发行检索从SQL查询结果PHP

Question

我的代码执行SQL查询，如下图所示：

$query = 'SELECT 
     cl.*,c.id,c.type,c.firstname,c.surname,c.job,c.company,c.directorycompany_id, dc.id, dc.name, es.id, GROUP_CONCAT(dirc.name), GROUP_CONCAT(dirp.name), GROUP_CONCAT(dirs.name) 
     FROM contactlist_contact cl '. 
     'INNER JOIN contact c ON cl.contact_id = c.id '. 
     'LEFT JOIN directorycompany dc ON dc.id = c.directorycompany_id '. 
     'LEFT JOIN expertsection es ON es.id = c.expertsection_id '. 
     'LEFT JOIN directorycolumn dirc ON dirc.directorycontact_id = c.id '. 
     'LEFT JOIN directoryprogramme dirp ON dirp.directorycontact_id = c.id '. 
     'LEFT JOIN directorysupplement dirs ON dirs.directorycontact_id = c.id '. 
     'WHERE cl.contactlist_id = '.$contactListId.' ORDER BY dc.surname'; 

$db->query($query,$t); 

我试图检索结果中的内容返回如图所示：

while($db->next($t)) 
{ 
    $contact = new DirectoryContact(0); 
    $contact->setId($db->get('c.id',$t)); 
    $contact->setContactType($db->get('c.type',$t)); 
    $contact->setFirstname($db->get('c.firstname',$t)); 
    $contact->setSurname($db->get('c.surname',$t)); 
    $contact->setJob($db->get('c.job',$t)); 
    $contact->setCompany($db->get('c.company',$t)); 
    $contact->setDisplayColumn($db->get('GROUP_CONCAT(dirc.name)', $t)); 
    $contact->setDisplaySupplement($db->get('GROUP_CONCAT(dirs.name)', $t)); 
    $contact->setDisplayProgramme($db->get('dirp.name', $t)); 

    $directoryCompany = new DirectoryCompany(0); 
    $directoryCompany->setId($db->get('dc.id',$t)); 
    $directoryCompany->setName($db->get('dc.name',$t)); 

    $expertSection = new ExpertSection(0); 
    $expertSection->setId($db->get('es.id',$t)); 

    $contact->setDirectoryCompany($directoryCompany); 
    $contact->setExpertSection($expertSection); 

    $this->contacts->add($contact); 
} 

对于一些原因，每一次，这些线路都返回false。

$contact->setDisplayColumn($db->get('GROUP_CONCAT(dirc.name)', $t)); 
$contact->setDisplaySupplement($db->get('GROUP_CONCAT(dirs.name)', $t)); 
$contact->setDisplayProgramme($db->get('GROUP_CONCAT(dirp.name)', $t)); 

有没有人知道我要去哪里错了？非常感谢。

Answer 1

我不知道它会解决问题，但也许你可以尝试一个别名; e.g：

$query = 'SELECT cl.*,c.id,c.type,c.firstname,c.surname,c.job,c.company,c.directorycompany_id, dc.id, dc.name, es.id, GROUP_CONCAT(dirc.name) AS gcname, GROUP_CONCAT(dirp.name) AS gpname, GROUP_CONCAT(dirs.name) AS gsname 
FROM contactlist_contact cl '. 
'INNER JOIN contact c ON cl.contact_id = c.id '. 
'LEFT JOIN directorycompany dc ON dc.id = c.directorycompany_id '. 
'LEFT JOIN expertsection es ON es.id = c.expertsection_id '. 
'LEFT JOIN directorycolumn dirc ON dirc.directorycontact_id = c.id '. 
'LEFT JOIN directoryprogramme dirp ON dirp.directorycontact_id = c.id '. 
'LEFT JOIN directorysupplement dirs ON dirs.directorycontact_id = c.id '. 
'WHERE cl.contactlist_id = '.$contactListId.' ORDER BY dc.surname'; 
$db->query($query,$t); 

然后，而不是：

$contact->setDisplayColumn($db->get('GROUP_CONCAT(dirc.name)', $t)); 
$contact->setDisplaySupplement($db->get('GROUP_CONCAT(dirs.name)', $t)); 
$contact->setDisplayProgramme($db->get('GROUP_CONCAT(dirp.name)', $t)); 

务必：

$contact->setDisplayColumn($db->get('gcname', $t)); 
$contact->setDisplaySupplement($db->get('gsname', $t)); 
$contact->setDisplayProgramme($db->get('gpname', $t));