Issue
I'm using scrapy and I'm trying to export files, I have some spider and I was wondering if it is possible to specify only the path, using FEED_URI worked, but only with the name of the file together.
FEED_URI = 'file:///C:/Users/Acer/Desktop/data.json'
If I leave it at that, the file will overwrite the previous one.
I wanted something like this,
FEED_URI ='file:///C:/Users/Acer/Desktop/'
but that I could specify when crawling, like this
scrapy crawl name -o filename.json
Solution
You can parameterize the feed URI.
Answered By - Gallaecio
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