查询当前返回
Bob President 5
Joe Programmer 3
Greg Programmer 2
Joe Janitor 6
Liz Programmer 6
但是我希望这样的每个人只有一行
Bob President 5
Joe Programmer (or Janitor) 9
Greg Programmer 2
Liz Programmer 6
我现在按名称和角色分组。我如何也只能获得唯一的名称并合并该值？
SELECT
...
FROM
vw_QuickbaseAudit v
INNER JOIN [staging].[QuickbaseSystemUsers] qsu ON v.[user] = qsu.UserId
WHERE
Date_Modified >= @StartDate
AND Date_Modified < DATEADD(dd, 1, @EndDate)
GROUP BY
qsu.UserFirstName + ' ' + qsu.UserLastName,
qsu.RoleName
ORDER BY
RoleName,
NumberOfEditTotal DESC;
答案 0 :(得分：2)
{
"id": 259609,
"parent_id": 0,
"number": "CW-104-270219",
"order_key": "xxxxxxxxxxxx",
"created_via": "checkout",
"version": "3.5.5",
"status": "completed",
"currency": "USD",
"date_created": "2019-02-27T15:11:58",
"date_created_gmt": "2019-02-27T20:11:58",
"date_modified": "2019-02-27T15:11:59",
"date_modified_gmt": "2019-02-27T20:11:59",
"discount_total": "0.00",
"discount_tax": "0.00",
"shipping_total": "0.00",
"shipping_tax": "0.00",
"cart_tax": "0.00",
"total": "9.68",
"total_tax": "0.00",
"prices_include_tax": false,
"customer_id": 2,
"customer_ip_address": "xxxxxxxxxxxx",
"customer_user_agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/72.0.3626.119 Safari/537.36",
"customer_note": "",
"billing": {
"first_name": "Valerie",
"last_name": "Criswell",
"company": "",
"address_1": "",
"address_2": "",
"city": "",
"state": "",
"postcode": "",
"country": "",
"email": "xxxxxxxxxx",
"phone": ""
},
"shipping": {
"first_name": "",
"last_name": "",
"company": "",
"address_1": "",
"address_2": "",
"city": "",
"state": "",
"postcode": "",
"country": ""
},
"payment_method": "authorize_net_aim",
"payment_method_title": "Credit Card",
"transaction_id": "xxxxxxxxxxxxx",
"date_paid": "2019-02-27T15:11:59",
"date_paid_gmt": "2019-02-27T20:11:59",
"date_completed": "2019-02-27T15:11:59",
"date_completed_gmt": "2019-02-27T20:11:59",
"cart_hash": "xxxxxxxxxxxxx",
"meta_data": [
{
"id": 8785641,
"key": "_wcson_order_number",
"value": "CWD-259609-022719-108"
},
{
"id": 8785642,
"key": "_order_number",
"value": "104"
},
{
"id": 8785643,
"key": "_order_number_formatted",
"value": "CW-104-270219"
},
{
"id": 8785644,
"key": "_order_number_meta",
"value": {
"prefix": "CW-",
"suffix": "-{DD}{MM}{YY}",
"length": "2"
}
},
{
"id": 8785645,
"key": "_wc_authorize_net_aim_retry_count",
"value": "0"
},
{
"id": 8785646,
"key": "_wc_authorize_net_aim_trans_id",
"value": "xxxxxxxx"
},
{
"id": 8785647,
"key": "_wc_authorize_net_aim_trans_date",
"value": "2019-02-27 15:11:59"
},
{
"id": 8785648,
"key": "_wc_authorize_net_aim_environment",
"value": "production"
},
{
"id": 8785649,
"key": "_wc_authorize_net_aim_account_four",
"value": "xxxxxxxx"
},
{
"id": 8785650,
"key": "_wc_authorize_net_aim_authorization_amount",
"value": "xxxxxxx"
},
{
"id": 8785651,
"key": "_wc_authorize_net_aim_authorization_code",
"value": "xxxxxxx"
},
{
"id": 8785652,
"key": "_wc_authorize_net_aim_charge_captured",
"value": "yes"
},
{
"id": 8785653,
"key": "_wc_authorize_net_aim_card_expiry_date",
"value": "21-08"
},
{
"id": 8785654,
"key": "_wc_authorize_net_aim_card_type",
"value": "mastercard"
},
{
"id": 8785659,
"key": "_pip_invoice_number",
"value": "CW-104-270219"
},
{
"id": 8785660,
"key": "_wc_pip_invoice_email_count",
"value": "1"
},
{
"id": 8785661,
"key": "_wc_pip_packing_list_email_count",
"value": "1"
}
],
"line_items": [
{
"id": 17,
"name": "I Give Myself Away - MP3 Download - 30 Day Access",
"product_id": 192127,
"variation_id": 192128,
"quantity": 1,
"tax_class": "",
"subtotal": "9.68",
"subtotal_tax": "0.00",
"total": "9.68",
"total_tax": "0.00",
"taxes": [],
"meta_data": [
{
"id": 182,
"key": "pa_format",
"value": "download-30-day"
}
],
"sku": "131599-DL30DAY",
"price": 9.68
}
],
"tax_lines": [],
"shipping_lines": [],
"fee_lines": [],
"coupon_lines": [],
"refunds": []
}
那个的一些变体 （编辑：错过了分组依据中的其他STUFF命令）
答案 1 :(得分：1)
基本上，您需要从RoleName子句中删除GROUP BY。您可以使用STRING_AGG()进行字符串聚合：
SELECT
qsu.UserFirstName + ' ' + qsu.UserLastName,
STRING_AGG(RoleName, ' or ') RoleNames,
...
FROM
vw_QuickbaseAudit v
INNER JOIN [staging].[QuickbaseSystemUsers] qsu ON v.[user] = qsu.UserId
WHERE
Date_Modified >= @StartDate
AND Date_Modified < DATEADD(dd, 1, @EndDate)
GROUP BY
qsu.UserFirstName,
qsu.UserLastName
ORDER BY
NumberOfEditTotal DESC;
答案 2 :(得分：1)
我被解释为：
Joe Programmer (or Janitor) 9
说这两个都可以：
Joe Programmer 9
Joe Janitor 9
查询如下：
SELECT qsu.UserFirstName + ' ' + qsu.UserLastName,
MIN(qsu.RoleName) as RoleName
SUM(NumberOfEditTotal) as NumberOfEditTotal
FROM vw_QuickbaseAudit v JOIN
[staging].[QuickbaseSystemUsers] qsu
ON v.[user] = qsu.UserId
WHERE Date_Modified >= @StartDate AND
Date_Modified < DATEADD(day, 1, @EndDate)
GROUP BY qsu.UserFirstName + ' ' + qsu.UserLastName
ORDER BY RoleName, NumberOfEditTotal DESC;
目前尚不清楚NumberOfEditTotal的计算方式。
编辑：
棘手地追加所有行。在最新版本的SQL Server中，可以将STRING_AGG()用作字符串聚合函数。那很简单。
我在早期版本中的建议是一个折衷方案。因此值可能是：
Janitor
Janitor and Programmer
Janitor, Programmer, and others
您可以执行以下操作：
(case when min(rolename) = max(rolename)
then min(rolename)
when count(distinct rolename) = 2
then min(rolename) + ' and ' + max(rolename)
else min(rolename) + ', ' + max(rolename) + ' others'
end)