Question

我有一个查询插入语句，如下所示

struct win32_struct
{
  DWORD stuff;
  char data[1];
};

#pragma pack(push, 1)
template <size_t DATA_SIZE>
struct my_struct
{
  win32_struct base;
  char rest_of_data[DATA_SIZE - (sizeof(win32_struct) - offsetof(win32_struct, data))];
};
#pragma pack(pop)

这里我们有case语句，它引用相同的表（TABLE_A）但在每个语句中检查不同的值（XXX，YYY等等）。这花了很多时间。我们正在尝试优化此insert into my_table select col1,col2,CASE WHEN EXISTS (SELECT 1 FROM Table_A m WHERE m.COLA='XXX' AND m.id=b.id ) THEN (SELECT m.COLB FROM Table_A m WHERE m.COLA='XXX' AND m.id=b.id ) ELSE 'Not Open' END XXX, CASE WHEN EXISTS (SELECT 1 FROM Table_A m WHERE m.COLA='YYY' AND m.id=b.id ) THEN (SELECT m.COLB FROM Table_A m WHERE m.COLA='YYY' AND m.id=b.id ) ELSE 'Not Open' END YYY, . . . . . -- ~150 case statement from Test b,table_main c where b.id=c.id查询。有人可以在这帮忙吗？

Answer 1

尝试透视查询的变体：

<checkbox-input checked.bind="selectAll"></checkbox-input>

toggleAll()
    {
        this.items.forEach( item =>
        {
            item.selected = this.selectAll;
        });
        this.showHideToolbar();
    }

Answer 2

使用decode和LEFT JOIN重写如下。

  SELECT col1, col2 , 
  decode ( m.COLA , 'XXX' , m.COLB , 'Not Open' ) XXX ,
  decode ( m.COLA , 'YYY',  m.COLB , 'Not Open' ) YYY ,
  ...
  ...
  FROM Test b LEFT JOIN Table_A m ON m.id=b.id JOIN table_main c
  ON b.id=c.id;

Answer 3

这需要时间，因为您要对表格进行多次调用以获取您的值。只需减少对Table_A的调用次数。

如果条件Table_A.id = m.id且m.COLA ='Any Value'将始终返回一行，那么您可以将表Table_A，Test和Table_main连接到单个连接中，如下所示

Insert into my_table(col1, col2) values
select 
decode(a.COLA,'XXX' , a.COLB , 'Not Open'),
decode(a.COLA,'YYY' , a.COLB , 'Not Open'),
...
...
from Table_A a, Test b, Table_main c
where a.id = b.id
and b.id = c.id

如果条件可能返回多个记录，那么您可以使用如下的With子句

Insert into my_table(col1, col2) values
with table_a as (select col1, col2 , a.COLA, a.COLB
from Table_A a, Test b, Table_main c
where a.id = b.id
and b.id = c.id 
group by col1, col2,a.COLA, a.COLB)
select col1,col2,
decode(x.COLA,'XXX' , x.COLB , 'Not Open'),
decode(x.COLA,'YYY' , x.COLB , 'Not Open'),
....
....
from table_a x

oracle中Case语句的替代方法

3 个答案: