在提交时传递ajax表单ID

Question

我在页面上有多个ajax表单,它们包含相同的元素,我在成功时使用jquery函数搜索表单和更新的childrean元素,但我需要以某种方式将表单的Id传递给控制器​​以及表单中的其他值,做这个的最好方式是什么？也许添加隐藏字段到窗体和页面加载得到parentid并将其放在这个领域与jQuery.但我不知道jquery做得那么好.

@using (Ajax.BeginForm("Index", "Controller",new AjaxOptions { OnFailure = "error", OnSuccess ="sucess" }))
           {
                <input type="submit" name="value" class="up" value="1" />
                <span class="result">0</span>
                <input type="submit" name="value" class="down" value="-1" />
           }


 public ActionResult Index(int value ,string formID)
        {
           //do something
        }

function sucess(arg) {

        var json = $.parseJSON(arg.responseText);

        var form = $("form#" + json.FormID);

        var up = form.children('input.up');

        var down = form.children('input.down');

        form.children('span.result').replaceWith(json.Result);

        if (json.Result == 1) {

            up.addClass('active');

        }

        if (json.Result == -1) {

            down.addClass('active');

        }
    }

更新:这是部分视图,加载n次,所以我无法找到formID,直到页面在浏览器中呈现.

Answer 1

您可以为表单指定一个唯一ID,并将此id传递给回调:

@{
    var formId = Guid.NewGuid().ToString();
}
@using (Ajax.BeginForm("Index", "Controller", null, new AjaxOptions { OnFailure = "error", OnSuccess = string.Format("success(data, '{0}')", formId) }, new { id = formId }))
{
    <input type="submit" name="value" class="up" value="1" />
    <span class="result">1</span>
    <input type="submit" name="value" class="down" value="-1" />
}

现在在success函数中,您将能够获取服务器返回的JSON结果和表单的id:

function success(json, formId) {
    // Remark you don't need to call $.parseJSON here.
    // inside the success callback the first argument will already
    // represent a parsed object that you can use directly. This obviously
    // assumes that your controller action returns JSON:
    // return Json(new { Result = 1 });

    var form = $("#" + formId);
    var up = form.children('input.up');
    var down = form.children('input.down');
    form.children('span.result').replaceWith(json.Result);
    if (json.Result == 1) {
        up.addClass('active');
    }
    if (json.Result == -1) {
        down.addClass('inactive');
    }
}