如何从POST请求

时间:2016-06-22 11:55:59

标签: javascript html http-post

我想获取HTML表单的一个字段,并使用它来修改操作的URL。我关注了this question,其中显示了GET请求是如何完成的,但我在修改POST请求的技术时遇到了问题。

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
    <head>
        <title>User location lookup</title>
    </head>

    <body>
        <!-- method: https://stackoverflow.com/questions/11811782/form-value-creates-a-url -->

        <script>
            function process()
            {
                var url="http://localhost:43/" + document.getElementById("url").value;
                location.href=url;
                return false;
            }
        </script>

        <form onSubmit="return process();" method="get">
            Username: <input type="text" name="url" id="url"></input>
            <input type="submit" value="go"></input>
        </form>

        <script>
            function process2()
            {
                var url="http://localhost:43/" + document.getElementById("url").value;
                location.href=url;
                return false;
            }
        </script>

        <form method="post"  onSubmit="return process2();">
            Username: <input type="text" name="url" id="url"> </input>
            Location: <input type="text" name="location" id="location"> </input>
            <input type="submit" value="go"></input>
        </form>
    </body>
</html>

GET请求正确发送:

GET /brian HTTP/1.1
Accept: text/html, application/xhtml+xml, */*
Accept-Language: en-GB
User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64; Trident/7.0; rv:11.0) like Gecko
Accept-Encoding: gzip, deflate
Host: localhost:43
DNT: 1
Connection: Keep-Alive

对于POST我得到了:

GET /brian HTTP/1.1
Accept: text/html, application/xhtml+xml, */*
Accept-Language: en-GB
User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64; Trident/7.0; rv:11.0) like Gecko
Accept-Encoding: gzip, deflate
Host: localhost:43
DNT: 1
Connection: Keep-Alive

即使该方法为POST,仍然是GET。

我想要产生的是:

POST /brian HTTP/1.1
Accept: text/html, application/xhtml+xml, */*
Accept-Language: en-GB
User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW64; Trident/7.0; rv:11.0) like Gecko
Accept-Encoding: gzip, deflate
Host: localhost:43
Content-Length: 11
DNT: 1
Connection: Keep-Alive

location=SO

有人可以用尽可能少的JavaScript代码指出我正确的方向吗?

1 个答案:

答案 0 :(得分:4)

您的代码未提交表单。它实际上是通过GET直接调用URL,即:此方法仅适用于GET。

请尝试:

<form id="myformid" method="post">
    Username: <input type="text" name="url" id="url2" onChange="updateAction()"> </input>
    Location: <input type="text" name="location" id="location"> </input>
    <input id="mysubmit" type="submit" value="go"></input>
</form>

<script>
    function updateAction() {
        var url = document.getElementById("url2").value;
        document.getElementById("myformid").setAttribute("action", "http://localhost:43/" + url);
    }
</script>