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What is the purpose of replacing the% 20 space before executing PHP rawurlencode ()?

This is a pretty silly question, sorry. There is a large and quite complex system that has a bug and I was able to trace it back to this part.

return str_replace('%2F', '/', rawurlencode(str_replace('%20', ' ', $key)));

There is a comment explaining why the slashes are replaced - to preserve the path structure, for example. encoded1 / encoded2 / etc .. However, there is no explanation why% 20 is replaced with space, and this part is the direct cause of the error. I'm tempted to just delete str_replace (), but it looks like it fit in there for some reason, and I get the feeling that I'll break something else by doing this. Has anyone come across something similar? Perhaps this is a dirty fix for some PHP bug? Any guesses and ideas are greatly appreciated!

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3 answers


This will result in the %20

(encoded space) not being encoded before %2F20

. However, it only serves to prevent double shielding spaces; other special characters will still be double-encoded.

This is a sign of bad code; strings that are passed to this function must not have rights to encoded characters.



I would recommend creating unit tests that cover all of the referenced code, and then refactoring that function to remove str_replace()

to make sure it doesn't break tests.

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The first thing that comes to mind is a method to prevent double encoding.



Not that I would recommend doing such a thing this way, as it would quickly become useless (and already wondering why only this entity, perhaps "they" have never encountered problems with others ... yet) ...

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This may be the result of misunderstanding rawurlencode()

vsurlencode()

urlencode()

replaces spaces with characters +

If the original author thought they rawurlencode()

did the same, they will try to pre-encode the spaces so they don't turn into +

s

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