意外的查询成功 [英] unexpected query success

问题描述

SELECT COUNT (*)
  FROM rps2_workflow
 WHERE     workflow_added > TO_DATE ('01.09.2011', 'dd.mm.yyyy')
       AND workflow_finished < TO_DATE ('wtf', 'dd.mm.yyyy')
       AND workflow_status IN (7, 12, 17)
       AND workflow_worker = 159

由于日期无效，我希望此查询失败，但是它返回0

I expect this query to fail, because of invalid date, but it returns 0

此查询的计划显示，在第8步处理了无效子句 :

The plan for this query shows that on 8th step the invalid clause is processed:

8 TABLE ACCESS BY INDEX ROWID TABLE RPS2.RPS2_WORKFLOW Object Instance: 1  Filter Predicates: ("WORKFLOW_STATUS"=7 OR "WORKFLOW_STATUS"=12 OR "WORKFLOW_STATUS"=17) AND SYS_EXTRACT_UTC("WORKFLOW_FINISHED")<SYS_EXTRACT_UTC(TO_DATE('wtf','dd.mm.yyyy'))  Cost: 11  Bytes: 33  Cardinality: 1  CPU Cost: 8 M  IO Cost: 10  Time: 1                     

如果我们注释掉AND workflow_status IN (7, 12, 17)条件-那么预期我们会得到ORA-01858: a non-numeric character was found where a numeric was expected

If we comment out AND workflow_status IN (7, 12, 17) condition - then expectedly we get ORA-01858: a non-numeric character was found where a numeric was expected

如果我们注释掉AND workflow_finished < TO_DATE ('wtf', 'dd.mm.yyyy')，那么我们将获得满足该条件的记录数量(> 0)

If we comment out AND workflow_finished < TO_DATE ('wtf', 'dd.mm.yyyy') then we get amount of records that fit that conditions (> 0)

这怎么可能?

UPD :

提示/*+no_index(rps2_workflow) */并没有任何改变(而在该计划中，我们看到执行了全扫描)

The hint /*+no_index(rps2_workflow) */ doesn't change anything (whereas in the plan we see that fullscan is performed)

SELECT STATEMENT  ALL_ROWSCost: 254  Bytes: 31  Cardinality: 1  CPU Cost: 34 M  IO Cost: 248  Time: 4       
2 SORT AGGREGATE  Bytes: 31  Cardinality: 1     
    1 TABLE ACCESS FULL TABLE RPS2.RPS2_WORKFLOW Object Instance: 1  Filter Predicates: "WORKFLOW_WORKER"=159 AND ("WORKFLOW_STATUS"=7 OR "WORKFLOW_STATUS"=12 OR "WORKFLOW_STATUS"=17) AND SYS_EXTRACT_UTC("WORKFLOW_ADDED")>SYS_EXTRACT_UTC(TIMESTAMP' 2011-09-01 00:00:00') AND SYS_EXTRACT_UTC("WORKFLOW_FINISHED")<SYS_EXTRACT_UTC(TO_DATE('wtf','dd.mm.yyyy'))  Cost: 254  Bytes: 31  Cardinality: 1  CPU Cost: 34 M  IO Cost: 248  Time: 4  

推荐答案

它可能发现满足所有其他条件的每个记录都有一个NULL workflow_finished字段.

It probably found that every record satisfying all the other conditions have a NULL workflow_finished field.

与NULL相比的任何事物都是未知的，因此不需要评估另一个操作数.

And anything compared to NULL is unknown so it doesn't need to evaluate the other operand.

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