返回没有特定值的结果。

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英文:

Return results that don't have a specific value

问题

表格信息:

返回没有特定值的结果。

有2列,personsnumber(人员编号)和their code(他们的代码)。

一个人可以有多个代码。

我的任务是找出没有9210的人。
当我写一个语句 WHERE code != 9210 时,它会返回每个人,即使那个人有9210,但我只需要那些没有9210的数字。

英文:

Table info:

返回没有特定值的结果。

There are 2 columns personsnumber and their code.

a person can have multiple codes.

I was tasked to find people that don't have the 9210
When I write a statement WHERE code != 9210 it returns every person even if that person has 9210, but I only need the numbers that don't had 9210.

答案1

得分: 1

你可以使用 not exists

select t.*
from table t
where not exists (select 1 
                  from table t1 
                  where t1.personsnumber = t.personsnumber and t1.code = 9210
                );

如果只需要 personsnumber,聚合也会有帮助:

select t.personsnumber
from table t
group by t.personsnumber
having sum(case when t.code = 9210 then 1 else 0 end) = 0;
英文:

You can use not exists :

select t.*
from table t
where not exists (select 1 
                  from table t1 
                  where t1.personsnumber = t.personsnumber and t1.code = 9210
                );

Aggregation would also help you if only personsnumber is needed. :

select t.personsnumber
from table t
group by t.personsnumber
having sum(case when t.code = 9210 then 1 else 0 end) = 0;

答案2

得分: 0

你可以使用不存在 (就像Yogesh在他们的回答中所示) 或者左连接 (就像这样):

SELECT *
FROM sometable base
LEFT JOIN sometable code on base.personnumber = code.personnumber and code.code = 9210
WHERE code.personnumber is null
英文:

You can use not exists (like Yogesh shows in their answer) or a left join (like this):

SELECT *
FROM sometable base
LEFT JOIN sometable code on base.personnumber = code.personnumber and code.code = 9210
WHERE code.personnumber is null

答案3

得分: 0

以下是您要翻译的内容:

有很多种方法可以做到这一点。我将一些(不是从图像中全部输入的)数据放入了一个表变量中:

DECLARE @MyTable TABLE (PersonNumber VARCHAR(50), Code VARCHAR(50));
INSERT INTO @MyTable SELECT '4050100', '9210';
INSERT INTO @MyTable SELECT '4050100', '1020';
INSERT INTO @MyTable SELECT '2650100', '1020';

以下是获得您想要的答案的五种方法:

SELECT DISTINCT m1.PersonNumber FROM @MyTable m1 LEFT JOIN @MyTable m2 ON m2.PersonNumber = m1.PersonNumber AND m2.Code = '9210' WHERE m2.PersonNumber IS NULL;
SELECT DISTINCT PersonNumber FROM @MyTable m1 WHERE NOT EXISTS (SELECT * FROM @MyTable m2 WHERE m2.PersonNumber = m1.PersonNumber AND m2.Code = '9210');
WITH x AS (SELECT DISTINCT PersonNumber FROM @MyTable WHERE Code = '9210') SELECT DISTINCT m.PersonNumber FROM @MyTable m LEFT JOIN x ON x.PersonNumber = m.PersonNumber WHERE x.PersonNumber IS NULL;
SELECT PersonNumber FROM @MyTable GROUP BY PersonNumber HAVING MAX(CASE WHEN Code = '9210' THEN 1 END) IS NULL;
SELECT PersonNumber FROM @MyTable GROUP BY PersonNumber HAVING MAX(CASE WHEN Code = '9210' THEN 1 ELSE 0 END) = 0;

简要地说,它们的工作原理如下:

  • 使用LEFT JOIN来检查是否没有相同人员编号的9210代码记录
  • NOT EXISTS - 基本上是相同的逻辑,但可能更容易理解?
  • 使用公共表达式首先识别9210,然后找到没有匹配项的情况
  • 使用聚合
  • 使用更易于理解的查询中的聚合

我进行了非常粗略的分析,发现聚合方法在性能上似乎稍微优于其他方法,其他方法性能大致相当。

英文:

There's many, many ways to do this. I put some (not typing all of that from an image) data into a table variable:

DECLARE @MyTable TABLE (PersonNumber VARCHAR(50), Code VARCHAR(50));
INSERT INTO @MyTable SELECT '4050100', '9210';
INSERT INTO @MyTable SELECT '4050100', '1020';
INSERT INTO @MyTable SELECT '2650100', '1020';

Here's five ways to get the answer you want:

SELECT DISTINCT m1.PersonNumber FROM @MyTable m1 LEFT JOIN @MyTable m2 ON m2.PersonNumber = m1.PersonNumber AND m2.Code = '9210' WHERE m2.PersonNumber IS NULL;
SELECT DISTINCT PersonNumber FROM @MyTable m1 WHERE NOT EXISTS (SELECT * FROM @MyTable m2 WHERE m2.PersonNumber = m1.PersonNumber AND m2.Code = '9210');
WITH x AS (SELECT DISTINCT PersonNumber FROM @MyTable WHERE Code = '9210') SELECT DISTINCT m.PersonNumber FROM @MyTable m LEFT JOIN x ON x.PersonNumber = m.PersonNumber WHERE x.PersonNumber IS NULL;
SELECT PersonNumber FROM @MyTable GROUP BY PersonNumber HAVING MAX(CASE WHEN Code = '9210' THEN 1 END) IS NULL;
SELECT PersonNumber FROM @MyTable GROUP BY PersonNumber HAVING MAX(CASE WHEN Code = '9210' THEN 1 ELSE 0 END) = 0;

Briefly, here's how they work:

  • LEFT JOIN to check there's no record with the 9210 code for the same person number
  • NOT EXISTS - basically the same logic, but might be easier to follow?
  • Using a common-table expression to identify the 9210s first, then find any cases where there's no match
  • Using aggregation
  • Using aggregation in an easier to follow query

I did some very, very rough analysis, and the aggregation methods seem to be slightly better for performance than the other methods, which are all about equal.

答案4

得分: 0

将"Code"列的数据类型转换为nvarchar(50)。

select * from personsnumber where convert(nvarchar(50),code) != '9210';
英文:

You may convert the data type of the Code column

select * from personsnumber where convert(nvarchar(50),code) != '9210'

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  • 本文由 发表于 2020年1月6日 21:43:45
  • 转载请务必保留本文链接:https://go.coder-hub.com/59613211.html
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