Question

我有一个使用CodeIgniter和数据表生成的查询。

查询如下：

SELECT `tbl_leads`.*, t2`.`username` as `namexx`
FROM `tbl_leads`
JOIN `tbl_users` AS `t2` ON `t2`.`user_id` = JSON_UNQUOTE(JSON_EXTRACT(JSON_KEYS(tbl_leads.permission), '$[0]'))
WHERE   (
`tbl_leads`.`lead_name` LIKE '%d%' ESCAPE '!'
OR  `tbl_leads`.`contact_name` LIKE '%d%' ESCAPE '!'
OR  `tbl_leads`.`email` LIKE '%d%' ESCAPE '!'
OR  `tbl_leads`.`phone` LIKE '%d%' ESCAPE '!'
OR  `tbl_leads`.`lead_status_id` LIKE '%d%' ESCAPE '!'
OR  `tbl_leads`.`permission` LIKE '%d%' ESCAPE '!'
OR  `t2`.`username` LIKE '%d%' ESCAPE '!'
OR  `tbl_leads`.`linkedin` LIKE '%d%' ESCAPE '!'
OR  `tbl_leads`.`leads_id` LIKE '%d%' ESCAPE '!'
 )
AND `converted_client_id` = '0'
ORDER BY `leads_id` DESC
 LIMIT 20

此查询是根据POST搜索请求生成的。

如果您还没有猜到，我会得到

“ where子句”中的未知列“ t2.username”

因为在where查询中无法识别列别名（tbl_users.username也不会更改）。

我正在通过next（数据表模型）生成它：

    if ($this->table == 'tbl_leads') {    
      $this->db->select ('tbl_leads.*, t2.username as namexx'); 
      $this->db->join("tbl_users AS t2", "t2.user_id = JSON_UNQUOTE(JSON_EXTRACT(JSON_KEYS(tbl_leads.permission), '$[0]'))", "LEFT"); 
    }
    $query = $this->db->get();

我还编辑了搜索功能，以便按如下方式获取表指针（我知道这不是一个好习惯，而且是多余的）：

        foreach ($this->column_search as $item) // loop column

    {

        if ($_POST['search']['value']) // if datatable send POST for search

        {

            if ($i === 0) // first loop

            {

                $this->db->group_start(); // open bracket. query Where with OR clause better with bracket. because maybe can combine with other WHERE with AND.
                    if($this->table=='tbl_leads'){
                    if( $item=='namexx'){
                    $this->db->like('tbl_users.username', $_POST['search']['value']);
                        }else{
                        $this->db->like($this->table.'.'.$item, $_POST['search']['value']);

                        }
                    }else{
                    $this->db->like($item, $_POST['search']['value']);

                }

            } else {
                if($this->table=='tbl_leads'){
                if( $item=='namexx'){
                $this->db->or_like('tbl_users.username', $_POST['search']['value']);

                }else{
                $this->db->or_like($this->table.'.'.$item, $_POST['search']['value']);
                }
                }else{
                $this->db->or_like($item, $_POST['search']['value']);

                }

            }

            if (count($this->column_search) - 1 == $i) //last loop

                $this->db->group_end(); //close bracket

        }

        $i++;

    }

Answer 1

修改您的查询，从连接语句中删除t2别名，如下所示：

if ($this->table == 'tbl_leads') {
    $this->db->select ('tbl_leads.*, tbl_users.username as namexx');
    $this->db->join("tbl_users", "tbl_users.user_id = JSON_UNQUOTE(JSON_EXTRACT(JSON_KEYS(tbl_leads.permission), '$[0]'))", "LEFT");
}
$query = $this->db->get();

PHP-WHERE Join中的MYSQL未知列

1 个答案: