Mysql 从以前的最新记录中选择列

考虑以下MySQL表：

应用程序（包含所有用户的所有应用程序）

我错过了什么？
TIA.

它应该可以工作：

SELECT user_id, date_of_application, status, coalesce(latest.score, 10) score
FROM applications
LEFT OUTER JOIN 
  (SELECT user_id, score 
   FROM applications a
   WHERE status = 'Ended' AND 
         date_of_application = (SELECT MAX(date_of_application)
                                FROM applications
                                WHERE status='Ended' AND user_id = a.user_id)) latest 
                                  ON latest.user_id = applications.user_id
WHERE status = 'Active'
ORDER BY date_of_application ASC

它应该在以下方面发挥作用：

SELECT user_id, date_of_application, status, coalesce(latest.score, 10) score
FROM applications
LEFT OUTER JOIN 
  (SELECT user_id, score 
   FROM applications a
   WHERE status = 'Ended' AND 
         date_of_application = (SELECT MAX(date_of_application)
                                FROM applications
                                WHERE status='Ended' AND user_id = a.user_id)) latest 
                                  ON latest.user_id = applications.user_id
WHERE status = 'Active'
ORDER BY date_of_application ASC

---

考虑到您希望分数基于最新的数据。 试试这个-

SELECT apps.user_id, apps.date_of_application, apps.status,
       IFNULL(
              (SELECT app.score 
              FROM applications app
              WHERE  app.user_id = apps.user_id
              AND app.status = 'Ended' 
              ORDER BY app.date_ended DESC
              LIMIT 1), 10) AS score
FROM applications apps
WHERE  apps.status = 'Active'
ORDER BY apps.score ASC,
         apps.date_of_application ASC 

---

考虑到您希望分数基于最新的数据。 试试这个-

SELECT apps.user_id, apps.date_of_application, apps.status,
       IFNULL(
              (SELECT app.score 
              FROM applications app
              WHERE  app.user_id = apps.user_id
              AND app.status = 'Ended' 
              ORDER BY app.date_ended DESC
              LIMIT 1), 10) AS score
FROM applications apps
WHERE  apps.status = 'Active'
ORDER BY apps.score ASC,
         apps.date_of_application ASC 

---

分数是最低的还是最晚的？@Juniad只考虑最晚的日期。分数可以是任何值（最高或最低）。我已经在这个基础上回答了。只需检查它是否适合你。分数应该是分数中最低的还是最晚的？@Juniad只考虑日期（最晚）。分数可以是任何值（最高或最低）。我已经在这个基础上回答了。检查一下它是否适合你。首先，感谢你的快速响应。稍微调整一下，你的答案就完成了预期的任务。唯一让我抓狂的是，当我运行你的代码时，分数的顺序表现得很奇怪（不是升序，也不是降序），而把“应用程序”按部分顺序留下。不过，把它拿出来（留给我的是按分数排序的ASC、应用程序ASC的日期）。首先，感谢您的快速响应。稍微调整一下，你的答案就完成了预期的任务。唯一让我抓狂的是，当我运行你的代码时，分数的顺序表现得很奇怪（不是升序，也不是降序），而把“应用程序”按部分顺序留下。但是，把它拿出来（留给我按分数排序的ASC，应用程序ASC的日期）解决了这个问题。

SELECT apps.user_id, apps.date_of_application, apps.status,
       IFNULL(
              (SELECT app.score 
              FROM applications app
              WHERE  app.user_id = apps.user_id
              AND app.status = 'Ended' 
              ORDER BY app.date_ended DESC
              LIMIT 1), 10) AS score
FROM applications apps
WHERE  apps.status = 'Active'
ORDER BY apps.score ASC,
         apps.date_of_application ASC