在ASP.NET MVC中使用jQuery渲染局部视图

Question

How do I render the partial view using jquery?

We can render the partial View like this:

<% Html.RenderPartial("UserDetails"); %>

How can we do the same using jquery?

Answer 1

You can't render a partial view using only jQuery. You can, however, call a method (action) that will render the partial view for you and add it to the page using jQuery/AJAX. In the below, we have a button click handler that loads the url for the action from a data attribute on the button and fires off a GET request to replace the DIV contained in the partial view with the updated contents.

$('.js-reload-details').on('click', function(evt) {
    evt.preventDefault();
    evt.stopPropagation();

    var $detailDiv = $('#detailsDiv'),
        url = $(this).data('url');

    $.get(url, function(data) {
        $detailDiv.replaceWith(data);         
    });
});

where the user controller has an action named details that does:

public ActionResult Details( int id )
{
    var model = ...get user from db using id...

    return PartialView( "UserDetails", model );
}

This is assuming that your partial view is a container with the id detailsDiv so that you just replace the entire thing with the contents of the result of the call.

Parent View Button

 <button data-url='@Url.Action("details","user", new { id = Model.ID } )'
         class="js-reload-details">Reload</button>

User is controller name and details is action name in @Url.Action() . UserDetails partial view

<div id="detailsDiv">
    <!-- ...content... -->
</div>

Answer 2

我使用ajax加载来执行此操作：

$('#user_content').load('@Url.Action("UserDetails","User")');

Answer 3

@tvanfosson rocks with his answer.

However, I would suggest an improvement within js and a small controller check.

When we use @Url helper to call an action, we are going to receive a formatted html. It would be better to update the content ( .html ) not the actual element ( .replaceWith ).

More about at: What's the difference between jQuery's replaceWith() and html()?

$.get( '@Url.Action("details","user", new { id = Model.ID } )', function(data) {
    $('#detailsDiv').html(data);
}); 

This is specially useful in trees, where the content can be changed several times.

At the controller we can reuse the action depending on requester:

public ActionResult Details( int id )
{
    var model = GetFooModel();
    if (Request.IsAjaxRequest())
    {
        return PartialView( "UserDetails", model );
    }
    return View(model);
}

Answer 4

Another thing you can try (based on tvanfosson's answer) is this:

<div class="renderaction fade-in" 
    data-actionurl="@Url.Action("details","user", new { id = Model.ID } )"></div>

And then in the scripts section of your page:

<script type="text/javascript">
    $(function () {
        $(".renderaction").each(function (i, n) {
            var $n = $(n),
                url = $n.attr('data-actionurl'),
                $this = $(this);

            $.get(url, function (data) {
                $this.html(data);
            });
        });
    });

</script>

This renders your @Html.RenderAction using ajax.

And to make it all fansy sjmansy you can add a fade-in effect using this css:

/* make keyframes that tell the start state and the end state of our object */
@-webkit-keyframes fadeIn { from { opacity:0; } to { opacity:1; } }
@-moz-keyframes fadeIn { from { opacity:0; } to { opacity:1; } }
@keyframes fadeIn { from { opacity:0; } to { opacity:1; } }

.fade-in {
    opacity: 0; /* make things invisible upon start */
    -webkit-animation: fadeIn ease-in 1; /* call our keyframe named fadeIn, use animattion ease-in and repeat it only 1 time */
    -moz-animation: fadeIn ease-in 1;
    -o-animation: fadeIn ease-in 1;
    animation: fadeIn ease-in 1;
    -webkit-animation-fill-mode: forwards; /* this makes sure that after animation is done we remain at the last keyframe value (opacity: 1)*/
    -o-animation-fill-mode: forwards;
    animation-fill-mode: forwards;
    -webkit-animation-duration: 1s;
    -moz-animation-duration: 1s;
    -o-animation-duration: 1s;
    animation-duration: 1s;
}

Man I love mvc :-)

Answer 5

You'll need to create an Action on your Controller that returns the rendered result of the "UserDetails" partial view or control. Then just use an Http Get or Post from jQuery to call the Action to get the rendered html to be displayed.

Answer 6

Using standard Ajax call to achieve same result

        $.ajax({
            url: '@Url.Action("_SearchStudents")?NationalId=' + $('#NationalId').val(),
            type: 'GET',
            error: function (xhr) {
                alert('Error: ' + xhr.statusText);

            },
            success: function (result) {

                $('#divSearchResult').html(result);
            }
        });




public ActionResult _SearchStudents(string NationalId)
        {

           //.......

            return PartialView("_SearchStudents", model);
        }

Answer 7

I did it like this.

$(document).ready(function(){
    $("#yourid").click(function(){
        $(this).load('@Url.Action("Details")');
    });
});

Details Method:

public IActionResult Details()
        {

            return PartialView("Your Partial View");
        }

Answer 8

If you need to reference a dynamically generated value you can also append query string paramters after the @URL.Action like so:

    var id = $(this).attr('id');
    var value = $(this).attr('value');
    $('#user_content').load('@Url.Action("UserDetails","User")?Param1=' + id + "&Param2=" + value);


    public ActionResult Details( int id, string value )
    {
        var model = GetFooModel();
        if (Request.IsAjaxRequest())
        {
            return PartialView( "UserDetails", model );
        }
        return View(model);
    }