我想知道是否有可能这样做(这并不管用):

select cast((exists(select * from theTable where column like 'theValue%') as bit)

似乎它应该是可行的,但很多事情,应该在SQL中工作不;)我已经看到了这种变通方法(SELECT 1,其中…Exists…),但似乎我应该能够仅仅将Exists函数的结果转换为位,并完成它。


当前回答

SELECT IIF(EXISTS(SELECT * FROM theTable WHERE theColumn LIKE 'theValue%'), 1, 0)

其他回答

不行,你得想个变通办法。

如果你必须返回一个条件位0/1,另一种方法是:

SELECT CAST(
   CASE WHEN EXISTS(SELECT * FROM theTable where theColumn like 'theValue%') THEN 1 
   ELSE 0 
   END 
AS BIT)

或者没有石膏:

SELECT
   CASE
       WHEN EXISTS( SELECT 1 FROM theTable WHERE theColumn LIKE 'theValue%' )
            THEN 1 
       ELSE 0 
   END

你可以使用IIF和CAST

SELECT CAST(IIF(EXISTS(SELECT * FROM theTable 
                       where theColumn like 'theValue%'), 1, 0) AS BIT)

您还可以执行以下操作:

SELECT DISTINCT 1
  FROM theTable
 WHERE theColumn LIKE 'theValue%'

如果没有以“theValue”开头的值,将返回null(没有记录),而不是位0

SELECT IIF(EXISTS(SELECT * FROM theTable WHERE theColumn LIKE 'theValue%'), 1, 0)

另一个解决方案是将ISNULL与SELECT TOP 1 1结合使用:

SELECT ISNULL((SELECT TOP 1 1 FROM theTable where theColumn like 'theValue%'), 0)