【发布时间】:2016-01-21 22:39:55
【问题描述】:
我正在构建一个带有选择框的站点搜索,以获取某些类别和地区。 如果选择我是否将其包含在查询中,但由于某种原因,使用 LIKE 是否会忽略查询的 WHERE 部分!?!?为什么会这样或者我做错了什么?
当我回显 Codeigniter 构建的查询时,我得到以下信息: (注意 ESCAPE '!')
查询回显:
SELECT SQL_CALC_FOUND_ROWS null as rows, ads.id AS id, location, provLabel, text,
adcat.id AS catid, ads.subcatid AS subcatid, ads.province R_rand, r_option, addate,
adcat.name AS catname, adsubcat.name AS subname, f_value, adtitle, ads.area, regionLabel,
adlink
FROM `ads`
JOIN `search_town` ON `search_town`.`townId`=`ads`.`townId`
JOIN `search_region` ON `search_region`.`regionId`=`ads`.`area`
JOIN `search_prov` ON `search_prov`.`provId`=`ads`.`province`
JOIN `adcat` ON `adcat`.`id`=`ads`.`catid`
JOIN `adsubcat` ON `adsubcat`.`id`=`ads`.`subcatid`
LEFT JOIN `adfields` ON `adfields`.`ad_id`=`ads`.`id`
WHERE `ads`.`catid` != 8 AND `ads`.`adactive` = 1 AND `scam` =0 AND `ads`.`province` = '1'
AND `ads`.`catid` = '3' AND `text` LIKE '%Nissan%' ESCAPE '!'
OR `f_value` LIKE '%Nissan%' ESCAPE '!'
GROUP BY `ads`.`id`
ORDER BY `addate` DESC
LIMIT 10
这是控制器中的实际查询:
public function get_search($fsearch, $fcategory, $fprovince, $farea, $limit, $start)
{
if($fcategory >=1){
$incl_cat=" AND ads.catid='$fcategory'";
}else{
$incl_cat='';
}
if($fprovince>=1){
$incl_prov=" AND ads.province='$fprovince'";
}else{
$incl_prov='';
}
if($farea >= 1){
$incl_area=" AND ads.area='$farea'";
}else{
$incl_area='';
}
$this->db->select('SQL_CALC_FOUND_ROWS null as rows, ads.id AS id, location, provLabel, text, adcat.id AS catid, ads.subcatid AS subcatid,ads.province
R_rand, r_option, addate, adcat.name AS catname, adsubcat.name AS subname, f_value, adtitle, ads.area, regionLabel,
adlink', FALSE);
$this->db->from('ads');
$this->db->join('search_town', 'search_town.townId=ads.townId');
$this->db->join('search_region', 'search_region.regionId=ads.area');
$this->db->join('search_prov', 'search_prov.provId=ads.province');
$this->db->join('adcat', 'adcat.id=ads.catid');
$this->db->join('adsubcat', 'adsubcat.id=ads.subcatid');
$this->db->join('adfields', 'adfields.ad_id=ads.id', 'left');
$where = "ads.catid!=8 AND ads.adactive=1 AND scam=0 $incl_prov $incl_cat $incl_area";
$this->db->where($where);
$this->db->like('text', $fsearch);
$this->db->or_like('f_value', $fsearch);
$this->db->group_by("ads.id");
$this->db->order_by('addate', 'DESC');
$this->db->limit($limit, $start);
$query = $this->db->get();
$return = $query->result_array();
echo $this->db->last_query();
$total_results=$this->db->query('SELECT FOUND_ROWS() count;')->row()->count;
$this->session->set_userdata('tot_search', $total_results);
return $return;
}
【问题讨论】:
-
ESCAPE '!'不会影响您的查询,是完全可以接受的 SQL 语法。这意味着,如果您想搜索通配符的文字实例,请在其前面加上“!”。它不会影响您的结果。 -
感谢@DFriend。非常感谢。
-
我认为您的查询没有返回结果?
-
事实上确实如此。下面的 Arth 准确地解释了我遇到的问题。
标签: php mysql codeigniter escaping sql-like