Question

我有以下SQL语句：

select
    DOCUMENT.DOCUMENT_ID,
    (case
        when DOCUMENT.CLASSIFICATION_CODE is not null
        then DOCUMENT.CLASSIFICATION_CODE
        else TEMPLATE.CLASSIFICATION_CODE end) CLASSIFICATION_CODE,
    CLASSIFICATION.NAME CLASSIFICATION_NAME
from
    DOCUMENT,
    TEMPLATE,
    CLASSIFICATION
where
    DOCUMENT.TEMPLATE_ID = TEMPLATE.TEMPLATE_ID and
    DOCUMENT.CLASSIFICATION_CODE = CLASSIFICATION.CLASSIFICATION_CODE(+)

我想做的是：

if DOCUMENT.CLASSIFICATION_CODE IS NOT NULL THEN 
  DOCUMENT.CLASSIFICATION_CODE = CLASSIFICATION.CLASSIFICATION_CODE(+),
ELSE
  TEMPLATE.CLASSIFICATION_CODE = CLASSIFICATION.CLASSIFICATION_CODE(+)

这可能吗？

Answer 1

您可以使用NVL或更标准的COALESCE：

COALESCE(DOCUMENT.CLASSIFICATION_CODE, TEMPLATE.CLASSIFICATION_CODE)
      = CLASSIFICATION.CLASSIFICATION_CODE(+)

在ANSI语法中：

select
    DOCUMENT.DOCUMENT_ID,
    COALESCE(DOCUMENT.CLASSIFICATION_CODE, TEMPLATE.CLASSIFICATION_CODE) 
        CLASSIFICATION_CODE,
    CLASSIFICATION.NAME CLASSIFICATION_NAME
from
    DOCUMENT
inner join 
    TEMPLATE on DOCUMENT.TEMPLATE_ID = TEMPLATE.TEMPLATE_ID
left join
    CLASSIFICATION on
         COALESCE(DOCUMENT.CLASSIFICATION_CODE, TEMPLATE.CLASSIFICATION_CODE) 
       = CLASSIFICATION.CLASSIFICATION_CODE

Answer 2

虽然您可以使用coalesce或nvl，但正如其他答案所示，我更愿意两次加入classification表。是的，引擎必须执行额外的连接，但是避免这些功能允许引擎使用索引。

SELECT document.document_id,
       COALESCE (d.classification_code, t.classification_code)
          AS classification_code,
       COALESCE (d_c.name, t_c.name) AS classification_name
FROM   document d
       JOIN template t ON d.template_id = t.template_id
       LEFT JOIN classification d_c
          ON d.classification_code = d_c.classification_code
       LEFT JOIN classification t_c
          ON t.classification_code = t_c.classification_code

Answer 3

您可以使用：

NVL( DOCUMENT.CLASSIFICATION_CODE, TEMPLATE.CLASSIFICATION_CODE )
COALESCE( DOCUMENT.CLASSIFICATION_CODE, TEMPLATE.CLASSIFICATION_CODE )
或CASE WHEN DOCUMENT.CLASSIFICATION_CODE IS NOT NULL THEN DOCUMENT.CLASSIFICATION_CODE ELSE TEMPLATE.CLASSIFICATION_CODE END（与SELECT子句中使用的完全相同）

作为连接条件，如下所示：

SELECT d.DOCUMENT_ID,
       NVL( d.CLASSIFICATION_CODE, t.CLASSIFICATION_CODE )
         AS CLASSIFICATION_CODE,
       c.NAME AS CLASSIFICATION_NAME
FROM   DOCUMENT d
       INNER JOIN TEMPLATE t
       ON ( d.TEMPLATE_ID = t.TEMPLATE_ID )
       LEFT OUTER JOIN CLASSIFICATION c
       ON ( NVL( d.CLASSIFICATION_CODE, t.CLASSIFICATION_CODE )
              = c.CLASSIFICATION_CODE )

Oracle SQL条件连接

3 个答案: