會話ID匹配表時獲取數據

Question

當會話StaffID匹配reports表中的StaffID時，我試圖從report檢索所有報告。

Report_Name, ReportDate and ReportID are part of the report table 

這是我覺得應該去

從report其中ReportID獲取ReportID，Report_Name和ReportDate從Read_Report比賽ReportID從report 當會話StaffID = StaffID在Read_Report

和th是我的查詢

function get_read_report() 
    { 
     $this->db->select('report.Report_Name, report.ReportDate, report.ReportID') 
      ->from('Read_Report') 
      ->join('Read_Report', 'report.ReportID = Read_Report.ReportID') 
      ->where('StaffID', $this->session->userdata("StaffID")); 
     return $result = $this->db->get(); 
    } 

我得到這個錯誤

錯誤編號：1066

Not unique table/alias: 'Read_Report' 

SELECT `report`.`Report_Name`, `report`.`ReportDate`, `report`.`ReportID` FROM `Read_Report` JOIN `Read_Report` ON 

report。 ReportID = Read_Report。 ReportID WHERE StaffID = '3'

Filename: models/report/Report_model.php 

Line Number: 91 

控制器代碼

function my_read_reports() 
    { 
     $data = array(); 

     if ($query = $this->report_model->get_read_report()) { 
      $data['reports'] = $query; 
     } 

     $this->template['middle'] = $this->load->view($this->middle = 'pages/read_reports_view', $data, true); 
     $this->layout(); 
    } 

Answer 1

您需要report表不一樣的表

更改加入你的表

->join('Read_Report', 'report.ReportID = Read_Report.ReportID') 

TO

->join('report', 'report.ReportID = Read_Report.ReportID') 

您需要查詢更改爲

$this->db->select('report.Report_Name, report.ReportDate, report.ReportID') 
      ->from('report') 
      ->join('Read_Report', 'report.ReportID = Read_Report.ReportID') 
      ->where('Read_Report.StaffID', $this->session->userdata("StaffID")); 
     $result = $this->db->get(); 
     return $result->result();// fetch data then return 

Answer 2

您的查詢應該是

$this->db->select('report.Report_Name, report.ReportDate, report.ReportID') 
     ->from('Read_Report') 
     ->join('report', 'report.ReportID = Read_Report.ReportID') 
     ->where('StaffID', $this->session->userdata("StaffID")); 
     return $result = $this->db->get();