Javascript 如何在angularjs';s提交的表格

Javascript 如何在angularjs';s提交的表格,javascript,angularjs,Javascript,Angularjs,我的表格是这样开始的: <!-- form --> <form ng-submit="form.submit()" class="form-horizontal" role="form" style="margin-top: 15px;"> <div class="form-group"> <label for="inputUrl" class="col-sm-2 control-l

我的表格是这样开始的:

    <!-- form -->
    <form ng-submit="form.submit()" class="form-horizontal" role="form" style="margin-top: 15px;">
            <div class="form-group">
                <label for="inputUrl" class="col-sm-2 control-label">URL</label>
                <div class="col-sm-10">
                    <input type="text" class="form-control" id="inputUrl" name="inputUrl" ng-model="form.url" />
                </div>
            </div>
</form>

控制台中的url变量未定义,好方法是什么?

您好,您错过了表单标记中的name属性检查它:

<!-- form -->
<form ng-submit="form.submit()" class="form-horizontal" role="form" style="margin-top: 15px;" name="form">
        <div class="form-group">
            <label for="inputUrl" class="col-sm-2 control-label">URL</label>
            <div class="col-sm-10">
                <input type="text" class="form-control" id="inputUrl" name="url" ng-model="form.url" />
            </div>
        </div>
</form>

统一资源定位地址
<!-- form -->
<form ng-submit="form.submit()" class="form-horizontal" role="form" style="margin-top: 15px;" name="form">
        <div class="form-group">
            <label for="inputUrl" class="col-sm-2 control-label">URL</label>
            <div class="col-sm-10">
                <input type="text" class="form-control" id="inputUrl" name="url" ng-model="form.url" />
            </div>
        </div>
</form>