我有如下 3 个表。我想加入他们并查看 View 中的数据。在我的代码中有一个错误叫做:
A PHP Error was encountered
Severity: Notice
Message: Undefined property: stdClass::$firstname1
Filename: views/boq_doc.php
Line Number: 12
project table
id | staff_id | client_id | location
staff table
id | firstname | lastname | address
client table
client_id | firstname | lastname | address
我写了一个模型如下从数据库中获取表数据:
function show_projects(){
$this->db->select("project.project_name, project.id, client.firstname AS firstname1, client.lastname AS lastname1,staff.firstname AS firstname2, staff.lastname AS lastname2, project.location,project.category, project.start_date, project.end_date");
$this->db->from('project');
$this->db->join('client', 'project.client_id = client.client_id');
$this->db->join('staff', 'staff.id = project.staff_id');
$query = $this->db->get();
return $query->result();
}
我想查看数据,我在 View 中有如下代码:
<?php foreach ($single_project as $project): ?>
<?php echo $project->location->location1; ?>
<?php echo $project->firstname1; ?>
<?php echo $project->lastname1; ?>
<?php echo $project->firstname2; ?>
<?php echo $project->lastname2; ?>
<?php endforeach; ?>
控制者
function show_project_id() {
$id = $this->uri->segment(3);
$data['projects'] = $this->project_list_model->show_projects();
$data['single_project'] = $this->project_list_model->show_project_id($id);
$this->load->view('boq_doc', $data);
}
最佳答案
在 Controller 代码中
$data['projects'] = $this->project_list_model->show_projects();
包含您的查询结果表单模型
所以在 View 中你可以使用
<?php foreach ($projects as $project): ?>
关于php - 如何查看在 codeigniter 中具有相同字段名称的连接表中的数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38091346/