0

各場所の毎日の合計を別のテーブルに取得しようとしています。

構造は次のとおりです。

reports_location_temp:

Table_Name:           Date:                              Total_Count:
London                2013-05-26 10:49:53                5000
London                2013-05-26 10:49:53                2000
Birmingham            2013-05-26 10:49:53                1000
London                2013-05-26 10:49:53                5000
Manchester            2013-05-26 10:49:53                50
Birmingham            2013-05-26 10:49:53                500

reports_location_total_daily:

Table_Name:           Date:                              Total_Count:
London                2013-05-26 23:55:00                12000
Manchester            2013-05-26 23:55:00                50
Birmingham            2013-05-26 23:55:00                1500

私はまだMysqlの使い方を学んでいます。

これは私が試したクエリですが、Unique Table_Nameごとに1つの列しか選択しませんでした:

UPDATE reports_Location_total_daily j1 INNER JOIN reports_location_temp l1 ON j1.Table_Name = l1.Table_Name SET j1.Total_Count = l1.Total_Count

ご協力いただきありがとうございます。

4

1 に答える 1

0 
CREATE TABLE sales
(id INT NOT NULL AUTO_INCREMENT,
location VARCHAR(40), 
today DATETIME NOT NULL,
sales INT NOT NULL,
PRIMARY KEY (id)
)
;


INSERT sales (location,today,sales) VALUES ('London','2013-05-26',2000);
INSERT sales (location,today,sales) VALUES ('Birm','2013-05-26',1000);
INSERT sales (location,today,sales) VALUES ('London','2013-05-26',1500);
INSERT sales (location,today,sales) VALUES ('London','2013-05-24',100);
INSERT sales (location,today,sales) VALUES ('Birm','2013-05-24',200);
INSERT sales (location,today,sales) VALUES ('London','2013-05-24',300);

CREATE TABLE daily_totals
(id INT NOT NULL AUTO_INCREMENT,
location VARCHAR(40), 
today DATETIME NOT NULL,
totalsales INT NOT NULL,
PRIMARY KEY (id)
)
;

DELETE FROM daily_totals;

INSERT INTO daily_totals (location,today,totalsales)
SELECT location,
DATE(today),
SUM(sales)
FROM sales
GROUP BY location,DATE(today)
于 2013-05-26T12:16:16.177 に答える