mysql - 选择给定年份的金额

Question

我有一张表格，上面有国家、年份和每年的金额。我正在尝试编写一个查询，将给定年份的总金额相加并将其作为一列提供给我。

例如，要具有如下所示的输出：

Country   1991     1992     1993
France     25       12       38

我尝试了这样的查询，但无济于事：

SELECT 
     country as "Recipient", 
     SUM(usd_amount) AS "1991" WHEN project_year = 1991,
     SUM(usd_amount) AS "1992" WHEN project_year = 1992,
     SUM(usd_amount) AS "1993" WHEN project_year = 1993

FROM tb1

GROUP BY project_year, country
order by country, project_year

有什么建议么？谢谢！克里斯

Answer 1

Give this a try:

SELECT 
     country as "Recipient", 
     SUM(if(project_year = 1991, usd_amount, 0)) AS "1991",
     SUM(if(project_year = 1992, usd_amount, 0)) AS "1992",
     SUM(if(project_year = 1993, usd_amount, 0)) AS "1993",
FROM tb1
GROUP BY country
ORDER BY country

PS: It makes no sense to order by project year as you're already aggregating on it.

Answer 2

尝试以下

SELECT 
     country as "Recipient", 
     SUM(case WHEN project_year = 1991 then usd_amount else 0 end) AS "1991" ,
     SUM(case WHEN project_year = 1992 then usd_amount else 0 end) AS "1992" ,
     SUM(case WHEN project_year = 1993 then usd_amount else 0 end) AS "1993" 

FROM tb1

GROUP BY country, project_year
order by country, project_year