Html 如何提交更改下拉列表的表单?

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时间:2020-08-29 10:20:19  来源:igfitidea点击:

How to submit form on change of dropdown list?

htmljsp

提问by John

I am creating a page in JSP where I have a dropdown list and once the user selects a value he has to click on the go button and then the value is sent to the Servlet.

我正在 JSP 中创建一个页面,其中有一个下拉列表,一旦用户选择了一个值,他必须单击 go 按钮,然后将该值发送到 Servlet。

            </select>
            <input type="submit" name="GO" value="Go"/>

How do I make it so that it does it on change? E.g. when the user selects John all his details are retrived from the DB and displayed. I want the system to do it without having to click the go button.

我如何制作它以便它在更改时执行?例如,当用户选择 John 时,他的所有详细信息都会从数据库中检索并显示出来。我希望系统无需单击执行按钮即可执行此操作。

回答by BalusC

Just ask assistance of JavaScript.

只需请求 JavaScript 的帮助。

<select onchange="this.form.submit()">
    ...
</select>

See also:

也可以看看:

回答by MD Sayem Ahmed

Simple JavaScript will do -

简单的 JavaScript 就可以了——

<form action="myservlet.do" method="POST">
    <select name="myselect" id="myselect" onchange="this.form.submit()">
        <option value="1">One</option>
        <option value="2">Two</option>
        <option value="3">Three</option>
        <option value="4">Four</option>
    </select>
</form>

Here is a link for a good javascript tutorial.

这是一个很好的javascript 教程的链接。

回答by sate wedos

other than using this.form.submit()you also submiting by id or name. example i have form like this : <form action="" name="PostName" id="IdName">

除了使用this.form.submit()您之外,您还可以通过 id 或 name 提交。例如我有这样的表格:<form action="" name="PostName" id="IdName">

  1. By Name : <select onchange="PostName.submit()">

  2. By Id : <select onchange="IdName.submit()">

  1. 按名字 : <select onchange="PostName.submit()">

  2. 通过 ID : <select onchange="IdName.submit()">

回答by nitsram

To those in the answer above. It's definitely JavaScript. It's just inline.

对于上面答案中的那些人。这绝对是 JavaScript。它只是内联。

BTW the jQuery equivalent if you want to apply to all selects:

顺便说一句,如果您想应用于所有选择,则等效于 jQuery:

$('form select').on('change', function(){
    $(this).closest('form').submit();
});