Question

我正在尝试将表格数据提供给scrapy.FormRequest对象。 formdata是以下结构的字典：

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

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)

，并且我得到以下错误：

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

我尝试了各种解决方案，包括整个字符串，带有各种转义符，以及为使其更易于接受而改变字典的方法，但是没有一种解决方案可以消除此错误（我得到400响应）。

我知道formdata以及我正在做的所有其他事情都是正确的，因为我已经成功地在curl中复制了它（formdata是通过-d formdata.txt提供的）。

有没有一种方法可以避免FormRequest无法处理复杂的字典结构？还是我错过了什么？

Answer 1

您可以尝试使用formdata参数代替body。示例：

FormRequest(url=url, method=method_post, body=json.dumps(formdata))

Scrapy FormRequest无法将复杂的字典作为FormData处理

1 个答案: