Question

我有问题。我有三个表，名为AUTHORS，BOOKS_AUTHORS和BOOKS，我试图从这三个表中选择Lev Tolstoi写的书。这就是我现在所拥有的，但它不起作用

select   "AUTHORS"."LASTNAME" as "LASTNAME",
     "AUTHORS"."FIRSTNAME" as "FIRSTNAME",
     "BOOKS"."NAME" as "NAME",
 from    "BOOKS_AUTHORS",
     "BOOKS",
     "AUTHORS" 
 where   "AUTHORS"."ID"="BOOKS_AUTHORS"."ID_AUTHOR"
 and     "BOOKS_AUTHORS"."ID_BOOK"="BOOKS"."ID"
 and     "AUTHORS"."LASTNAME"="Tolstoi" 
 and     "AUTHORS"."FIRSTNAME"="Lev"

我在这个名为iacademy3.oracle.com的网站上有表格，错误是

＆＃34; ORA-00936：缺少表达＆＃34;

Answer 1

为什么使用旧的$conn = mysqli_connect("localhost", "root","","ATFlogin");语法，使用正确的显式JOIN语法使其更具可读性，并在需要时更容易扩展到其他连接。

JOIN

Answer 2

在这里看起来你有一个额外的逗号：＆＃34; BOOKS＆＃34;。＆＃34; NAME＆＃34; as＆＃34; NAME＆＃34;， - 我将其删除

select   "AUTHORS"."LASTNAME" as "LASTNAME",
     "AUTHORS"."FIRSTNAME" as "FIRSTNAME",
     "BOOKS"."NAME" as "NAME"
 from    "BOOKS_AUTHORS",
     "BOOKS",
     "AUTHORS" 
 where   "AUTHORS"."ID"="BOOKS_AUTHORS"."ID_AUTHOR"
 and     "BOOKS_AUTHORS"."ID_BOOK"="BOOKS"."ID"
 and     "AUTHORS"."LASTNAME"="Tolstoi" 
 and     "AUTHORS"."FIRSTNAME"="Lev"

从具有多个条件的多个表中选择

2 个答案: