Question

如何通过加入以下查询来进行单个查询：

查询1：

$sql = "SELECT total_ce FROM mfb_agent_status_summary AS agent LEFT JOIN mfb_servicelog AS service ON agent.sl_id = service.sl_id WHERE service.h_id LIKE ('".$value[$i]."')  AND service_add_date >= DATE_FORMAT('" . $from . "', '%Y-%m-%d') AND service_add_date <=  DATE_FORMAT('" . $to . "', '%Y-%m-%d')";

查询2：

$query="SELECT DATE_FORMAT(service_add_date, '%d/%b/%y'), DATE_FORMAT(scan_date, '%d/%b/%y'), DATE_FORMAT(ftp_date, '%d/%b/%y'),no_of_cases_reported,no_of_cases_received,case_count_diff, DATEDIFF(ftp_date, service_add_date) FROM mfb_servicelog WHERE h_id LIKE ('".$value[$i]."') AND service_add_date >= DATE_FORMAT('" . $from . "', '%Y-%m-%d') AND service_add_date <=  DATE_FORMAT('" . $to . "', '%Y-%m-%d')";

谢谢！

Answer 1

我在这里对你的架构做了一些假设，但尝试这样的事情：

SELECT agent.total_ce,
       DATE_FORMAT(service.service_add_date, '%d/%b/%y'),
       DATE_FORMAT(service.scan_date, '%d/%b/%y'),
       DATE_FORMAT(service.ftp_date, '%d/%b/%y'),
       service.no_of_cases_reported,
       service.no_of_cases_received,
       service.case_count_diff,
       DATEDIFF(service.ftp_date, service.service_add_date)

FROM mfb_agent_status_summary AS agent
     LEFT JOIN mfb_servicelog AS service ON agent.sl_id = service.sl_id

WHERE service.h_id LIKE ('".$value[$i]."') AND
      agent.service_add_date >= DATE_FORMAT('" . $from . "', '%Y-%m-%d') AND
      agent.service_add_date <=  DATE_FORMAT('" . $to . "', '%Y-%m-%d')

我还想提一下，您应该真正关注prepared statements，而不是直接将$from和$to注入您的查询。你正在打开潜在的SQL injection攻击。

我强烈建议您使用PHP PDO library而非mysqli_*函数。

Answer 2

所以我的猜测是：

$sql = "SELECT 
      agent.total_ce,
      service.*
   FROM mfb_agent_status_summary AS agent 
   LEFT JOIN (
      SELECT 
        DATE_FORMAT(service_add_date, '%d/%b/%y'), 
        DATE_FORMAT(scan_date, '%d/%b/%y'), 
        DATE_FORMAT(ftp_date, '%d/%b/%y'),
        no_of_cases_reported,
        no_of_cases_received,
        case_count_diff, 
        DATEDIFF(ftp_date, service_add_date) 
      FROM mfb_servicelog 
      WHERE h_id LIKE ('".$value[$i]."') 
        AND service_add_date >= DATE_FORMAT('" . $from . "', '%Y-%m-%d') 
        AND service_add_date <=  DATE_FORMAT('" . $to . "', '%Y-%m-%d')
   ) AS service 
   ON agent.sl_id = service.sl_id";

我认为你可以用＆＃39; INNER JOIN＆＃39;;

替换LEFT JOIN

如何在PHP中加入两个MySQL查询

2 个答案: