php - 传递 javascript/ajax 参数时不会显示 codeigniter View

Question

我正在将 codeigniter 与 phpgrid 集成，但在通过 javascript 和 codeigniter Controller 将行值从 phpgrid( View A 中)传递到另一个 View ( View B)时遇到问题

我在 PHPGRID 中有一个像这样的虚拟列( View A):

$col_formatter = <<<COLFORMATTER
function(cellvalue, options, rowObject, rowid){ 
    var sessid = rowObject[0];

    return '<input type="button" value="View" onclick="btnView('+sessid+')">';
} 
COLFORMATTER;

以及 VIEW A 中的 javascript:

function btnView(sessid){
    var dataRow = {
        sessid:sessid,
    };
    $.ajax({
        type:"POST",
        url: "<?php echo base_url()."index.php/main/tes"; ?>",
        data: dataRow,
        success: function(msg){
        }
    });
    return false;
}

在 Codeigniter Controller 中:

public function tes(){
    $data['sessid'] = $_POST['sessid']; 
    $this->load->view('view_b', $data);
}

我似乎无法加载 View 。我使用 Mozilla 的 Firebug 来了解响应，确实响应是我的 view_b View 的代码，但是我如何切换到该 View ？

Answer 1

//Your are using ajax for some operation and want to reload the view page the you can test these options:
1) take a div in current view page and assing ajax retrun message to that div

function btnView(sessid){
var dataRow = {
    sessid:sessid,
};
$.ajax({
    type:"POST",
    url: "<?php echo base_url()."index.php/main/tes"; ?>",
    data: dataRow,
    success: function(msg){
         $("#divid").html(msg);
    }
});
return false;
}


//Or 2)just redirect to your view page again

function btnView(sessid){
var dataRow = {
    sessid:sessid,
};
$.ajax({
    type:"POST",
    url: "<?php echo base_url()."index.php/main/tes"; ?>",
    data: dataRow,
    success: function(msg){
        window.location.href=path to your view page;//<?php echo base_url()."index.php/controller/function"; ?>
    }
});
return false;
}