Comments (1)
debugging credit - Dave Santucci from Scopus.
This problem is due to url encoding. If using the cursor parameter to navigate pages, it would be necessary to url encode the cursor string before passing it for the next query.
import urllib.parse as urlencode
Assuming the returned cursor link is stored in a cursor
variable and your page return is stored in a page
variable
next_cursor = urlencode.quote(cursor)
This should solve the problem.
Another potential way is to extract the next page URI and pass it directly in the next call. You can access this link as the value of page['search-results']['link']['@href'] if page['search-results']['link']['@ref'] == "next"
The correct code for this would be something like:
next_URI = [page['search-results']['link']['@href'] if page['search-results']['link']['@ref'] == "next" for item in page['search-results']['link']]
FYI - I have not yet tested the code above.
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