SQL : How can I get count from multiple tables into one query? - mysql

Question

I have 3 tables, one for website connection errors, one for successful website connections and another with name/location of each specific website.
Table 1 has WebsiteClass_ID, Website_ID and Error_Date
Table 2 has WebsiteClass_ID, Website_ID and Success_Date
Table 3 has WebsiteClass_ID, Website Name and Location
I need to return the rate of error by WebsiteClass_ID by Website_ID per day. To do this, I need the count of errors per WebsiteClass_ID, Website_ID and Date from Table 1 and the count of successful connections per WebsiteClass_ID, Website_ID and Date from Table 2. I still need to return Website Name and Location from table 3 as well. The date field is different in Table 1 than it is in Table 2.
I can easily get the count for each in two different queries but would prefer to accomplish this in one query to avoid extra work in Excel. I created the two individual queries below but do not know how to merge them.
#QUERY
#TITLE-WEBSEROR
#SUBJECT-WEBSITE ERRORS PER DAY BY CLASS AND ID
SELECT
A.WEBSITE_CLASS_ID AS WEBSITE_CLASS_ID
,A.WEBSITE_ID AS WEBSITE_ID
,A.ERROR_DATE AS DATE_OF_ERROR
,COUNT(A.EVENT_NAME) AS NUMBER_OF_ERRORS
,B.NAME AS WEBSITE_NAME
,B.LOCATION AS COMPANY_LOCATION
FROM
&DATABASE..ERRORS A
,&DATABASE..DETAILS B
WHERE
A.WEBSITE_ID = B.WEBSITE_ID
GROUP BY A.WEBSITE_CLASS_ID, A.WEBSITE_ID, A.ERROR_DATE, B.NAME, B.LOCATION
#QUERY
#TITLE-WEBSCNFM
#SUBJECT-SUCCESSUL CONNECTIONS PER DAY BY CLASS AND ID
SELECT
C.WEBSITE_CLASS_ID AS WEBSITE_CLASS_ID
,C.WEBSITE_ID AS WEBSITE_ID
,DATE(C.SUCCESS_DATE) AS SUCCESSFUL_CONNECTION
,COUNT(C.SUCCESS) AS COUNT_SUCCESS_CNCTN
,B.NAME AS WEBSITE_NAME
,B.LOCATION AS COMPANY_LOCATION
FROM
&DATABASE..SUCCESS C
,&DATABASE..DETAILS B
WHERE
C.WEBSITE_ID = B.WEBSITE_ID
GROUP BY C.WEBSITE_CLASS_ID, C.WEBSITE_ID, DATE(C.SUCCESS_DATE), B.NAME, B.LOCATION
Data Sample:
Table 1: Errors
Table 2: Success
Table 3: Details
Expected Results :
Website_Class_ID
Website_ID
Date of Error or Success
Count of Errors
Count of Success
Website Name
Website Location
ClassB
ID 2
12/1/2019
3
5
Website #1
USA
ClassC
ID 3
12/2/2019
1
6
Website #2
Canada
SELECT
`Errors$`.WEBSITE_CLASS_ID
,`Errors$`.WEBSITE_ID
,`Errors$`.ERROR_DATE
,COUNT(`Errors$`.EVENT_NAME)
,`Details$`.NAME
,`Details$`.LOCATION
FROM
`D:\mike\SnapCommerce Case Study\Data.xlsx`.`Errors$` `Errors$`,
INNER JOIN `D:\mike\SnapCommerce Case Study\Data.xlsx`.`Details$`
`Details$`
ON `Details$`.WEBSITE_ID = `Errors$`.WEBSITE_ID
GROUP BY `Errors$`.WEBSITE_CLASS_ID, `Errors$`.WEBSITE_ID,
`Errors$`.ERROR_DATE, `Details$`.NAME, `Details$`.LOCATION
UNION
SELECT
`Success$`.WEBSITE_CLASS_ID
,`Success$`.WEBSITE_ID
,DATE(`Success$`.SUCCESS_DATE)
,COUNT(`Success$`.SUCCESS)
,`Details$`.NAME
,`Details$`.LOCATION
FROM
`D:\mike\SnapCommerce Case Study\Data.xlsx`.`Success$` `Success$`,
INNER JOIN `D:\mike\SnapCommerce Case Study\Data.xlsx`.`Details$`
`Details$`
ON `Details$`.WEBSITE_ID = `Success$`.WEBSITE_ID
GROUP BY `Success$`.WEBSITE_CLASS_ID, `Success$`.WEBSITE_ID,
`Success$`.SUCCESS_DATE, `Details$`.NAME, `Details$`.LOCATION

Answer 1

Vertically, You can use UNION, this would eliminate doubles as well.
If you need them use UNION ALL
SELECT
A.WEBSITE_CLASS_ID AS WEBSITE_CLASS_ID
,A.WEBSITE_ID AS WEBSITE_ID
,A.ERROR_DATE AS DATE_OF_ERROR
,COUNT(A.EVENT_NAME) AS NUMBER_OF_ERRORS
,B.NAME AS WEBSITE_NAME
,B.LOCATION AS COMPANY_LOCATION
FROM
&DATABASE..ERRORS A
INNER JOIN &DATABASE..DETAILS B
ON A.WEBSITE_ID = B.WEBSITE_ID
GROUP BY A.WEBSITE_CLASS_ID, A.WEBSITE_ID, A.ERROR_DATE, B.NAME, B.LOCATION
UNION
SELECT
C.WEBSITE_CLASS_ID AS WEBSITE_CLASS_ID
,C.WEBSITE_ID AS WEBSITE_ID
,DATE(C.SUCCESS_DATE) AS SUCCESSFUL_CONNECTION
,COUNT(C.SUCCESS) AS COUNT_SUCCESS_CNCTN
,B.NAME AS WEBSITE_NAME
,B.LOCATION AS COMPANY_LOCATION
FROM
&DATABASE..SUCCESS C
INNER JOIN
&DATABASE..DETAILS B
ON C.WEBSITE_ID = B.WEBSITE_ID
GROUP BY C.WEBSITE_CLASS_ID, C.WEBSITE_ID, DATE(C.SUCCESS_DATE), B.NAME, B.LOCATION