【发布时间】:2012-07-09 05:57:05
【问题描述】:
我正在使用 codeigniter,我想在 WHERE 子句中使用 IF 条件。但它没有正确执行。
$where_string="( a.FirstName LIKE '".$search_phrase."%' OR ( IF(c.PrivacySettingElephantiUser=1,'a.LastName LIKE \'".$search_phrase."%\' OR CONCAT(a.FirstName, \' \', a.LastName) LIKE \'".$search_phrase."%\' ','')))";
$this->db->where($where_string);
我在这里搜索。由 first name 、 last name 和 first name concat last name 。
我检查了 c.PrivacySettingsElephantiUser 的值。如果它是真的,我正在检查姓氏和名字姓氏连接值,如果c.PrivacySettingsElephantiUser=1 ANDc.PrivacySettingsElephantiUser=0,
这部分没有执行。
OR ( IF(c.PrivacySettingElephantiUser=1,'a.LastName LIKE \'".$search_phrase."%\' OR CONCAT(a.FirstName, \' \', a.LastName) LIKE \'".$search_phrase."%\' ','')
它总是只按名字搜索,忽略 if 条件,
如何正确写?
有没有使用AND ,OR 逻辑的简单方法?
更新
这是我的完整查询
public function search_people($profile_id,$search_phrase=NULL,$country=NULL,$state=NULL,$city=NULL,$neighborhood=NULL,$type=NULL,$limit=NULL,$offset=NULL)
{
$this->db->select('SQL_CALC_FOUND_ROWS a.ProfileID,a.FirstName,a.LastName,a.StateName,a.CityName,a.ShowLastName,a.UserType,a.ProfileImg,b.FriendStatus,b.RequesterID,b.AccepterID',FALSE);
$this->db->from($this->mastables['xxxxxxxx'].' as a');
$this->db->join($this->mastables['yyyyyyyyyyy'].' as b'," b.RequesterID=a.ProfileID AND b.AccepterID=".$profile_id." OR b.AccepterID=a.ProfileID AND b.RequesterID=".$profile_id,'left');
$this->db->where('a.ProfileID !=',$profile_id);
$this->db->where('a.UserType',2);
if($type=="friend")
{
$this->db->join($this->mastables['profile_privacy_settings'].' as c'," c.EntityID=a.ProfileID AND c.UserType=2 AND c.ProfilePrivacySettingDefaultID=1 ",'inner');
$where_string="( a.FirstName LIKE '".$search_phrase."%' OR ( IF(c.PrivacySettingFriend=1,'a.LastName LIKE \'".$search_phrase."%\' OR CONCAT(a.FirstName, \' \', a.LastName) LIKE \'".$search_phrase."%\' ','')))";
$this->db->where($where_string);
$this->db->where('b.FriendStatus',1);
}
else if($type=="other")
{
$this->db->join($this->mastables['profile_privacy_settings'].' as c'," c.EntityID=a.ProfileID AND c.UserType=2 AND c.ProfilePrivacySettingDefaultID=1 ",'inner');
$where_string="( a.FirstName LIKE '".$search_phrase."%' OR ( IF(c.PrivacySettingElephantiUser=1,'a.LastName LIKE \'".$search_phrase."%\' OR CONCAT(a.FirstName, \' \', a.LastName) LIKE \'".$search_phrase."%\' ','')))";
$this->db->where($where_string);
$this->db->where ('(b.FriendStatus IS NULL OR b.FriendStatus=0)');
} else {
$this->db->join($this->mastables['profile_privacy_settings'].' as c'," c.EntityID=a.ProfileID AND c.UserType=2 AND c.ProfilePrivacySettingDefaultID=1 ",'inner');
$where_string="( a.FirstName LIKE '".$search_phrase."%' OR ( IF(c.PrivacySettingElephantiUser=1 OR c.PrivacySettingFriend=1,'a.LastName LIKE \'".$search_phrase."%\' OR CONCAT(a.FirstName, \' \', a.LastName) LIKE \'".$search_phrase."%\'','')))";
//$where_string="( a.FirstName LIKE '".$search_phrase."%' OR ( IF(c.PrivacySettingElephantiUser=1 OR c.PrivacySettingFriend=1,'a.LastName LIKE b%','')))";
$this->db->where($where_string);
}
if($country)
{
$this->db->where('a.CountryID',$country);
}
if($state)
{
$this->db->where('a.StateID',$state);
}
if($city)
{
$this->db->where('a.CityID',$city);
}
if($neighborhood)
{
//$this->db->where('',$neighborhood);
}
if($limit)
{
$this->db->limit($limit,$offset);
}
$query = $this->db->get();
//echo $this->db->last_query();die;
$result['result'] = $query->result_array();
$query = $this->db->query('SELECT FOUND_ROWS() AS `Count`');
$result["totalrows"] = $query->row()->Count;
if(!empty($result))
{
return $result;
}
}
【问题讨论】:
-
我认为你不能在 if 条件下使用
sql expressions,可能这就是@manurajhada 所说的使用CASE的原因 -
@junaid,请告诉我,如何,我尝试过,但报错
-
试试这个-
$where_string="( a.FirstName LIKE '".$search_phrase."%' OR ( CASE WHEN c.PrivacySettingElephantiUser = 1 THEN 'a.LastName LIKE \'".$search_phrase."%\' OR CONCAT(a.FirstName, \' \', a.LastName) LIKE \'".$search_phrase."%\' ' ELSE '' )))";
标签: mysql sql codeigniter where-clause