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urlencode python

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Dangerous Deer answered on April 19, 2021 Popularity 10/10 Helpfulness 2/10

answer urlencode python

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## amazon.py
queries = ['tshirt for men', ‘tshirt for women’]
class AmazonSpider(scrapy.Spider):
    def start_requests(self):
        for query in queries:
            url = 'https://www.amazon.com/s?' + urlencode({'k': query})
            yield scrapy.Request(url=url, callback=self.parse_keyword_response)

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Contributed on Mar 31 2022

Dangerous Deer

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import urllib.parse
query = 'Hellö Wörld@Python'
print(urllib.parse.quote(query))
 >> 'Hell%C3%B6%20W%C3%B6rld%40Python'

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Source: www.urlencoder.io

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Contributed on Apr 19 2021

Precious Pigeon

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urlencode python

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import urllib.request 
import urllib.parse 
import re 
   
url = 'https://www.geeksforgeeks.org/'
values = {'s':'python programming', 
          'submit':'search'} 
   
data = urllib.parse.urlencode(values) 
data = data.encode('utf-8') 
req = urllib.request.Request(url, data) 
resp = urllib.request.urlopen(req) 
respData = resp.read() 
   
paragraphs = re.findall(r'<p>(.*?)</p>',str(respData)) 
   
for eachP in paragraphs: 
    print(eachP)

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Source: fr.acervolima.com

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Contributed on Mar 31 2022

Dangerous Deer

0 Answers Avg Quality 2/10

urlencode python

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