Question

我正在尝试基于userid加入两个表。我有两个表pr_resignation_requests（列名：userids）和pr_users_details（列名：userid）。我想在userid = ".$val."的两个表中选择所有的entires。我收到了这个错误：

错误号码：1248

每个派生表都必须有自己的别名

SELECT * FROM pr_resignation_requests LEFT JOIN（SELECT firstname，lastname FROM pr_users_details WHERE userid = 81）

我的控制器代码是：

＆＃13;

function view_exit_details(){
			global $SITE,$USER;
			$data = array();
			$data['row'] = new stdClass();
			$data['row'] = $this->admin_init_elements->set_post_vals($this->input->post());
			$data['error_message'] = '';
			$data['row']->id = $data['id'] = $this->uri->segment(3);	
			
			$data['action'] = 'add';
			$data['heading'] = 'Add';			
			$data['msg_class'] = 'sukses';
			
			
			$data['path']=$path;		
			$post_action = $this->input->post('action');
			if($post_action=='add' || $post_action =='update' ){
				$post_array = $this->input->post();
				$action = ($post_action == 'add')?'inserted':'updated';
				$data['error_message'] = $this->exit_common->add_edit_attendance_issue($post_array,$action);
				
			}
			if($data['id']>0){
					$data['rows']=$this->exit_common->get_cat_array2('pr_resignation_requests','id',$data['id'],'');
					
					$data['action'] = 'update';
				}
			$this->data['maincontent'] = $this->load->view('maincontents/view_resignation_action', $data,true); 
			$this->load->view('layout', $this->data);
			
		}

＆＃13;

我的模型代码是：

＆＃13;

		function get_cat_array2($table_name,$col=NULL,$val=NULL,$orderby=NULL){
			
			$result=array();
			$sql = "SELECT * FROM `".$table_name."` LEFT JOIN(SELECT firstname,lastname FROM pr_users_details WHERE userid = `".$val."` )";
			

			$query=$this->db->query($sql);	
			
			if($query->num_rows() > 0){
				$result=$query->result();
			}
			return $result;
	
			
			
		}

＆＃13;

Answer 1

使用下一个代码段：

SELECT *
FROM  pr_resignation_requests
LEFT JOIN 
    (
        SELECT firstname,lastname FROM pr_users_details WHERE userid = 81;
    ) AS a ON  (add u r code of primary key checking)

获取错误：每个派生表都必须有自己的别名

1 个答案: