弹出确认框选择更改

Question

我有SELECT字段的HTML表单，当我选择选项时，我想启动CONFIRM消息，如果我选择＆＃34;是＆＃34;它应该提交表格。

我该怎么做？

<form action="index.php?page=<?PHP echo $_GET['page'] ?>" method="post" >

    <select name="planID" onChange="this.form.submit()" class="form-control">
            <option value="">Choose plan1</option>
            <option value="">Choose plan2</option>
            <option value="">Choose plan3</option>
    </select>
</form>

Answer 1

HTML：

<form action="index.php?page=<?PHP echo $_GET['page'] ?>" method="post" id="myForm">

    <select name="planID" onchange="if(confirm('Are you sure?')) document.getElementById('myForm').submit();" class="form-control">
            <option value="">Choose plan1</option>
            <option value="">Choose plan2</option>
            <option value="">Choose plan3</option>
    </select>
</form>