2015-10-22 75 views
2

我正在建立一個站點搜索與選擇框,以獲得某些類別&地區。 如果選擇我將其包括在查詢中,但由於某種原因,使用LIKE它會忽略查詢的WHERE部分!?!?爲什麼那個或我做錯了什麼?CodeIgniter 3查詢ESCAPE'!'忽略WHERE語句?

當我呼應笨內置查詢我得到的fololowing: (注意ESCAPE '!')

查詢回聲:

 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 

這裏是在控制器中的實際查詢:

 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; 

    } 
+0

的'ESCAPE「!」'不影響您的查詢,是完全可以接受的SQL語法。這意味着如果你想搜索一個通配符的文字實例,你可以在它之前加上'!'。這不會影響你的結果。 – DFriend

+0

謝謝@DFriend。非常感激。 – Howzit

+0

我認爲你的查詢沒有返回結果? – DFriend

回答

2

你的WHERE條件

a AND b AND c AND d AND ... AND x OR y 

如果y是真的整個WHERE條件爲真。

也許你的意思是:

a AND b AND c AND d AND ... AND (x OR y) 

轉義字符默認爲\,但它看起來像笨改變了這!你(大概是因爲\已經在PHP中逃生,所以有時你需要數倍這可能會讓人困惑)。

目前,這是不相關的查詢。如果你需要匹配一個%_!%!_(默認\%\_)這隻會被使用。

+0

啊!感謝您指出@Arth有時候感覺像一個白癡錯過了明顯! – Howzit

+0

還有一個問題@Arth查詢怎麼會是這樣確保 $這個 - > DB->像(「文本」,$ fsearch); $ this-> db-> or_like('f_value',$ fsearch);是(x或y)格式? – Howzit

+0

我不知道代碼點火器,但我想它是在文檔的某處 – Arth