如何查看POST请求的原始内容. [英] How to view raw content of a POST request.

问题描述

我正在尝试向在线SMS服务伪造" HTTP POST请求，因此我不必每次想发送SMS时都加载其页面.这是一个非常简单的页面，其中包含文件上传的输入和一些其他字段.我怀疑服务器端的PHP页面正在渲染我看到的网页，以及正在处理的回发".为了了解如何构造POST请求的FormData，我想在Firebug中查看这样的请求，但是当前视图将POST数据显示为:

I''m trying to ''fake'' an HTTP POST request to my online SMS service, so I don''t have to load their page everytime I want to send SMS''s. It''s quite a simple page, with a file uploaded input and some sundry fields. I suspect a server side PHP page is rendering the web page I see, as well has handling ''postbacks''. To find out how to structure the FormData of the POST request, I would like to view such a request in Firebug, but the current view shows the POST data as:

Content-Type: multipart/form-data; boundary=---------------------------231991966824484
Content-Length: 1493

-----------------------------231991966824484
Content-Disposition: form-data; name="rm"

memberssmsbatch
-----------------------------231991966824484
Content-Disposition: form-data; name="Action"

Send

我确定这不是请求的实际内容.我该如何查看:

I''m sure this isn''t what the actual body of the request looks like. How can I view that:

推荐答案

尝试使用 Fiddler [ ^ ].

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