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Syntax error in code. Please advise #129

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Morpheyous opened this issue Jul 31, 2022 · 1 comment

Morpheyous commented Jul 31, 2022

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 Nov 12, 2022

Run on python3..

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