Codeigniter传递数据控制器来查看

nik*_*las 3 php codeigniter

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
}
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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

Nir*_*aju 9

要加载视图,您应该这样做.

$this->load->view("filename");
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如果要传递数据进行查看,则应该这样做.

$this->load->view("filename",$data);
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$data 应该具有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");
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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);
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$data['insertdata'] = $insertdata; 
$this->load->view('print_view',$data);
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有关详细信息,请查看此链接