Question

我有一个包含交易的SQLite数据库，每个交易都有价格和 transDate 。

我想检索按月分组的交易总和。检索到的记录应如下所示：

Price    month
230        2
500        3
400        4

Answer 1

当你按MONTH分组时它总是好的，它也应该检查YEAR

select SUM(transaction) as Price, 
       DATE_FORMAT(transDate, "%m-%Y") as 'month-year' 
       from transaction group by DATE_FORMAT(transDate, "%m-%Y");

FOR SQLITE

select SUM(transaction) as Price, 
       strftime("%m-%Y", transDate) as 'month-year' 
       from transaction group by strftime("%m-%Y", transDate);

Answer 2

SELECT
    SUM(Price) as Price, strftime('%m', myDateCol) as Month
FROM
    myTable
GROUP BY
    strftime('%m', myDateCol)

Answer 3

您可以在月初分组：

select  date(DateColumn, 'start of month')
,       sum(TransactionValueColumn)
from    YourTable
group by 
        date(DateColumn, 'start of month')

Answer 4

尝试以下方法：

SELECT SUM(price), strftime('%m', transDate) as month
FROM your_table
GROUP BY strftime('%m', transDate);

使用SQLite文档中的corresponding page以供将来参考。

Answer 5

在Sqlite中，如果您以unixepoch格式存储日期，则以秒为单位：

select count(myDate) as totalCount, 
strftime('%Y-%m', myDate, 'unixepoch', 'localtime') as yearMonth
from myTable group by strftime('%Y-%m', myDate, 'unixepoch', 'localtime');

如果您以unixepoch格式存储日期（以毫秒为单位）除以1000：

select count(myDate/1000) as totalCount, 
strftime('%Y-%m, myDate/1000, 'unixepoch', 'localtime') as yearMonth
from myTable group by strftime('%Y-%m, myDate/1000, 'unixepoch', 'localtime');

Answer 6

另一种形式：

SELECT SUM(price) AS price,
STRFTIME('%Y-%m-01', created_at) as created_at
FROM records
GROUP BY STRFTIME('%Y-%m-01', created_at);

Answer 7

另一种方法是从列中对年份和月份进行子串并按它们分组。假设格式类似于 2021-05-27 12:58:00，您可以减去前 7 位数字：

SELECT 
    substr(transDate, 1, 7) as YearMonth
    SUM(price) AS price
FROM 
    records
GROUP BY 
    substr(transDate, 1, 7);

在SQLite中按月分组

7 个答案: