM t t2
                         WHERE t1.latitude = t2.latitude AND
                               t1.longitude = t2.longitude)
                    );

但我得到一个无效的标识符错误。我觉得我把它复杂化了，正确的方法是什么？

语法应该是：

UPDATE <table1>
set (<column1>, <column2>) = (select <column1>,<column2> 
                              from <table2> 
                              where <where clause table2 > )
where <where clause table1>

UPDATE t1
SET (city, state) = (SELECT closestcity(latitude, longitude) newcity
                            ,state
                       FROM t t2
                      WHERE t1.latitude = t2.latitude 
                        AND t1.longitude = t2.longitude)
where city IS NULL

---

出现该错误的原因是Oracle试图限定

SELECT newcity.city，newcity.state

SELECT语句中引用的

newcity

。它找不到对象

newcity

，因此抛出错误。如果确实需要，还可以将内联视图别名为

t1

。作为@Rene，您可以成功地将其替换为表名。为此，您可以按如下方式重写update语句：

UPDATE (SELECT * FROM t t1 WHERE city IS NULL) t1
   SET (city, state) = (
                         SELECT closestcity(latitude, longitude).city 
                              , closestcity(latitude, longitude).state
                           FROM t t2
                          WHERE t1.latitude = t2.latitude 
                            AND t1.longitude = t2.longitude 
                        );

或者简单地说：

UPDATE t t1 
   SET city  = closestcity(latitude, longitude).city 
     , state = closestcity(latitude, longitude).state
where t1.city is null

更新#1

update (select nd.newdata.city  as newcity
             , nd.newdata.state as newstate
             , city
             , state
          from (                             
                 SELECT closestcity(latitude, longitude) newdata
                      , city
                      , state
                   FROM t
                  where city is null
                ) nd
              ) q  
set q.city  = q.newcity
  , q.state = q.newstate 

只要一个变化

UPDATE (SELECT * FROM t WHERE city IS NULL) **t1**
SET (city, state) = (
                     SELECT newcity.city, newcity.state FROM
                       ( SELECT closestcity(latitude, longitude) newcity
                         FROM t t2
                         WHERE t1.latitude = t2.latitude AND
                               t1.longitude = t2.longitude)
                    );

---

我得到了

ORA-00904:“T1”。“经度”：无效标识符

这是可行的，但我最终调用了closestcity两次-这是一个非常重的函数State是newcity的一个变量（就像city一样），我无法直接选择它