Question

以下是我的SELECT语句的缩小版本。

SELECT
  start_date,
  IFNULL(@stop_date := HAIRY_CALCULATION(this_date, that_date)), '') as stop_date,
  IF(@stop_date = '', 0, DATEDIFF(@stop_date, start_date)) as elapsed_days
FROM mytable

我想将elapsed_days作为第二列，但这取决于@stop_date变量的存在。

我怎么能做下面这样的事情，我在没有显示的情况下分配变量，所以我可以在后面的列计算中使用变量？

SELECT
  DO_NOT_DISPLAY(@stop_date := IFNULL(HAIRY_CALCULATION(this_date, that_date), '')),
  start_date,
  IF(@stop_date = '', 0, DATEDIFF(@stop_date, start_date)) as elapsed_days,
  @stop_date as stop_date
FROM mytable

Answer 1

您始终可以SELECT SELECT来缩小要传播的数据范围：

SELECT stop_date, elapsed_days FROM
  (SELECT
    start_date,
    IFNULL(@stop_date := HAIRY_CALCULATION(this_date, that_date)), '') as stop_date,
    IF(@stop_date = '', 0, DATEDIFF(@stop_date, start_date)) as elapsed_days
  FROM mytable) as subset

注意什么都没有＆＃34;显示＆＃34;除了你的客户。您正在谈论操纵结果集中的内容。

如果这是您经常做的事情，并且其背后的逻辑不太可能随意改变，您可能希望将其包装在视图中，以便您可以执行以下操作：

SELECT stop_date, elapsed_days FROM mytable_hairy_view

修改：这是一种可能的双程方法：

SELECT
  start_date,
  IF(stop_date = '', 0, DATEDIFF(stop_date, start_date)) AS elapsed_days
  FROM
    (SELECT
      start_date,
      HAIRY_CALCULATION(this_date, that_date) AS stop_date
      FROM mytable) as stop_dates

Answer 2

我找到了一个解决方法。也许不是最好的，但是：

SELECT
  CONCAT(
-- -------- The following are just variable assignments for later use; they're not displayed in this column.
    SUBSTRING(
      (@stop_date := IFNULL(HAIRY_CALCULATION(this_date, that_date), '')),
      0
    ),
-- -------- end of variable assignments for later use
    start_date
  ) as start_date,
  IF(@stop_date = '', 0, DATEDIFF(@stop_date, start_date)) as elapsed_days,
  @stop_date as stop_date
FROM mytable

如何分配变量而不显示？

2 个答案: