Foreach内で複数のボタンを生成する(itemはforeachの要素となります) <form method="post"> <script src="<%= Url.Content("~/Scripts/jquery-1.7.1.min.js") %>" type="text/javascript"></script> …省略… <% foreach (var item in Model) { %> <input id="buttonwrong<%= item.ID %>" type="submit" name="action:WrongSubmit" onclick="DoAjaxPostWrong(this)" value="×" /> JQueryコード function DoAjaxPostWrong(btnClicked) { var $form = $(btnClicked).parents('form'); var id = btnClicked.id; var postData = id.substring(11); $.ajax({ cache: false, async: true, type: "POST", url: '<%=Url.Action("WrongSubmit","Examine") %>', data: { id: postData }, error: function (xhr, status, error) { alert("Error"); return false; }, success: function (response) { //do something with response alert("Success"); return false; } }); return false; } コントローラ側 [HttpPost] public ContentResult WrongSubmit(int id) { string result = string.Empty; WordModels wordmodels = db.Find(id); if (wordmodels != null) { wordmodels.ErrorTimes++; db.Edit(wordmodels); db.SaveChanges(); result = "Success"; } else { result = "Failure"; } return new ContentResult { Content = result }; } セシウム137を97.7%吸着 コメント: |