这个问题已经在这里有了答案 :
在同一网页上显示两个表的数据 (1个答案)
7年前关闭。
我在CodeIgniter中使用联接查询,但无法使其正常工作。它仅显示一个表数据,而不显示其他表数据。我是CodeIgniter的新手,无法解决此问题。请别人帮我。提前谢谢。
看法
<?php foreach ($query as $row): ?> <?php echo $row->answerA;?><br> <?php echo $row->answerB;?><br> <?php echo $row->answerC;?><br> <?php echo $row->comment;?><br> <?php echo $row->name; ?> <?php endforeach; ?>
控制器
function getall() { $this->load->model('result_model'); $data['query'] = $this->result_model->result_getall(); // print_r($data['query']); // die(); $this->load->view('result_view', $data); }
模型
function result_getall() { $this->db->select('tblanswers.*,credentials.*'); $this->db->from('tblanswers'); $this->db->join('credentials', 'tblanswers.answerid = credentials.cid', 'right outer'); $query = $this->db->get(); return $query->result(); }
编辑
的结果
print_r($data['query']); die();
是一个如下数组:
Array ( [0] => stdClass Object ( [answerid] => [userid] => [questionid] => [answerA] => [answerB] => [answerC] => [comment] => [cid] => 83 [name] => Edvinas [second_name] => liutvaits [phone] => [email] => ledvinas@yahoo.ie ) [1] => stdClass Object ( [answerid] => [userid] => [questionid] => [answerA] => [answerB] => [answerC] => [comment] => [cid] => 84 [name] => Edvinas [second_name] => liutvaits [phone] => [email] => ledvinas@yahoo.ie ) [2] => stdClass Object ( [answerid] => [userid] => [questionid] => [answerA] => [answerB] => [answerC] => [comment] => [cid] => 85 [name] => Edvinas [second_name] => Liutvaitis [phone] => [email] => ledvinas@yahoo.ie ) [3] => stdClass Object ( [answerid] => [userid] => [questionid] => [answerA] => [answerB] => [answerC] => [comment] => [cid] => 86 [name] => EdvinasQ [second_name] => LiutvaitisQ [phone] => 12345678 [email] => ledvinas@yahoo.ie ) [4] => stdClass Object ( [answerid] => [userid] => [questionid] => [answerA] => [answerB] => [answerC] => [comment] => [cid] => 87 [name] => Edvinas [second_name] => Liutvaitis [phone] => 123456 [email] => ledvinas@yahoo.ie ) )
表结构
证书
cid(PRIMARY),姓名,second_name,电话,电子邮件
tblanswers
answerid(PRIMARY),userid,questionid,answerA,answerB,answerC。
试试这个:
$this->db->join('credentials', 'tblanswers.answerid = credentials.cid');
或者
$this->db->join('credentials', 'tblanswers.answerid = credentials.cid', 'inner');
并打印结果以查看是否要