Question

我正在尝试加入公司及其详细信息，以及交易，即使它们不存在。

我正在计算交易数量，以确定有多少用户参加课程，如果没有交易，我仍然想加入公司和详细信息，但计数将为0，在我的查询中，正在选择training_company表，但是由于某种原因没有选择training_details：

SELECT training.*, count(distinct training_transactions.training_transaction_course) as completed_training_payments 
 FROM training
LEFT JOIN training_company
    ON training.course_main = training_company_id
LEFT JOIN training_details
    ON training.course_main = training_details_company
LEFT JOIN training_transactions
    ON training.course_user = training_transactions.training_transaction_user
WHERE course_id = ?
        AND       training_transactions.training_transaction_status = 'complete'
        AND       training_transactions.training_transaction_payment_status = 'complete'
        AND course_enabled = 'enabled'

training_company：

CREATE TABLE IF NOT EXISTS `training_company` (
  `training_company_id` int(11) NOT NULL,
  `training_company_name` varchar(100) NOT NULL,
  `training_company_user` int(11) NOT NULL,
  `training_company_enabled` varchar(50) NOT NULL DEFAULT 'enabled',
  `training_company_has_avatar` int(5) NOT NULL DEFAULT '0',
  `training_company_has_banner` int(5) NOT NULL DEFAULT '0'
) ENGINE=InnoDB AUTO_INCREMENT=11 DEFAULT CHARSET=latin1;

--
-- Dumping data for table `training_company`
--

INSERT INTO `training_company` (`training_company_id`, `training_company_name`, `training_company_user`, `training_company_enabled`, `training_company_has_avatar`, `training_company_has_banner`) VALUES
(1, '123', 1, 'enabled', 0, 0),

training_details：

CREATE TABLE IF NOT EXISTS `training_details` (
  `training_details_id` int(11) NOT NULL,
  `training_details_user` int(11) NOT NULL,
  `training_details_company` int(11) NOT NULL,
  `training_details_registration_number` varchar(10) NOT NULL,
  `training_details_type` varchar(100) NOT NULL,
  `training_details_name` varchar(100) NOT NULL,
  `training_details_street` varchar(100) NOT NULL,
  `training_details_town` varchar(100) NOT NULL,
  `training_details_county` varchar(100) NOT NULL,
  `training_details_postcode` varchar(100) NOT NULL,
  `training_details_country` varchar(100) NOT NULL,
  `training_details_company_name` varchar(100) NOT NULL,
  `training_details_company_street` varchar(100) NOT NULL,
  `training_details_company_town` varchar(100) NOT NULL,
  `training_details_company_county` varchar(100) NOT NULL,
  `training_details_company_postcode` varchar(100) NOT NULL,
  `training_details_company_country` varchar(100) NOT NULL,
  `training_details_total_employees` varchar(100) NOT NULL,
  `training_details_fax` varchar(100) NOT NULL,
  `training_details_landline` varchar(100) NOT NULL,
  `training_details_mobile` varchar(50) NOT NULL,
  `training_details_email` varchar(50) NOT NULL,
  `training_details_website` varchar(250) NOT NULL,
  `company_differs_address` int(11) NOT NULL DEFAULT '0'
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=latin1;

--
-- Dumping data for table `training_details`
--

INSERT INTO `training_details` (`training_details_id`, `training_details_user`, `training_details_company`, `training_details_registration_number`, `training_details_type`, `training_details_name`, `training_details_street`, `training_details_town`, `training_details_county`, `training_details_postcode`, `training_details_country`, `training_details_company_name`, `training_details_company_street`, `training_details_company_town`, `training_details_company_county`, `training_details_company_postcode`, `training_details_company_country`, `training_details_total_employees`, `training_details_fax`, `training_details_landline`, `training_details_mobile`, `training_details_email`, `training_details_website`, `company_differs_address`) VALUES
(1, 0, 1, '0', '', '123', '123', '123', '123456', 'WN8', 'Australia', '123', '123', '123', '', 'WN8', 'Australia', '', '', '', '', '', '', 4),

训练：

CREATE TABLE IF NOT EXISTS `training` (
  `course_id` int(11) NOT NULL,
  `course_user` int(11) NOT NULL,
  `course_main` int(11) NOT NULL,
  `course_type` varchar(255) NOT NULL,
  `course_name` varchar(255) NOT NULL,
  `course_description` text 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',
  `course_location_name` varchar(255) NOT NULL,
  `course_location_street` varchar(255) NOT NULL,
  `course_location_town` varchar(255) NOT NULL,
  `course_location_county` varchar(255) NOT NULL,
  `course_location_postcode` varchar(255) NOT NULL,
  `course_location_country` varchar(255) NOT NULL,
  `course_certificate` varchar(250) NOT NULL,
  `course_certificate_valid` int(30) NOT NULL
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=latin1;

