python - 如何使用 WSGIREF 捕获 POST

Question

我正在尝试从简单的表单中捕获 POST 数据。

这是我第一次使用 WSGIREF，我似乎找不到正确的方法。

This is the form:
<form action="test" method="POST">
<input type="text" name="name">
<input type="submit"></form>

这个函数显然缺少正确的信息来捕获帖子:

def app(environ, start_response):
    """starts the response for the webserver"""
    path = environ[ 'PATH_INFO']
    method = environ['REQUEST_METHOD']
    if method == 'POST':
        if path.startswith('/test'):
            start_response('200 OK',[('Content-type', 'text/html')])
            return "POST info would go here %s" % post_info
    else:
        start_response('200 OK', [('Content-type', 'text/html')])
        return form()

Answer 1

您应该读取来自服务器的响应。

来自nosklo's answer类似的问题:“PEP 333 说 you must read environ['wsgi.input'] 。”

测试代码(改编自 this answer ):
警告:此代码仅用于演示目的。
警告:尽量避免硬编码路径或文件名。

def app(environ, start_response):
    path    = environ['PATH_INFO']
    method  = environ['REQUEST_METHOD']
    if method == 'POST':
        if path.startswith('/test'):
            try:
                request_body_size = int(environ['CONTENT_LENGTH'])
                request_body = environ['wsgi.input'].read(request_body_size)
            except (TypeError, ValueError):
                request_body = "0"
            try:
                response_body = str(request_body)
            except:
                response_body = "error"
            status = '200 OK'
            headers = [('Content-type', 'text/plain')]
            start_response(status, headers)
            return [response_body]
    else:
        response_body = open('test.html').read()
        status = '200 OK'
        headers = [('Content-type', 'text/html'),
                    ('Content-Length', str(len(response_body)))]
        start_response(status, headers)
        return [response_body]