Question

jquery的

var js={"employer":$employer,"district":$district,"state":$state};

     $.post(url+"/beneficarydetailsdisp",js,function(data){
                                        $("#noofbeneficiaryplacedtable").html(data.result);
            },"json");

我可以加载视图页面，但它显示在firebug中而不是浏览器页面上？

萤火虫的回应

{"result":"<!DOCTYPE html PUBLIC \"-\/\/W3C\/\/DTD XHTML 1.0 Transitional\/\/EN\" \"http:\/\/www.w3.org\/TR\/xhtml1\/DTD\/xhtml1-transitional.dtd\">\r\n<html crossriderapp3491=\"true\" xmlns=\"http:\/\/www.w3.org\/1999\/xhtml\"><head id=\"ctl00_Head1\">\r\n<meta http-equiv=\"content-type\" content=\"text\/html; charset=UTF-8\"><title>\r\n\tJob Placement\r\n<\/title>\r\n<!-- Page layout -->\r\n<link href=\"http:\/\/localhost\/******\/\/css\/StyleSheet.css\" rel=\"stylesheet\" type=\"text\/css\ .....contd"

控制器

$employer ="Hass Enterprises";
    $district ="ADILABAD";
    $state ="ANDHRA PRADESH";
    $data['noofbeneficiaryplaceddetails']=$this->homemodel->getnoofbeneficiaryplaceddetails($employer,$district,$state);    

     $str=$this->load->view('noofbeneficiaryplaced',$data,true);
     $value=array(
            'result'=>$str

        );
        echo json_encode($value);}

我无法加载视图？代码有什么问题

通过post显示视图页面的jquery响应

0 个答案: