插入數據

Question

我一直停留在這個問題上一會兒 ，這裏是有方向

客戶誰已被邀請沒有「蒙娜麗莎」 gala_night現在就邀請到5 -jan-2014蒙娜麗莎gala_night。顯示插入命令和生成的Invite表。

這裏是兩個表

SQL> select * from invite;

GALA_DATE PAINTING_NAME   CUSTID 
----------- ---------------------- ---------- 
10-nov-2013 Watercolors     1430 
15-nov-2013 Woman      1502 
15-nov-2013 Woman      1619 
05-dec-2013 Watercolors     1207 
22-dec-2013 Sunflowers     1806 
22-dec-2013 Sunflowers     1904 
31-dec-2013 Fiddler      1236 
31-dec-2013 Fiddler      1280 
05-jan-2014 Mona_Lisa     1111 
05-jan-2014 Mona_Lisa     1502 
25-jan-2014 Madonna      1806 
25-jan-2014 Madonna      1822 
25-jan-2014 Madonna      1904 
18-feb-2014 Maya       1619 
18-feb-2014 Maya       1822 
18-feb-2014 Maya       1904 
28-feb-2014 Mona_Lisa     1502 
30-apr-2014 Lovers      1207 
30-apr-2014 Lovers      1280 
30-apr-2014 Lovers      1822 
30-apr-2014 Lovers      1904 

SQL> select * from customer;

CUSTID CUSTNAME CUSTBDATE CUST_TYPE BENEFACTOR DOCENT 

---

1301 Disney  01-nov-1980 NM  No  No 
    1806 Garcia  31-dec-2000 VIP  Yes  No 
    1502 LaGardia 15-jan-1960 VIP  Yes  Yes 
    1207 Perry  20-jan-1960 VIP  Yes  Yes 
    1280 Beecham  31-dec-1979 VIP  Yes  No 
    1822 Becker  30-jan-1967 VIP  Yes  Yes 
    1140 Klim  05-apr-1990 NM  No  No 
    1509 Roberts  21-jul-1989 VIP  Yes  No 
    1619 Robins  20-feb-1990 VIP  Yes  Yes 
    1111 Bardot  28-feb-1970 VIP  No  No 
    1515 David  18-apr-1980 NM  No  No 
    1701 Martin  20-aug-1972 RM  No  No 
    1904 Gross  30-sep-1975 VIP  Yes  Yes 
    1236 Brooks  23-sep-1975 VIP  Yes  No 
    1430 Todd  15-jul-1982 VIP  Yes  Yes 

我試着做創建

insert into invite(gala_date,painting_name,custid) 
select invite.gala_date,invite.painting_name,invite.custid 
from invite,customer 
where invite.gala_date='05 Jan 2014' and invite.painting_name='Mona_Lisa' 
and customer.custid=invite.custid 
and customer.custid not in 
(select custid from invite where gala_date in('05 Jan 2014') 
and painting_name in('Mona_Lisa')); 

0行。

但是，正如你所看到的結果產量「0行創建」 任何想法？ 謝謝！

Answer 1

你只需要使用一個不存在得到你想要的結果。

insert into invite(gala_date,painting_name,custid) 
select '01/05/2014', 'Mona_Lisa', customer.custid 
From customer 
Where Not Exists (select '' from invite where invite.custid = customer.custid and painting_name = 'Mona_Lisa') 

Answer 2

試試這個

insert into invite(gala_date,painting_name,custid) 
select '01/05/2014', 'Mona_Lisa', a.custid 
from customer a 
left join invite b on (a.custid = b.custid and b.PaintingName = 'Mona_Lisa' 
where b.custid is null 

我認爲這將讓你關閉，可能需要調整它，因爲我沒有檢查