Question

我正在尝试计算一个课程完成了多少交易，我试图加入training_transactions，其中包含training_transaction_course = course_id和training_transaction_status =＆＃39;已完成的所有行的计数。这是我到目前为止的代码：

SELECT    training.*, 
          Count(DISTINCT training_transactions.training_transaction_course) AS completed_training_payments
left JOIN users 
ON        training.course_user = users.user_id 
LEFT JOIN training_transactions 
ON        training.course_user = training_transactions.training_transaction_user 
FROM      training 
WHERE     course_id = ? 
AND       training_transactions.training_transaction_status = 'complete'
AND       course_enabled = 'enabled'

我的桌子：

培训交易

CREATE TABLE IF NOT EXISTS `training_transactions` (
  `training_transaction_id` int(11) NOT NULL,
  `training_transaction_user` int(11) NOT NULL,
  `training_transaction_course` int(11) NOT NULL,
  `training_transaction_status` varchar(50) NOT NULL,
  `training_transaction_enabled` varchar(50) NOT NULL DEFAULT 'enabled',
  `training_transaction_date` datetime NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

训练

CREATE TABLE IF NOT EXISTS `training` (
  `course_id` int(11) NOT NULL,
  `course_user` int(11) NOT NULL,
  `course_type` varchar(255) NOT NULL,
  `course_name` varchar(255) NOT NULL,
  `course_location` varchar(255) NOT NULL,
  `course_duration` varchar(255) NOT NULL,
  `course_fitness_type` varchar(255) NOT NULL,
  `course_instructor_name` varchar(255) NOT NULL,
  `course_price` int(15) NOT NULL,
  `course_start_date` date NOT NULL,
  `course_max_attendees` int(8) NOT NULL,
  `course_accommodation` varchar(255) NOT NULL,
  `course_accommodation_price` varchar(255) NOT NULL,
  `course_status` varchar(50) NOT NULL,
  `course_enabled` varchar(10) NOT NULL DEFAULT 'enabled'
) ENGINE=InnoDB AUTO_INCREMENT=24 DEFAULT CHARSET=latin1;

正如您所看到的，我正在尝试将完成的交易计数作为从课程_max_attendees中扣除的计数，然后我可以检查是否还有任何地方。

Answer 1

您想选择培训。所以从training中选择。您希望使用它显示事务计数，您可以在select子句的子查询中执行此操作：

select
  t.*,
  (
    select count(*) 
    from training_transactions tt
    where tt.training_transaction_user = t.course_user
    and tt.training_transaction_status = 'complete'
  ) as completed_training_payments
from training t
where t.course_id = ? 
and t.course_enabled = 'enabled';

以下是联接：

select
  t.*, coalesce(tt.cnt, 0) as completed_training_payments
from training t
left join
(
  select training_transaction_status, count(*) as cnt
  from training_transactions
  where training_transaction_status = 'complete'
  group by training_transaction_status
) tt on tt.training_transaction_user = t.course_user
where t.course_id = ? 
and t.course_enabled = 'enabled';

Answer 2

首先，如果您想知道每个课程已经完成了多少已完成的交易，您就无法获得所涉及的用户表。您将汇总任何用户信息。

然后，您必须从课程表开始，该表似乎已命名为get_planet_name。现在，您想要计算每个课程的每个已完成的交易。左连接对此非常合适：

training

这个问题是它会给出计数值＆＃34; 1＆＃34;对于每个已完成交易的课程，以及那些根本没有完成交易的课程！所以每一行都有＆＃34; 1＆＃34;会怀疑的。解决方案是计算密钥，而不是行。这是使用select t.Name, count( * ) as completed_training_payments from training t left join training_transactions tt on tt.user = t.course_user and tt.status = 'complete' where t.course_status = 'enabled' group by t.Name;而不是sum函数完成的。

count

由于select t.Name, sum( case when tt.course_user is null then 0 else 1 end ) as completed_training_payments from training t left join training_transactions tt on tt.user = t.course_user and tt.status = 'complete' where t.course_status = 'enabled' group by t.Name;只有tt.course_user，当根本没有完成的交易时，该课程将显示＆＃34; count＆＃34; ＆＃34; 0＆＃34;。

来自另一个表的LEFT JOIN计数

2 个答案: