在where/having子句之外使用布尔表达式。

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英文:

Using boolean expression outside of a where/having clause

问题

SQL Server returns the result rowlett@bikes.shop for the expression email=rowlett@bikes.shop because it is treating the expression as a comparison, not a mathematical operation. In SQL, when you use the = operator in a SELECT statement, it is used for comparison, not for mathematical equality.

So, when you write email=rowlett@bikes.shop, SQL Server interprets it as "Is the email column equal to the string 'rowlett@bikes.shop'?" Since it's a comparison, it returns the value of the email column for each row where the condition is true, which in this case is just 'rowlett@bikes.shop'.

SQL Server doesn't convert it to something like 1 because it doesn't have a boolean type, and it treats the result as a string literal that matches the condition you specified.

英文:

I have the following data:

store_id	store_name	phone	email	street	city	state	zip_code
1	Santa Cruz Bikes	(831) 476-4321	santacruz@bikes.shop	3700 Portola Drive	Santa Cruz	CA	95060
2	Baldwin Bikes	(516) 379-8888	baldwin@bikes.shop	4200 Chestnut Lane	Baldwin	NY	11432
3	Rowlett Bikes	(972) 530-5555	rowlett@bikes.shop	8000 Fairway Avenue	Rowlett	TX	75088

If I write a query such as:

select * from sales.stores where email='rowlett@bikes.shop'
-- store_id	store_name	phone	email	street	city	state	zip_code
-- 3	Rowlett Bikes	(972) 530-5555	rowlett@bikes.shop	8000 Fairway Avenue	Rowlett	TX	75088

I get the expected result with one row. However, if I use the exact same expression in the select list, I get what seems like gibberish:

SELECT email='rowlett@bikes.shop' from sales.stores where email='rowlett@bikes.shop'
-- email
-- rowlett@bikes.shop

Why does SQL Server return the result rowlett@bikes.shop for the expression email=rowlett@bikes.shop ? I understand it doesn't have the boolean type, but I'd think it would be converted to something like 1. Why does that occur?

答案1

得分: 2

根据@Martin Smith在评论中指出的,无论您使用:

SELECT email = 'rowlett@bikes.shop' ...

还是

SELECT 'rowlett@bikes.shop' AS email ...

您都会得到相同的查询结果 - 即,在您的查询结果集中将会有一个名为'email'的列,每一行的值都是'rowlett@bikes.shop'。

|email               |column2|column3|
 -------------------- ------- -------
|'rowlett@bikes.shop'|value1 |another value1|
|'rowlett@bikes.shop'|value2 |another value2|
等等。

如果您想进一步回答您的问题(假设您实际上希望在查询结果中返回一个真/假值),那么您可以使用CASE表达式来实现 - 例如:

SELECT
CONVERT(bit,
CASE 
  WHEN email = 'rowlett@bikes.shop' 
  THEN 1 
  ELSE 0 
END
)
FROM sales.stores;

并扩展到查看一些潜在的结果:

SELECT
email,
CONVERT(bit,
CASE 
  WHEN email = 'rowlett@bikes.shop' 
  THEN 1 
  ELSE 0 
END
) AS EmailCheck
FROM sales.stores;

您将会看到类似以下的结果:

|email                |EmailCheck |
|'rowlett@bikes.shop' |1 |
|'someone@xyz.com'    |0 |
|'someoneelse@xyz.com'|0 |
英文:

As @Martin Smith pointed out in the comments, whether you use:

SELECT email = 'rowlett@bikes.shop' ...

or

SELECT 'rowlett@bikes.shop' AS email ...

you'll get the same query result - that is, you'll get a column named 'email' in your query resultset, with 'rowlett@bikes.shop' as the value for each row.

|email               |column2|column3|
 -------------------- ------- -------
|'rowlett@bikes.shop'|value1 |another value1|
|'rowlett@bikes.shop'|value2 |another value2|
etc.

To go one step further beyond your question (making the assumption that you actually do want to have a true/false value returned in your query results), then you can use a CASE expression to achieve that - eg.

SELECT
CONVERT(bit,
CASE 
  WHEN email = 'rowlett@bikes.shop' 
  THEN 1 
  ELSE 0 
END
)
FROM sales.stores;

And expanding that to see some potential results:

SELECT
email,
CONVERT(bit,
CASE 
  WHEN email = 'rowlett@bikes.shop' 
  THEN 1 
  ELSE 0 
END
) AS EmailCheck
FROM sales.stores;

you'd see something like:

|email                |EmailCheck |
|'rowlett@bikes.shop' |1 |
|'someone@xyz.com'    |0 |
|'someoneelse@xyz.com'|0 |

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  • 本文由 发表于 2023年6月29日 12:31:20
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