我一直坚持以下查询，并且无法弄清楚如何做到这一点（以下说明）：
SELECT
a.id, a.name, t.agency_id, t.initial_amount,
DATE_FORMAT(t.created_at, "%d.%m.%Y") as date,
DATE_FORMAT(t.created_at, "%Y-%m-%d") as date_created,
t.created_at, t.collection, t.pay, (t.initial_amount + t.collection - t.pay) as total_amount
FROM (
SELECT
hh.agency_id, hh.amount, h0.initial_amount, hh.created_at,
SUM(IF(hh.payment_type=0,hh.amount,0)) AS collection,
SUM(IF(hh.payment_type=1,hh.amount,0)) AS pay
FROM house_register AS hh
LEFT JOIN (
SELECT h1.initial_amount, h1.agency_id, h1.created_at
FROM house_register AS h1
INNER JOIN (
SELECT min(h2.id) AS min_id, h2.agency_id, h2.created_at
FROM house_register AS h2
GROUP BY h2.agency_id
) AS min_id_ag
ON h1.agency_id = min_id_ag.agency_id
AND h1.id = min_id_ag.min_id
) AS h0
ON hh.agency_id = h0.agency_id
WHERE hh.created_at BETWEEN '2015-11-01' AND '2015-11-30'
GROUP BY hh.created_at, hh.agency_id WITH ROLLUP
) AS t
INNER JOIN agencies as a
on a.id = t.agency_id
所以，我在这里尝试实现的是从单个表中获取某段时间的报告，在此表中我的操作具有类型状态（付款或收款）并且每个操作对应到agency_id（商店）。
我的报告必须是这样的：
列：
agency_id date X initial_amount_of_first_transaction_of_day_in_loop sum(payments) sum(collection)
行看起来像这样：
agency_1 2015-01-01 500 100 0
agency_2 2015-01-01 600 100 0
.... next date
agency_1 2015-01-02 600 0 150
agency_2 2015-01-02 450 0 150
etc
我想要的是获取按min(id)和agency_id分组的每个日期的第一个事务的initial_amount（因此created_at）。例如，如果我在2015-01-01的日期有50笔交易，我想获得该日期第一笔交易的第initial_amount栏的价值。
现在它只选择整个日期间隔的最小值。我和所有日期的最小值相同。
希望我足够清楚。
谢谢你的帮助！
更新结构：
CREATE TABLE IF NOT EXISTS `house_register` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`created_at` date DEFAULT NULL,
`created_time` time DEFAULT NULL,
`initial_amount` decimal(14,2) DEFAULT NULL,
`document_type` tinyint(3) unsigned DEFAULT NULL,
`document_id` int(10) unsigned DEFAULT NULL,
`document_number` varchar(100) DEFAULT NULL,
`description` varchar(200) DEFAULT NULL,
`amount` decimal(14,2) DEFAULT NULL,
`final_amount` decimal(14,2) DEFAULT NULL,
`agency_id` int(11) DEFAULT NULL,
`user_id` int(10) DEFAULT NULL,
`payment_type` tinyint(1) DEFAULT NULL COMMENT '0 - incasare, 1 - plata',
`number` varchar(50) DEFAULT NULL,
`debit_account` varchar(45) DEFAULT NULL,
`credit_account` varchar(45) DEFAULT NULL,
`short_desc` varchar(45) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=34;
使用结果集更新查询细分：
完整的一个返回： Results complete sql
子查询1：Subquery 1 results
SELECT
hh.agency_id, hh.amount, h0.initial_amount, hh.created_at,
SUM(IF(hh.payment_type=0,hh.amount,0)) AS collection,
SUM(IF(hh.payment_type=1,hh.amount,0)) AS pay
FROM house_register AS hh
LEFT JOIN (
SELECT h1.initial_amount, h1.agency_id, h1.created_at
FROM house_register AS h1
INNER JOIN (
SELECT min(h2.id) AS min_id, h2.agency_id, h2.created_at
FROM house_register AS h2
GROUP BY h2.agency_id
) AS min_id_ag
ON h1.agency_id = min_id_ag.agency_id
AND h1.id = min_id_ag.min_id
) AS h0
ON hh.agency_id = h0.agency_id
WHERE hh.created_at BETWEEN '2015-11-01' AND '2015-11-30'
GROUP BY hh.created_at, hh.agency_id WITH ROLLUP;
子查询2：参见评论
SELECT h1.initial_amount, h1.agency_id, h1.created_at
FROM house_register AS h1
INNER JOIN (
SELECT min(h2.id) AS min_id, h2.agency_id, h2.created_at
FROM house_register AS h2
GROUP BY h2.agency_id
) AS min_id_ag
ON h1.agency_id = min_id_ag.agency_id
AND h1.id = min_id_ag.min_id;
子查询3：见评论
SELECT min(h2.id) AS min_id, h2.agency_id, h2.created_at
FROM house_register AS h2
GROUP BY h2.agency_id;
答案 0 :(得分：0)
我相信您需要将日期（created_at）添加到组中，并将日期之间的where子句添加到组中。
WHERE h2.created_at BETWEEN '2015-11-01' AND '2015-11-30'
GROUP BY h2.agency_id, date_format(h2.created_at, "%Y-%b-%d")
第一行将搜索条件缩减到您想要的日期。 第二行按agency_id对所有内容进行分组，然后是当天（通过对created_at格式化的YYY-MM-DD进行分组。
即。类似的东西（我没有测试过但是......）
SELECT
a.id, a.name, t.agency_id, t.initial_amount,
DATE_FORMAT(t.created_at, "%d.%m.%Y") as date,
DATE_FORMAT(t.created_at, "%Y-%m-%d") as date_created,
t.created_at, t.collection, t.pay, (t.initial_amount + t.collection - t.pay) as total_amount
FROM (
SELECT
hh.agency_id, hh.amount, h0.initial_amount, hh.created_at,
SUM(IF(hh.payment_type=0,hh.amount,0)) AS collection,
SUM(IF(hh.payment_type=1,hh.amount,0)) AS pay
FROM house_register AS hh
LEFT JOIN (
SELECT h1.initial_amount, h1.agency_id, h1.created_at
FROM house_register AS h1
INNER JOIN (
SELECT min(h2.id) AS min_id, h2.agency_id, h2.created_at
FROM house_register AS h2
WHERE h2.created_at BETWEEN '2015-11-01' AND '2015-11-30'
GROUP BY h2.agency_id, date_format(h2.created_at, "%Y-%b-%d")
) AS min_id_ag
ON h1.agency_id = min_id_ag.agency_id
AND h1.id = min_id_ag.min_id
) AS h0
ON hh.agency_id = h0.agency_id
WHERE hh.created_at BETWEEN '2015-11-01' AND '2015-11-30'
GROUP BY hh.created_at, hh.agency_id WITH ROLLUP
) AS t
INNER JOIN agencies as a
on a.id = t.agency_id
答案 1 :(得分：0)
对不起，是的，我不得不重新阅读你的问题几次，但我想我已经把它排序了。这能为您提供所需的结果吗？
对于min(h2.id)选择，您还需要group by h2.created_at，而不仅仅是第{18}行的h2.agency_id - 完整群组= GROUP BY h2.agency_id, h2.created_at。
SELECT
a.id, a.name, t.agency_id, t.initial_amount,
DATE_FORMAT(t.created_at, "%d.%m.%Y") as date,
DATE_FORMAT(t.created_at, "%Y-%m-%d") as date_created,
t.created_at, t.collection, t.pay, (t.initial_amount + t.collection - t.pay) as total_amount
FROM (
SELECT
hh.agency_id, hh.amount, h0.initial_amount, hh.created_at,
SUM(IF(hh.payment_type=0,hh.amount,0)) AS collection,
SUM(IF(hh.payment_type=1,hh.amount,0)) AS pay
FROM house_register AS hh
LEFT JOIN (
SELECT h1.initial_amount, h1.agency_id, h1.created_at
FROM house_register AS h1
INNER JOIN (
SELECT min(h2.id) AS min_id, h2.agency_id, h2.created_at
FROM house_register AS h2
GROUP BY h2.agency_id, h2.created_at
) AS min_id_ag
ON h1.agency_id = min_id_ag.agency_id
AND h1.id = min_id_ag.min_id
) AS h0
ON hh.agency_id = h0.agency_id
WHERE hh.created_at BETWEEN '2015-11-01' AND '2015-11-30'
GROUP BY hh.created_at, hh.agency_id WITH ROLLUP
) AS t
INNER JOIN agencies as a
on a.id = t.agency_id
答案 2 :(得分：0)
你回答几乎是正确的，肯定让我在正确的轨道上修复它。 非常感谢你的帮助和耐心@wkdmarty！
这是我想要的查询。
SELECT a.id,
a.NAME,
t.agency_id,
t.initial_amount,
Date_format(t.created_at, "%d.%m.%Y") AS date,
Date_format(t.created_at, "%Y-%m-%d") AS date_created,
t.created_at,
t.collection,
t.pay,
(t.initial_amount + t.collection - t.pay) AS total_amount
FROM (
SELECT hh.agency_id,
hh.amount,
h0.initial_amount,
hh.created_at,
sum(IF(hh.payment_type=0,hh.amount,0)) AS collection,
sum(IF(hh.payment_type=1,hh.amount,0)) AS pay
FROM house_register AS hh
LEFT JOIN
(
SELECT h1.initial_amount,
h1.agency_id,
h1.created_at
FROM house_register AS h1
INNER JOIN
(
SELECT min(h2.id) AS min_id,
h2.agency_id,
h2.created_at
FROM house_register AS h2
GROUP BY h2.agency_id,
h2.created_at ) AS min_id_ag
ON h1.agency_id = min_id_ag.agency_id
AND h1.id = min_id_ag.min_id
WHERE h1.created_at BETWEEN '2015-11-01' AND '2015-11-30' ) AS h0
ON hh.agency_id = h0.agency_id
AND hh.created_at = h0.created_at
WHERE h0.created_at BETWEEN '2015-11-01' AND '2015-11-30'
GROUP BY hh.created_at,
hh.agency_id WITH rollup ) AS t
INNER JOIN agencies AS a
ON a.id = t.agency_id
ORDER BY t.agency_id,
t.created_at