区分ID的

Question

这是我的查询：

$tsql = "SELECT * 
FROM 
    dbo.messages m INNER JOIN 
    dbo.contacts c 
    ON m.contactID = c.ID 
WHERE 
    assigned='{$_COOKIE['member_name']}' 
ORDER BY 
    m.status DESC, m.ID DESC"; 

这将返回接触ID而不是消息ID。我如何得到它显示：

<?php echo $row['ID'] ?> 

但随着HTML表单的消息ID和不接触ID？

Answer 1

您可以使用访问CIDÇ表的ID，和M表的id作为中期

像$行[ 'CID']和$行[ '中']

$tsql = "SELECT m.*, c.*, c.id as cid, m.id as mid 
    FROM 
     dbo.messages m INNER JOIN 
     dbo.contacts c 
     ON m.contactID = c.ID 
    WHERE 
     assigned='{$_COOKIE['member_name']}' 
    ORDER BY 
     m.status DESC, m.ID DESC";