Question

我有2个sql查询，其中一个工作，但另一个给出错误。以下查询效果很好

select /*ordered*/ coupon_address.coupon,merchant_address.id  
from merchant_address, 
     coupon_address, 
     customers c  
WHERE merchant_address.id = coupon_address.merchant_address 
  and c.CUSTOMER_ID = 'temp1' 
  AND sdo_within_distance(c.cust_geo_location,merchant_address.store_geo_location,'distance = 1 unit=MILE') = 'TRUE';

但是以下查询不起作用并发出错误

select /*ordered*/ coupon_address.coupon,merchant_address.id  
from  coupon_address, 
      customers c 
  JOIN merchant_address ON merchant_address.id=coupon_address.merchant_address
 WHERE c.CUSTOMER_ID = 'temp1' 
   AND sdo_within_distance (c.cust_geo_location,merchant_address.store_geo_location, 'distance = 1 unit=MILE') = 'TRUE';

错误是

第1行的错误：ORA-00904：“COUPON_ADDRESS”。“MERCHANT_ADDRESS”：标识符无效

Answer 1

不要将显式和隐式连接语法混合在一起！它们总会导致混乱和错误：

select /*ordered*/ coupon_address.coupon,merchant_address.id
FROM coupon_address
JOIN merchant_address
  ON merchant_address.id=coupon_address.merchant_address
JOIN customers c ON c.CUSTOMER_ID = 'temp1'
WHERE  sdo_within_distance (c.cust_geo_location,
                            merchant_address.store_geo_location, 
                           'distance = 1 unit=MILE') = 'TRUE';

它没有工作的原因是解析器评估查询的顺序。可能未知栏目的表尚未进行评估。

Answer 2

JOIN的优先级高于,子句中的FROM。 ON是JOIN的一部分，所以它只会看到哪些开始加入，包括之前的联接。

所以：

select /*ordered*/ coupon_address.coupon,merchant_address.id  
from  coupon_address, 
      customers c 
  JOIN merchant_address
    ON merchant_address.id=coupon_address.merchant_address

本质上是：

select /*ordered*/ coupon_address.coupon,merchant_address.id  
from  coupon_address, 
      (customers c 
  JOIN merchant_address
    ON merchant_address.id=coupon_address.merchant_address)

和coupon_address尚不在范围内。

正如其他人所说，最好坚持一种联接方式，SQL-92 join关键字或from子句中较早的逗号，并加入where子句中的条件。我更喜欢明确的join语法。

Answer 3

Sagi的回答是正确的。您的版本不起作用的原因是因为FROM子句中逗号的范围规则。哦，我提到过：从不在FROM子句中使用逗号。始终使用明确的JOIN语法。

您的FROM子句评估为：

from coupon_address, 
     (customers c JOIN
      merchant_address
      ON merchant_address.id = coupon_address.merchant_address
     )

我认为这很明显为什么你会收到错误。

2个sql查询有什么区别？

3 个答案: