[英]SQL count occurrence by column
經過一番研究,我沒有找到我需要的東西,我想我會在這里問。 我目前正在嘗試為應用程序開發高級搜索模式,而我的工作陷於困境。 也許你可以幫我。 因此,假設我有下表:
ID | Name | Surname
1 | John | Mim
2 | Johnny | Crazy
3 | Mike | Something
4 | Milk | Milk
5 | Peter | IDontknow
6 | Mitch | SomeName
然后在我的前端,有一個輸入字段。 該字段的輸入將以這種方式通過查詢:
SELECT name, surname FROM people WHERE name LIKE 'input%' OR surname LIKE 'input%'
現在,假設我的輸入為“ Mi”,因此在“名稱”列中將有3列匹配項,在姓氏中將有2列匹配項。 這就是我想要的。
計算結果如下:
Column | Count
Name | 3
Surname | 2
是否只有一種查詢可以實現此目的?
到目前為止,我已經嘗試過:
我實際上在數據庫的本地主機上創建了上面的表,並嘗試了不同的查詢。 嘗試使用SELECT count(name), count(surname)
,但是兩個計數都將輸出3。 因此,我什至不確定僅在一個查詢中是否可行。
union all
使用union all
SELECT 'name' as col, count(name) as cnt FROM people WHERE name LIKE 'input%'
union all
SELECT 'surname', count(surname) FROM people WHERE surname LIKE 'input%'
在以下情況下使用案例進行自定義分組
SELECT (case when name LIKE 'input%' then 'name'
else 'surname' end) as Column, count(*) as cnt
FROM people WHERE name LIKE 'input%' OR surname LIKE 'input%'
group by Column
嘗試這個:
SELECT "Name" as Column, count(*) as Count FROM people WHERE name LIKE 'mi%'
UNION
SELECT "Surname" as Column, count(*) as Count FROM people WHERE surname LIKE 'mi%'
在Mysql中,布爾值被評估為1或0,因此您可以執行以下操作:
select 'Name' Column, sum(name LIKE 'input%') Count from people
union all
select 'Surname', sum(surname LIKE 'input%') from people
對於Mysql 8.0+,可以避免使用CTE對表進行兩次掃描:
with cte as (
select
sum(name LIKE 'input%') namecounter,
sum(surname LIKE 'input%') surnamecounter
from people
)
select 'Name' Column, namecounter Count from cte
union all
select 'Surname', surnamecounter from cte
沒有people
表的UNION[ ALL]
的解決方案:
SELECT
CASE cj.x WHEN 1 THEN 'Name' ELSE 'Surname' END AS `Column`,
CASE cj.x
WHEN 1 THEN COUNT(CASE WHEN Name LIKE concat(@input, '%') THEN 1 end)
ELSE COUNT(CASE WHEN Surname LIKE concat(@input, '%') THEN 1 END)
END `Count`
FROM people CROSS JOIN (SELECT 1 AS x UNION ALL SELECT 2) AS cj
WHERE Name LIKE concat(@input, '%') OR Surname LIKE concat(@input, '%')
GROUP BY cj.x;
Mi
輸入的輸出:
| Column | Count |
+---------+-------+
| Name | 3 |
| Surname | 2 |
使用SQL Fiddle在線進行測試。
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