[FIXED] Scrapy - Passing Parameters In URLs

Issue

If we use requests and we need Passing Parameters In URLs we can use params

import requests


params = (
    ('q', 'scrapy'),
)

response = requests.get('https://github.com/search', params=params)

response.url will be

In [4]: response.url                                                                                                                                            
Out[4]: 'https://github.com/search?q=scrapy'

https://2.python-requests.org//en/latest/user/quickstart/#passing-parameters-in-urls
But how Passing Parameters In URLs if we use Scrapy?

Solution

Need use FormRequest with GET method and pass params in formdata

        return FormRequest(url='https://github.com/search',
                           method='GET',
                           headers=headers,
                           formdata=params,
                           callback=self.parse_data)

Answered By - Serhii