--
-- Dumping data for table `training`
--

INSERT INTO `training` (`course_id`, `course_user`, `course_main`, `course_type`, `course_name`, `course_description`, `course_location`, `course_duration`, `course_fitness_type`, `course_instructor_name`, `course_price`, `course_start_date`, `course_max_attendees`, `course_accommodation`, `course_accommodation_price`, `course_status`, `course_enabled`, `course_location_name`, `course_location_street`, `course_location_town`, `course_location_county`, `course_location_postcode`, `course_location_country`, `course_certificate`, `course_certificate_valid`) VALUES
(1, 3, 1, 'Hazardous', '123', 'dddddddddddddd', 'other', '14', 'lol', 'lol', 123, '2015-11-09', 4, '0', '', 'pending', 'enabled', '123', '123', '123', '123456', '123', 'Australia', '123', 2),

那么我怎样才能让我的联盟处理细节问题，因为即使所有内容都匹配，目前它还没有加入

Answer 1

正如Gordon Linoff在第一条评论中所建议的那样，将training_transactions条款中涉及WHERE的条件移至LEFT JOIN。

当然，您还应该明确列出列而不是*，并为GROUP BY添加相应的COUNT。

SELECT 
    training.course_id
    ,training.course_user
    ...
    ,count(distinct training_transactions.training_transaction_course) as completed_training_payments 
FROM
    training
    LEFT JOIN training_company ON training.course_main = training_company_id
    LEFT JOIN training_details ON training.course_main = training_details_company
    LEFT JOIN training_transactions
        ON  training_transactions.training_transaction_user = training.course_user
        AND training_transactions.training_transaction_status = 'complete'
        AND training_transactions.training_transaction_payment_status = 'complete'
WHERE
    training.course_id = ?
    AND training.course_enabled = 'enabled'
GROUP BY
    training.course_id
    ,training.course_user
    ...

在此查询中，您还应该注意一些事项。

WHERE
    course_id = ?
    AND course_enabled = 'enabled'

对于阅读该问题的新人（因为每个人都在阅读这个问题以及将要维护您的代码的下一个人，这可能是您在两年时间内）这些字段属于哪个表格，目前尚不清楚。始终尝试明确指出表格，如下所示：

WHERE
    training.course_id = ?
    AND training.course_enabled = 'enabled'

使用别名也很有帮助。

以类似的方式，不清楚JOIN：

中发生了什么

    LEFT JOIN training_company ON training.course_main = training_company_id
    LEFT JOIN training_details ON training.course_main = training_details_company

在简化此问题的查询时，这只是一个疏忽，或者这是您真正的代码吗？无论如何，请包括表名（或别名）。

LEFT JOIN training_company ON training.course_main = training_company.training_company_id
LEFT JOIN training_details ON training.course_main = training_details.training_details_company

Answer 2

我建议完全不同的路线，不要使用左连接。由于您需要来自训练表的所有数据，并且您希望从不同的表中聚合（在您的情况下计数），我建议将聚合分解为单独的子查询。这就是我的意思（注意我在表格中添加了别名以简化操作）：

SELECT t.*,
       (SELECT count(distinct tt.training_transaction_course)
          FROM training_company tc        
          JOIN training_details td ON td.course_main = tc.training_details_company
          JOIN training_transactions tt ON t.course_user = tt.training_transaction_user
         WHERE t.course_main = tc.training_company_id
           AND tt.training_transaction_status = 'complete'
           AND tt.training_transaction_payment_status = 'complete')
        as completed_training_payments 
 FROM training t
WHERE t.course_id = ?
  AND t.course_enabled = 'enabled'

这将保证训练表中与输入的course_id匹配的每一行返回一行。此外，如果子查询没有返回任何行，则它将返回0.

Answer 3

这有一些问题。首先，如果没有training_transactions的模式，很难重现。

此外，从您的描述和架构中可以看出，您希望在第一个表上进行内部联接。如果我读得对，training总会有一家公司。 training_details似乎与查询无关，如果它们不是，则会提供交叉产品，因此完全不考虑。

我假设用户加入交易是错误的，因为你有一个training_transaction_course我认为是正确的外键。

计数是错误的 - 它不是如何工作的，如果你想要计数，你需要一个GROUP BY子句。

WHERE子句对我来说是正确的，因为元素没有在任何连接中使用。

这会给你这样的东西：

SELECT training.course_id, -- everything you put here needs to go in the group by clause
    count(training_transactions.training_transaction_course) as completed_training_payments
   FROM training
   JOIN training_company
       ON training.course_main = training_company_id
   LEFT JOIN training_transactions
       ON training.course_id = training_transactions.training_transaction_course
   WHERE course_id = ?
       AND training_transactions.training_transaction_status = 'complete'
       AND training_transactions.training_transaction_payment_status = 'complete'
       AND course_enabled = 'enabled'
   GROUP BY training.course_id, training_transactions.training_transaction_course

〜

Answer 4

问题是您的查询从不会产生多行。

使用Groupby或嵌套查询，如其他答案中所述。

当我从另一个表中选择计数时，左连接不起作用

4 个答案: