php – Codeigniter传递数据控制器来查看
发布时间:2020-12-13 22:11:21 所属栏目:PHP教程 来源:网络整理
导读:这是我的控制器: class CommonController extends CI_Controller { public function __construct() { parent::__construct(); $this-load-model('common_model'); //load your model my model is "common model" } public function add_work(){ $names = $_
这是我的控制器:
class CommonController extends CI_Controller { public function __construct() { parent::__construct(); $this->load->model('common_model'); //load your model my model is "common model" } public function add_work(){ $names = $_POST['name']; $works = $_POST['work']; $allValues = array(); // array to contains inserted rows foreach($names as $key => $name){ $name= "your specified name"; $insertdata = array(); $insertdata['work'] = $works[$key]; $insertdata['name'] = $name; $this->common_model->insert($insertdata); array_push($allValues,$insertdata); //$insert = mysql_query("INSERT INTO work(name,work) values ( '$name','$work')"); } foreach($allValues as $insertRow){ echo $insertRow['work']; echo $insertRow['name'];//this shows data well. but how to pass data in view.php } //view code will add here to show data in browser } 基本上我想将所有数据传递给view.php进行打印或导出.我该怎么办 解决方法
要加载视图,您应该这样做.
$this->load->view("filename"); 如果要传递数据进行查看,则应该这样做. $this->load->view("filename",$data); $data应包含要在视图中打印的所有参数. 语法是这样的. $this->load->view("filename","data to view","Returning views as data(true / false"); 如果第三个参数为true,则视图将作为数据.它不会作为输出转到浏览器. 编辑: 更改 $this->load->view('print_view',$insertdata); 至 $data['insertdata'] = $insertdata; $this->load->view('print_view',$data); 有关详细信息,请查看this link (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |