Question

我有一个服务器端数据表，该表从名为leads的表中获取一些结果。

在此销售线索表中，我有一列名为permission的列，该列已进行json_encoded；

示例：{"13":["view"]}

数字'13'与我的accounts表中的user_id相关联。

搜索正在对其他所有字符串字段进行操作，因此我决定进行某种JOIN语句，以便我也可以将该字段作为字符串读取。

当我意识到我无法真正将JSON数据作为创建ID的简单ID处理时，我陷入了困境。因此，类似：

 if (!empty($this->db->field_exists('permission', $this->table))) {

     $this->db->join("accounts AS t2", "t2.user_id = leads.permission"); 

 }

不起作用，因为t2.user_id确实是一个数字，但是leads.permission是json编码的字符串

如果您需要更多信息，我们将很乐意编辑我的帖子。

谢谢！

模棱两可的错误

 SELECT `tbl_leads`.`email` as `email`, `tbl_leads`.`contact_name` as `contact_name`, `tbl_leads`.`phone` as `phone`, `tbl_leads`.`lead_status_id` as `lead_status_id`, `tbl_leads`.`permission` as `permission`, `tbl_leads`.`linkedin` as `linkedin`, `tbl_leads`.`leads_id` as `leads_id`, `tbl_leads`.`converted_client_id` as `converted_client_id`, `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   (
`lead_name` LIKE '%w%' ESCAPE '!'
OR  `contact_name` LIKE '%w%' ESCAPE '!'
OR  `email` LIKE '%w%' ESCAPE '!'
OR  `phone` LIKE '%w%' ESCAPE '!'
OR  `lead_status_id` LIKE '%w%' ESCAPE '!'
OR  `permission` LIKE '%w%' ESCAPE '!'
OR  `namexx` LIKE '%w%' ESCAPE '!'
OR  `linkedin` LIKE '%w%' ESCAPE '!'
OR  `leads_id` LIKE '%w%' ESCAPE '!'
 )
AND `converted_client_id` = '0'
ORDER BY `leads_id` DESC
 LIMIT 20

这是由

创建的

   if ($this->table == 'tbl_leads') {

        //  $this->db->join("tbl_users AS t2", "t2.user_id = tbl_leads.permission"); 


         $this->db->select ( 
          'tbl_leads.email as email,
          tbl_leads.contact_name as contact_name,
          tbl_leads.phone as phone,
          tbl_leads.lead_status_id as lead_status_id,
          tbl_leads.permission as permission,
          tbl_leads.linkedin as linkedin,
          tbl_leads.leads_id as leads_id,
          tbl_leads.converted_client_id as converted_client_id,
          t2.username as namexx'); 

          $this->db->join("tbl_users AS t2", "t2.user_id = JSON_UNQUOTE(JSON_EXTRACT(JSON_KEYS(tbl_leads.permission), '$[0]'))"); 



        }
        $query = $this->db->get();

Answer 1

您可以结合使用MySQL JSON search functions：

if (!empty($this->db->field_exists('permission', $this->table))) {

    $this->db->join("accounts AS t2", "t2.user_id = JSON_UNQUOTE(JSON_EXTRACT(JSON_KEYS(leads.permission), '$[0]'))"); 

}

首先JSON_KEYS将json键作为数组格式的字符串返回，然后JSON_EXTRACT将其提取为字符串，最后JSON_UNQUOTE将删除引号。

CodeIgniter Datatable联合搜索（json_encoded数据）

1 个答案: