Question

需要查找距离给定日期时间过去10分钟的记录。从下面的查询中，它从给定日期开始提前10分钟获取记录，该日期又选择前一天或前一天的所有记录。

因为我期望从之前10分钟的给定日期检索所有记录。但并非所有记录，即提到的给定日期的前几天。

$sql = "SELECT otb.*
FROM oc_table_book otb
LEFT JOIN oc_order o
ON (o.order_id = otb.order_id)
WHERE otb.preorder_status_id = '3' AND 
otb.booked_date <= '" . $this->db->escape($table_data['booked_date']) . 
"' - INTERVAL 10 MINUTE ORDER BY otb.booked_date";

尝试使用DATEADD（）无法正常工作

$sql = "SELECT otb.*
FROM oc_table_book otb
LEFT JOIN oc_order o
ON (o.order_id = otb.order_id)
WHERE otb.preorder_status_id = '3' AND 
otb.booked_date <= dateadd(minute, -10, '" . $this->db->escape($table_data['booked_date']) . 
")";

结果

Fatal error: Uncaught Exception: Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''2018-01-19 07:30:00)' at line 6<br />Error No: 1064<br />SELECT otb.* FROM oc_table_book otb LEFT JOIN oc_order o ON (o.order_id = otb.order_id) WHERE otb.preorder_status_id = '3' AND otb.booked_date <= dateadd(minute, -10, '2018-01-19 07:30:00) in /var/www/html/silverspoon/system/library/db/mysqli.php on line 40

Answer 1

您似乎想要：

SELECT otb.*
FROM oc_table_book otb JOIN  -- I don't see why LEFT JOIN is necessary
     oc_order o
     ON o.order_id = otb.order_id
WHERE otb.preorder_status_id = 3 AND -- probably an integer
      otb.booked_date <= ? - INTERVAL 10 MINUTE AND
      otb.booked_date >= DATE(?)
ORDER BY otb.booked_date;

请注意?的使用。这是一个参数占位符。您应该使用它将常量值传递给查询。

Answer 2

SQL语法中有错误

尝试此查询：

SELECT otb.* FROM oc_table_book otb
LEFT JOIN oc_order o
ON o.order_id = otb.order_id
WHERE otb.preorder_status_id = '3' AND 
otb.booked_date <= dateadd(minute, -10,".$this->db->escape($table_data['booked_date']).")

在给定日期时间前10分钟检索记录

2 个答案: