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在 Oracle 11.2 中工作

我已经创建了一个电话号码类型以与员工相关联......我正在尝试学习语法，不想因数据库设计的好坏而受到批评......

这是我的 *.sql 文件

CREATE TYPE AddressType AS OBJECT(streetNumber NUMBER(5), StreetName VARCHAR2(30), city VARCHAR2(20));
/
CREATE TYPE empName AS OBJECT(firstname VARCHAR2(10), middle VARCHAR2(10), lastname VARCHAR2(10));
/
CREATE TYPE PhoneNumbers AS OBJECT(phNumb NUMBER(10), numType VARCHAR2(10));
/
CREATE TYPE ContactNumbers AS VARRAY(5) OF PhoneNumbers;
/

CREATE TABLE Workers(eid NUMBER(5), name empName, loc AddressType,    contactNums ContactNumbers);

INSERT INTO Workers VALUES( 1,
                        empName('Ramos', null, 'Phil'), 
                        AddressType(123, 'A Street', 'San Diego'), 
                        ContactNumbers(
                                    PhoneNumbers(1234567890, 'cell'),
                                    PhoneNumbers(2345678901, 'home')
                                    )
                      );


--Display all members of workers
SELECT * FROM Workers; -- i get something here I'm ok with 

--Display all the names
--SELECT (w.name.firstname||' '||w.name.middle||' '||w.name.lastname) as     "Name" FROM Workers w; --this works

--Display their location
--SELECT (w.loc.streetNumber||' '||w.loc.StreetName||', '||w.loc.city) as "Location" FROM Workers w; -- this works

--Display their phone numbers
SELECT (w.contactNums(1).phNumb||' <'||w.ContactNumbers(1).numType||'>') as "Phone Numbers" FROM Workers w; -- this line I can't figure out

问：我如何写最后一个SELECT以获得电话号码列表？

我希望我的输出看起来像：

phone Numbers
-------------
1234567890 <cell>
2345678901 <home>
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1 回答 1

8

在 a 的情况下，您将使用TABLE运算符VARRAY：

SELECT e.phNumb||' <'||e.numType||'>' as "Phone Numbers"
 FROM Workers w, TABLE(w.contactNums) e;

Phone Numbers                                       
-----------------------------------------------------
1234567890 <cell>                                    
2345678901 <home>

您可以在Database Oject-Relational Developer's Guide中找到有关它的更多信息。

于 2016-08-15T23:25:37.177 回答