2014-10-28 50 views
1

出于某种原因,我的巨大查询中的成员id字段(auto inc。)正在返回null.I已尝试过每种选择方式... m.member_id AS member_id等。我无法弄清楚它为什么是当表中的该字段有值时返回null。为什么我的成员id列返回null?

<?php  
public function get_info($criteria = 0){    
     if(is_numeric($criteria)){ 
      $where = "WHERE m.member_id = ".$criteria; 
     } else { 
      $where = "WHERE email_address = '".$criteria."'"; 
     } 
     $query_member = " 
      SELECT 
       m.member_id AS member_id, m.display_name, m.email_address, m.group_id, m.status, m.activation_code, UNIX_TIMESTAMP(m.date_joined) AS date_joined, 
       m.gender, m.location, m.biography, m.mantra, m.birth_date, m.results_per_page, m.admin_emails, m.member_emails, m.last_active, m.avatar_id, 
       m.banner_id, m.signature, m.newsletter_subscription, m.recruiting_status, m.facebook_username, m.website, m.steam_username, m.xboxlive_gamertag, m.psn_id, 
       g.group_id, g.title, g.description, 
       a.attachment_id, a.file_name,  
       f.message_id, f.author_id, COUNT(f.message_id) AS forum_count, 
       b.attachment_id AS banner_id, b.file_name AS banner_file, 
       mr.request_id, mr.author_id, mr.recipient_id, mr.status, COUNT(mr.request_id) AS total_friends, 
       tm.team_member_id, tm.member_id, tm.team_id 
      FROM members AS m 
      LEFT JOIN member_groups AS g ON (m.group_id = g.group_id) 
      LEFT JOIN attachments AS a ON (m.avatar_id = a.attachment_id)   
      LEFT JOIN forum_messages AS f ON (m.member_id = f.author_id) 
      LEFT JOIN attachments AS b ON (m.banner_id = b.attachment_id) 
      LEFT JOIN member_requests AS mr ON (m.member_id = mr.author_id OR m.member_id = mr.recipient_id) AND mr.status = 1 
      LEFT JOIN team_members AS tm ON (m.member_id = tm.member_id) AND date_left = '' 
      ".$where." 
      GROUP BY m.member_id 
      LIMIT 1";  

     //show_error($query_member);  
     if($query_member = $this->db->query($query_member)){    
      if($query_member->num_rows() > 0){      
       var_dump($query_member->row_array()); 

回答

0

因为您选择了两个同名的字段。所以MySQL会返回最后一个的结果。添加别的:

SELECT m.member_id AS member_id_1, tm.member_id AS member_id_2 ...