實現ajax提交表單,頁面不跳轉
阿新 • • 發佈:2019-02-19
表單程式碼
js程式碼:<form id="save" class="form-horizontal" role="form" method="post" action="${request.getContextPath()}/yuyue/saveBingli.do"> <div class="form-group"> <label id="username" class="col-sm-2 control-label"> </label> <div class="col-sm-6"> <input id="username" type="text" name="username" class="form-control" value="${editBingliObj.username}" readonly="readonly"/> </div> </div> <div class="form-group"> <label for="symptom" class="col-sm-2 control-label"></label> <div class="col-sm-6"> <textarea id="symptom" name="symptom" class="form-control" rows="2"></textarea> </div> </div> <div class="form-group"> <label for="content" class="col-sm-2 control-label"></label> <div class="col-sm-6"> <textarea id="content" name="content" class="form-control" rows="4"></textarea> </div> </div> <div class="form-group"> <div class="col-sm-offset-2 col-sm-6"> <button type="button" class="btn btn-primary" onclick="save();">儲存</button> </div> </div> </form>
為表單的按鈕繫結事件,點選時驗證表單,滿足條件下提交表單,為表單繫結submit事件,提交的時候返回false阻止預設事件,ajax提交表單。function save(){ if($("#symptom").val()!=""&&$("#content").val()!=""){ $("#save").submit(); } } var contextPath; $(function(){ contextPath=$("#contextPath").val(); $("#save").submit(function(){ $.ajax({ type : "post", url : contextPath+"/yuyue/save.do", data : $("#save").serialize(), success : function(){ } }); return false; }); });