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
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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至
$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