大家好,我確實(shí)在MySql查詢中苦苦掙扎,我有一個(gè)名為“ info”的表,并且其中有一個(gè)名為“ rating”的列,評分在1-10之間.

現(xiàn)在,我需要生成一個(gè)百分比值,該百分比值表示1-6、7-8和9-10中有多少評級,但我需要它們拼命顯示,之后我需要第二個(gè)查詢,該查詢可以從中減去結(jié)果的百分比值1-6和9-10.

下面的查詢與我從所有研究中獲得的查詢一樣接近,但是我不知道如何僅獲得1-6的百分比,而不是全部,以及如何獲得第二個(gè)查詢以減去1-6和9-10個(gè)評分百分比.

SELECT rating,

COUNT(*) AS Count,

(COUNT(*) / _total ) * 100 AS Percentege

FROM info,

(SELECT COUNT(*) AS _total FROM info) AS myTotal

GROUP BY rating

解決方法:

select if(rating between 1 and 6, '1-6',

if( rating between 7 and 8, '7-8',

'9-10' )

) as rating_range,

count(1) as num

from info

group by rating_range

編輯：添加舍入和計(jì)算

可以用作子查詢.給定分組依據(jù),您需要單獨(dú)獲取總金額：

select Q.rating_range,

Q.num,

round(Q.num * 100 / Q.total, 2) as percent

from (

select R.*,

(select count(1) from info) as total

from (

select if(rating between 1 and 6, '1-6',

if( rating between 7 and 8, '7-8',

'9-10' )

) as rating_range,

count(1) as num

from info

group by rating_range ) R

) Q

group by Q.rating_range

就相對價(jià)值而言,如果我有一個(gè),我可能會(huì)在外部應(yīng)用程序中做到這一點(diǎn).否則,您可以執(zhí)行我想進(jìn)行的特定查詢：

select Q.rating_range,

Q.num,

round(Q.num * 100 / Q.total, 2) as percent,

round( (Q.num - Q.total_nine_ten) * 100 / Q.total, 2) as diff_from_nine_ten

from (

select R.*,

(select count(1) from info) as total,

(select count(1) from info where rating > 8 ) as total_nine_ten

from (

select if(rating between 1 and 6, '1-6',

if( rating between 7 and 8, '7-8',

'9-10' )

) as rating_range,

count(1) as num

from info

group by rating_range ) R

) Q

group by Q.rating_range

不是很優(yōu)雅,但是可以用

標(biāo)簽：survey,mysql,php

來源： https://codeday.me/bug/20191119/2037466.html

總結(jié)

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