1

我正在尝试将 formdata 提供给一个 scrapy.FormRequest 对象。formdata 是以下结构的字典：

{
  "param1": [
    {
      "paramA": "valueA",
      "paramB": "valueB"
    }
  ]
}

通过等效于以下代码，在scrapy shell中运行：

from scrapy import FormRequest

url = 'www.example.com'
method_post = 'POST'
formdata = <the above dict>

fr = FormRequest(url=url, method=method_post, formdata=formdata)

fetch(fr)

作为回应，我收到以下错误：

Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "/Users/chhk/.local/share/virtualenvs/project/lib/python3.6/site-packages/scrapy/http/request/form.py", line 31, in __init__
    querystr = _urlencode(items, self.encoding)
  File "/Users/chhk/.local/share/virtualenvs/project/lib/python3.6/site-packages/scrapy/http/request/form.py", line 66, in _urlencode
    for k, vs in seq
  File "/Users/chhk/.local/share/virtualenvs/project/lib/python3.6/site-packages/scrapy/http/request/form.py", line 67, in <listcomp>
    for v in (vs if is_listlike(vs) else [vs])]
  File "/Users/chhk/.local/share/virtualenvs/project/lib/python3.6/site-packages/scrapy/utils/python.py", line 119, in to_bytes
    'object, got %s' % type(text).__name__)
TypeError: to_bytes must receive a unicode, str or bytes object, got dict

我尝试了多种解决方案，包括将整个内容作为字符串，使用各种转义字符以及 dict 的变体以使其更容易接受，但是消除此错误的解决方案都不适用于请求（我得到 400回复）。

我知道 formdata 以及我所做的一切都是正确的，因为我已经在 curl 中成功复制了它（formdata 是通过 提供的-d formdata.txt）。

有没有办法解决 FormRequest 无法处理复杂的 dict 结构？还是我错过了什么？

4

1 回答 1

6

而不是formdata你可以尝试使用body参数。例子：

FormRequest(url=url, method=method_post, body=json.dumps(formdata))
于 2019-04-09T14:49:18.363 回答