Question

CREATE TABLE Landmark
(
   Indexc NUMBER
   Birth DATE
   Address VARCHAR2(50 BYTE)
   City VARCHAR2(30 BYTE)
   State CHAR(2 BYTE)
   Zip VARCHAR2(15 BYTE)
   ...
   Other properties
)  

ALTER TABLE Landmark ADD (
   CONSTRAINT Landmark_PK PRIMARY KEY (Indexc, Birth));  

CREATE TABLE MapPoint
(
   Indexc NUMBER
   Longitude FLOAT(126)
   Latitude FLOAT(126)
   BuildDate DATE
)  

ALTER TABLE MapPoint ADD (
   CONSTRAINT MapPoint_FK FOREIGN KEY (Indexc) REFERENCES Landmark (Indexc));

Landmark表包含可以不时移动的地标列表。 MapPoint表保存这些地标的经度和纬度。只要地标的属性发生变化，就会在地标表中插入一个新行，该地标的新出生日期为地址。

我有一个脚本会在MapPoint表中为没有物理移动的所有地标将BuildDate设置为null，并且我正在尝试编写将重用最后一行地址的UDPATE语句。以下是我想要的，但不起作用，因为内部最多的子查询将不与update语句相关联。我得到ORA-00904：“lm2”。“索引”：标识符无效。

UPDATE Landmark lm1
   SET (Address, City, State, Zip) = (
   SELECT Address, City, State, Zip
     FROM Landmark lm2
    WHERE lm2.Indexc = lm1.Indexc
      AND lm2.birth = (
      SELECT MIN(Birth)
        FROM (
         SELECT Birth
           FROM Landmark lm3
          WHERE lm3.Indexc = lm2.Indexc
          ORDER BY Birth DESC)
         AND ROWNUM < 3))
 WHERE Indexc IN (
   SELECT Indexc
     FROM MapPoint
    WHERE BuildDate IS NULL);

之前的样本：

Indexc                    Birth        Address          City   State     Zip
------   ----------------------   ------------   -----------   -----   -----
    45    8/16/2009  4:46:21 AM    123 Main St   Springfield      PA   84679
    45    8/18/2009  7:20:27 PM    123 Main St   Springfield      PA   84679
    45    8/20/2009  9:01:44 PM   456 Smith Ln   Springfield      PA   84153
    45   10/31/2009 12:29:07 AM

示例之后：

Indexc                    Birth        Address          City   State     Zip
------   ----------------------   ------------   -----------   -----   -----
    45    8/16/2009  4:46:21 AM    123 Main St   Springfield      PA   84679
    45    8/18/2009  7:20:27 PM    123 Main St   Springfield      PA   84679
    45    8/20/2009  9:01:44 PM   456 Smith Ln   Springfield      PA   84153
    45   10/31/2009 12:29:07 AM   456 Smith Ln   Springfield      PA   84153

Answer 1

这完美无缺。虽然它对于第n个最新项目来说不是那么可扩展。

UPDATE Landmark lm1
   SET (Address, City, State, Zip) = (
   SELECT Address, City, State, Zip
     FROM Landmark lm2
    WHERE lm2.Indexc = lm1.Indexc
      AND lm2.birth = (
      SELECT MAX(Birth)
        FROM Landmark lm3
       WHERE lm3.Indexc = lm2.Indexc
         AND lm3.birth < (
         SELECT MAX(Birth)
           FROM Landmark lm4
          WHERE lm4.Indexc = lm3.Indexc)))
 WHERE Indexc IN (
   SELECT Indexc
     FROM MapPoint
    WHERE BuildDate IS NULL)
   AND Birth = (
   SELECT MAX(Birth)
     FROM Landmark lm2
    WHERE lm2.Indexc = lm1.Indexc);

Answer 2

此解决方案使用分析功能。代码看起来稍微多余，但成本较低，至少在我的测试数据库中，可能是因为子查询较少：

update landmark L1 set (address, city, state, zip) =  
    (select address, city, state, zip from 
       (select  last_value(address ignore nulls) over (partition by cindex order by birth) address, 
                last_value(city ignore nulls) over (partition by cindex order by birth) city, 
                last_value(state ignore nulls) over (partition by cindex order by birth) state, 
                last_value(zip ignore nulls) over (partition by cindex order by birth) zip, 
                cindex, birth
          from landmark) L2
      where l1.cindex = l2.cindex and l2.birth = l1.birth)       
where CIndex IN (
 SELECT CIndex
     FROM MapPoint
    WHERE BuildDate IS NULL)
    and address is null

基本上它是说，从表中获取所有行，并且对于每一行，如果它们不为空，则选择地址值，如果是，则选择最后一个非空值，然后为需要的行更新这些值。更新。

Answer 3

我将“index”替换为“indexc”，因为索引在oracle中不是有效的列名。我认为这可以做你想要的事情：

UPDATE Landmark lm1
 SET (Address, City, State, Zip) = (
 SELECT Address, City, State, Zip
   FROM Landmark lm2
  WHERE lm2.Indexc = lm1.Indexc
    AND lm2.birth = (
    SELECT MAX(Birth)
      FROM Landmark lm3
        WHERE lm3.Indexc = lm2.Indexc
        and lm3.birth < lm2.birth))
 WHERE  (indexc,birth) in (
         select indexc, max(birth)
         from   Landmark
         where  Indexc IN (
           SELECT Indexc
           FROM   MapPoint
           WHERE  BuildDate IS NULL));

使用第二行到最后一行的数据更新最后一行

3 个答案: