mysql - 如何以uid為基準查詢所有一分鐘之內有兩條數據以上的數據
問題描述
如下表
uid order moneytime1 100 10 2016-08-08 12:00:001 101 6 2016-08-08 12:00:582 102 8 2016-08-08 12:02:002 103 10 2016-08-08 12:02:332 104 15 2016-08-08 12:03:003 105 10 2016-08-08 12:03:011 106 10 2016-08-08 12:05:00
根據題意需要找到的數據:
uid order moneytime1 100 10 2016-08-08 12:00:001 101 6 2016-08-08 12:00:582 102 8 2016-08-08 12:02:002 103 10 2016-08-08 12:02:332 104 15 2016-08-08 12:03:00
問題解答
回答1:寫法一:
SELECT * FROM table AS aWHERE EXISTS ( SELECT 1 FROM table AS b WHERE a.uid = b.uid AND b.time >= date_sub(now(), INTERVAL 1 minute) GROUP BY b.uid HAVING count(1) > 1);
寫法二:
SELECT * FROM tableWHERE uid IN ( SELECT uid FROM table WHERE time >= date_sub(now(), INTERVAL 1 minute) GROUP BY uid HAVING count(1) > 1);
建議將date_sub(now(), INTERVAL 1 minute)用程序運算出來再代替進去。
相關文章:
1. javascript - ionic1的插件如何遷移到ionic2的項目中2. java - 如何在Fragment中調用Activity的onNewIntent?3. javascript - h5上的手機號默認沒有識別4. mysql里的大表用mycat做水平拆分,是不是要先手動分好,再配置mycat5. css - 關于input標簽disabled問題6. python - 獲取到的數據生成新的mysql表7. 怎么用css截取字符?8. window下mysql中文亂碼怎么解決??9. javascript - jquery hide()方法無效10. python的文件讀寫問題?
