Sean
I have realized that the method I proposed can be expressed entirely within the
Zinc system, which may make it a bit neater and easier to follow. There
probably is no completely general solution, but there is a completely general
way of finding a solution for your problem domain.
It is important to realize that String>>urlEncoded is defined as:
ZnPercentEncoder new encode: self.
ZnPercentEncoder does not attempt to parse the input string as a url. It scans
the entire string, and percent encodes any character that is not in its safe
set (see the comment to ZnPercentEncoder>>encode:). Sven has given as default a
minimum safe set, which does not include slash, but there is a setter method to
redefine the safe set.
So the general way to find a solution for your domain is to collect a
representative set of the url strings, apply String>>urlEncoded to each, and
work out which characters have been percent encoded wrongly for your domain.
For any url cases this is likely to include ':/?#', as well as '%' if it
includes things already percent encoded, but there may be others specific to
your domain. Now construct an instance of ZnPercentEncoder with the safe set
extended to include these characters - note that the default safe set is given
by the class method ZnPercentEncoder class>> rfc3986UnreservedCharacters. Apply
this instance to encode all your test incoming url strings and verify that they
work. Iterate, extending the safe set, until everything passes.
If you want to keep the neatness being able to write something like
'incomingString urlEncoded asZnUrl', you can add a method to String; for the
case of the common url characters mentioned above:
String>> urlEncodedMyWay
"As urlEncoded, but with the safe set extended to include characters commonly
found in a url"
^ ZnPercentEncoder new safeSet: ':/?#%', (ZnPercentEncoder
rfc3986UnreservedCharacters);
encode: self
This works in much the same way as the snippet I posted originally, because my
code simply reproduces the essentials of ZnPercentEncoder>>encode:.
I seem to be trying to monopolize this thread, so I shall shut up now.
HTH
Peter Kenny
-----Original Message-----
From: Pharo-users <pharo-users-boun...@lists.pharo.org> On Behalf Of PBKResearch
Sent: 24 March 2019 15:36
To: 'Any question about pharo is welcome' <pharo-users@lists.pharo.org>
Subject: Re: [Pharo-users] ZnURL and parsing URL with diacritics
Well it didn't take long to find a potential problem in what I wrote, at least
as a general solution. If the input string contains something which has already
been percent encoded, it will re-encode the percent signs. In this case,
decoding will recover the once-encoded version, but we need to decode twice to
recover the original text. Any web site receiving this version will almost
certainly decode once only, and so will not see the right details.
The solution is simple - just include the percent sign in the list of excluded
characters in the third line, so it becomes:
url asString do: [ :ch|(':/?%' includes: ch )
-----Original Message-----
From: Pharo-users <pharo-users-boun...@lists.pharo.org> On Behalf Of PBKResearch
Sent: 24 March 2019 12:11
To: 'Any question about pharo is welcome' <pharo-users@lists.pharo.org>
Subject: Re: [Pharo-users] ZnURL and parsing URL with diacritics
Sean, Sven
Thinking about this, I have found a simple (maybe too simple) way round it. The
obvious first approach is to apply 'urlEncoded' to the received url string, but
this fails because it also encodes the slashes and other segment dividers. A
simple-minded approach is to scan the received string, copy the slashes and
other segment dividers unchanged and percent encode everything else. I cobbled
together the following in a playground, but it could easily be turned into a
method in String class.
urlEncodedSegments := [ :url||outStream|
outStream := String new writeStream.
url asString do: [ :ch|(':/?' includes: ch )
ifTrue: [ outStream nextPut: ch ]
ifFalse:[outStream nextPutAll: ch asString urlEncoded ] ].
outStream contents].
urlEncodedSegments value: 'https://fr.wiktionary.org/wiki/péripétie'
=> https://fr.wiktionary.org/wiki/p%C3%A9rip%C3%A9tie
This may fail if a slash can occur in a url other than as a segment divider. I
am not sure if this is possible - could there be some sort of escaped slash
within a segment? Anyway, if the received url strings are well-behaved, apart
from the diacritics, this approach could be used as a hack for Sean's problem.
HTH
Peter Kenny
Note to Sven: The comment to String>>urlEncoded says: ' This is an encoding
where characters that are illegal in a URL are escaped.' Slashes are escaped
but are quite legal. Should the comment be changed, or the method?
-----Original Message-----
From: Pharo-users <pharo-users-boun...@lists.pharo.org> On Behalf Of Sven Van
Caekenberghe
Sent: 23 March 2019 20:03
To: Any question about pharo is welcome <pharo-users@lists.pharo.org>
Subject: Re: [Pharo-users] ZnURL and parsing URL with diacritics
> On 23 Mar 2019, at 20:53, Sean P. DeNigris <s...@clipperadams.com> wrote:
>
> Peter Kenny wrote
>> And when I inspect the result, it is the address of a non-existent
>> file in my image directory.
>
> Ah, no. I see the same result. By "worked" I meant that it created a
> URL that safari accepted, but I see now it's not the same as correctly
> parsing it.
>
>
> Peter Kenny wrote
>> Incidentally, I tried the other trick Sven cites in the same thread.
>> The same url as above can be written:
>> 'https://fr.wiktionary.org/wiki' asUrl / 'péripétie'.
>
> Yes, this works if you are assembling the URL, but several people
> presented the use case of processing URLs from elsewhere, leaving one
> in a chicken-and-egg situation where one can't parse due to the
> diacritics and can't escape the diacritics (i.e. without incorrectly
> escaping other things) without parsing :/
Yes, that is pretty close to a catch 22. Strictly speaking, such URLs are
incorrect and can't be parsed.
I do understand that sometimes these URLs occur in the wild, but again,
strictly speaking they are in error.
The fact that browser search boxes accept them is a service on top of the
strict URL syntax, I am not 100% sure how they do it, but it probably involves
a lot of heuristics and trial and error.
The parser of ZnUrl is just 3 to 4 methods. There is nothing preventing
somebody from making a new ZnLoseUrlParser, but it won't be easy.
> -----
> Cheers,
> Sean
> --
> Sent from: http://forum.world.st/Pharo-Smalltalk-Users-f1310670.html
>
