Question

我正在构建一个可以与手机同步的版本系统例如，手机在版本1和服务器上运行有版本2，所以这意味着服务器将返回版本2的数据。

我有以下表格：

CREATE TABLE `exercise` (
  `id_exercise` int(11) NOT NULL,
  `name` varchar(45) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `exercise_audit` (
  `id_exercise_audit` int(11) NOT NULL,
  `name` varchar(45) NOT NULL COMMENT 'The name will take the new value if inserted ,or the old value if edited.And the latest value if deleted.',
  `action_peformed` varchar(45) NOT NULL,
  `version` int(11) NOT NULL,
  `reference_id` int(11) NOT NULL,
  `change_date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

CREATE TABLE `version` (
  `id_version` int(11) NOT NULL,
  `date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `state` enum('C','P') NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

练习表有3个不同的触发器，因此它可以处理exercise_audit表。

我对以下查询的当前回复

SELECT ea.action_peformed,
    ea.version,
    ea.change_date,
    e.name,
    e.id_exercise 
FROM exercise_audit AS ea 
LEFT JOIN exercise AS e 
ON e.id_exercise = ea.reference_id 
WHERE version > :version

看起来像这样：

[
{
    "action_peformed": "insert",
    "version": "1",
    "change_date": "2017-06-22 16:42:03",
    "audit_name": "Push Ups",
    "current_name": "Push Ups",
    "id_exercise": "1"
},
{
    "action_peformed": "insert",
    "version": "1",
    "change_date": "2017-06-22 16:42:06",
    "audit_name": "Squat",
    "current_name": "Squat",
    "id_exercise": "2"
},
{
    "action_peformed": "insert",
    "version": "1",
    "change_date": "2017-06-22 16:42:09",
    "audit_name": "Chin Ups",
    "current_name": "Chin Ups",
    "id_exercise": "3"
},
{
    "action_peformed": "insert",
    "version": "2",
    "change_date": "2017-06-22 16:44:25",
    "audit_name": "Pull Ups",
    "current_name": "Pull Ups",
    "id_exercise": "4"
},
{
    "action_peformed": "insert",
    "version": "2",
    "change_date": "2017-06-22 16:45:08",
    "audit_name": "Sit Up",
    "current_name": "Sit Up",
    "id_exercise": "5"
},
{
    "action_peformed": "insert",
    "version": "2",
    "change_date": "2017-06-22 16:45:28",
    "audit_name": "Pike Push Up",
    "current_name": "Pike Push Ups",
    "id_exercise": "6"
},
{
    "action_peformed": "update",
    "version": "3",
    "change_date": "2017-06-22 16:47:28",
    "audit_name": "Pike Push Up",
    "current_name": "Pike Push Ups",
    "id_exercise": "6"
}

问题在于这一部分：

{
    "action_peformed": "insert",
    "version": "2",
    "change_date": "2017-06-22 16:45:28",
    "audit_name": "Pike Push Up",
    "current_name": "Pike Push Ups",
    "id_exercise": "6"
},
{
    "action_peformed": "update",
    "version": "3",
    "change_date": "2017-06-22 16:47:28",
    "audit_name": "Pike Push Up",
    "current_name": "Pike Push Ups",
    "id_exercise": "6"
}

请注意！名称在两者中相同的原因是因为在插入中 name是NEW.name，在更新中名称是OLD.name，所以我可以跟踪历史数据。

正如您所看到的，我在id_exercise为6的情况下进行了同样的练习，它已插入版本2并在版本3中更新。

我的问题是如何在sql中将它们组合在一起。

例如，要实现以下目标：

{
        "action_peformed": "insert",
        "version": "3",
        "change_date": "2017-06-22 16:47:28",
        "audit_name": "Pike Push Up",
        "current_name": "Pike Push Ups",
        "id_exercise": "6"
    }

它结合了更新和插入，以便它使用最后更新的数据作为插入数据。

原因是用户在移动设备上有版本1，并且他暂时没有与网络同步。例如，网络版本为6，版本中添加了练习3，它也在版本6中更新。所以当手机请求数据时它不会插入和更新同一件事。而不是将它组合成一个具有最新值的东西作为插入而不是更新

我可以用一些PHP代码解决它，但它会让它自我复杂......

Answer 1

不确定是什么决定了action_performed的值，因此该部分是猜测。

SELECT IF('insert' IN GROUP_CONCAT(ea.`action_peformed`),'insert','update') as `action_performed`,
    MAX(ea.`version`) as `version`,
    MAX(ea.`change_date`) as `change_date`,
    e.`name`,
    e.`id_exercise` 
FROM `exercise_audit` AS ea 
LEFT JOIN `exercise` AS e 
ON e.`id_exercise` = ea.`reference_id` 
WHERE `version` > :version
GROUP BY e.`id_exercise`

如何将两个单独的结果组合成一个组合结果

1 个答案: