python - 提交表格并上传带有请求的文件

Question

我正在努力提交带有 python requests 的特定表单.我想使用的网站上的其他表格工作正常，我可以提交登录表格等。这只是我遇到的文件上传问题。显然提交表格工作正常，因为我从网站上收到一条消息说“请返回并选择至少一个图像文件上传。”但文件未发布到网站。

如果我使用 Firefox 插件 Tamperdata 查看请求，在我提交空表单后它看起来像这样:

-----------------------------201075691119887339851216603987\r
Content-Disposition: form-data; name="userfile[]"; filename=""\r
Content-Type: application/octet-stream\r
\r
\r
-----------------------------201075691119887339851216603987\r
Content-Disposition: form-data; name="userfile[]"; filename=""\r
Content-Type: application/octet-stream\r
\r
\r
-----------------------------201075691119887339851216603987\r
Content-Disposition: form-data; name="userfile[]"; filename=""\r
Content-Type: application/octet-stream\r
\r
\r
-----------------------------201075691119887339851216603987\r
Content-Disposition: form-data; name="userfile[]"; filename=""\r
Content-Type: application/octet-stream\r
\r
\r
-----------------------------201075691119887339851216603987\r
Content-Disposition: form-data; name="userfile[]"; filename=""\r
Content-Type: application/octet-stream\r
\r
\r
-----------------------------201075691119887339851216603987\r
Content-Disposition: form-data; name="upload_to"\r
\r
0\r
-----------------------------201075691119887339851216603987\r
Content-Disposition: form-data; name="upload_type"\r
\r
standard\r
-----------------------------201075691119887339851216603987--\r

(我用换行符替换了所有的\n 以使其更具可读性)

我的 python 上传代码如下所示:

index = s.get("https://example.com/")
file = {'userfile[]': open('tmp/cover/cover.jpg', 'rb')}

res = s.post(url='https://example.com/upload.php',
                data=file,
                headers={'Content-Type': 'application/octet-stream'})

HTML 表单如下所示:

<form action="upload.php" method="post" id="upload_form" enctype="multipart/form-data">
    <p>
        <input name="userfile[]" type="file" size="30"> <br>
        <input name="userfile[]" type="file" size="30"> <br>
        <input name="userfile[]" type="file" size="30"> <br>
        <input name="userfile[]" type="file" size="30"> <br>
        <input name="userfile[]" type="file" size="30"> <br>

        <span id="more_file_inputs"></span>
        <br>
        <span id="upoptions_hidden">
            Uploading Options: <a href="javascript:void(0);" onclick="toggle('upoptions_hidden'); toggle('upoptions_shown');">Show</a>
        </span>

        <span id="upoptions_shown" style="display: none;">
            Uploading Options: <a href="javascript:void(0);" onclick="toggle('upoptions_hidden'); toggle('upoptions_shown');">Hide</a>
            <br><br>

                                Upload to: 
                <select name="upload_to">

                    <option value="0">Root Album</option>

                </select>
                <br><br>


           Output Layout: <input type="radio" name="upload_type" value="standard" checked="checked"> <span onclick="toggle_lightbox('index.php?layoutprev=std', 'upload_layout_preview_lightbox');" title="Click to preview" class="help">Standard</span> <input type="radio" name="upload_type" value="normal-boxed"> <span onclick="toggle_lightbox('index.php?layoutprev=bx', 'upload_layout_preview_lightbox');" title="Click to preview" class="help">Boxed</span>
        </span>
        <br><br>

        <input class="button1" type="button" value="Add File" onclick="new_file_input();"> 
        <input class="button1" style="font-weight:bold;" type="button" value="Start Upload" onclick="toggle_lightbox('index.php?act=upload_in_progress', 'progress_bar_lightbox'); $('form[id=upload_form]').submit();">
    </p>
</form>

有什么想法或更好的调试方法吗？谢谢!

编辑:我刚刚意识到我并没有真正提交 upload_type 和 upload_to 字段，但即使我将它们设置为:

file = {'userfile[]': open('tmp/cover/cover.jpg', 'rb'), 'upload_to': '0', 'upload_type': 'standard'}

还是不行。

Answer 1

使用 files 关键字进行上传，而不是 data。也不要设置内容类型 header ；表单作为多部分表单数据发布，请求应该使用适合您的正确内容类型:

res = s.post(
    url='https://example.com/upload.php',
    data={'upload_type': 'standard', 'upload_to': '0'},
    files=file)

并将其余的表单字段放入 data 的映射中。

如果您想以相同的名称上传多个文件，请使用文件的序列:

files = {
    'userfile[]': [
        open('tmp/cover/cover1.jpg', 'rb'),
        open('tmp/cover/cover2.jpg', 'rb'),
    ]
}

res = s.post(
    url='https://example.com/upload.php',
    data={'upload_type': 'standard', 'upload_to': '0'},
    files=files)