由于某种原因,我巨大的查询中的成员ID字段(汽车公司)返回null。我尝试了选择它的每种方式... m.member_id AS member_id,等等,我不知道为什么它返回null当表格中该字段有值时。
<?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());
因为您选择了两个具有相同名称的字段。因此,MySQL将返回最后一个的结果。添加别名:
SELECT m.member_id AS member_id_1, tm.member_id AS member_id_2 ...
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我来说两句