Syntax error in code. Please advise
Morpheyous opened this issue · 1 comments
Morpheyous commented
if "Referer" in fuzzing_headers:
fuzzing_headers["Referer"] = f'https://{fuzzing_headers["Referer"]}'
return fuzzing_headers
This section of code produces a syntax error which is annoying, can you please help me fix this.
Peter
Pruthviraj-Deshmukh commented
Run on python3..