[FIXED] I want to store my url in variable name "url" to saved the url in excel sheet csv

Issue

I want to store my url in the variable name "url" to save the url in an excel sheet CSV but giving me unboundlocalerror local variable 'url' referenced before the assignment.

class NewsSpider(scrapy.Spider): name = "articles"

def start_requests(self):
    url = input("Enter the article url: ")
    
    yield scrapy.Request(url, callback=self.parse_dir_contents)

def parse_dir_contents(self, response):
    url = url
    yield{
    'Category':Category,
    'Headlines':Headlines,
    'Author': Author,
    'Source': Source,
    'Publication Date': Published_Date,
    'Feature_Image': Feature_Image,
    'Skift Take': skift_take,
    'Article Content': Content
    }
        # =============== Data Store +++++++++++++++++++++
    Data = [[Category,Headlines,Author,Source,Published_Date,Feature_Image,Content,url]]
    try:
        df = pd.DataFrame (Data, columns = ['Category','Headlines','Author','Source','Published_Date','Feature_Image','Content','URL'])
        print(df)
        with open('C:/Users/Public/pagedata.csv', 'a') as f:
            df.to_csv(f, header=False)
    except:
        df = pd.DataFrame (Data, columns = ['Category','Headlines','Author','Source','Published_Date','Feature_Image','Content','URL'])
        print(df)
        df.to_csv('C:/Users/Public/pagedata.csv', mode='a')

Solution

You can just invoke response.url instead of url = url

 url = response.url

#OR

def parse_dir_contents(self, response):
    
    yield{
    'Category':Category,
    'Headlines':Headlines,
    'Author': Author,
    'Source': Source,
    'Publication Date': Published_Date,
    'Feature_Image': Feature_Image,
    'Skift Take': skift_take,
    'Article Content': Content,
    'url': response.url
    }

Answered By - Fazlul