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Re: howto run curl again with quoted argument if it failed due to parsing error?



hello, you should quote the "offending" argument, since it contains shell
metacharacters

curl
https://www.google.com/search?q=test&ie=utf-8&oe=utf-8&client=firefox-b-ab

should become

curl "
https://www.google.com/search?q=test&ie=utf-8&oe=utf-8&client=firefox-b-ab";

you can use single or double quotes, or you can escape (prefixing with a
blackslash) only the & and the ? character, since they both are shell
metacharacters, like this:

curl
https://www.google.com/search\?q=test\&ie=utf-8\&oe=utf-8\&client=firefox-b-ab

you can read more about shell metacharacters here:

http://faculty.salina.k-state.edu/tim/unix_sg/shell/metachar.html

I hope to have been helpful, best regards

Il giorno dom 30 set 2018 alle ore 15:55 chiasa.men <chiasa.men@xxxxxx> ha
scritto:

> if you call e.g. curl with a link that contains an & zsh says:
> zsh: parse error near `&'.
>
> Is there an zsh idiom to fix that?
> I came up with
> curl ^@|ctrl+shift+v|esc'.
>
> I also tried sth like
> !!:*:q
> but that doesnt return the whole link (only a part of it)
> curl
> https://www.google.com/search?q=test&ie=utf-8&oe=utf-8&client=firefox-b-ab
> zsh: parse error near `&'
> echo !!:*:q
> echo 'https://www.google.com/search?q=test&ie=utf-8&;'
> https://www.google.com/search?q=test&ie=utf-8&;
>
> Is that related to my zshrc entry:
> WORDCHARS='*?_-.[]~=&;!#$%^(){}<>'
> or intended behavior? It breaks at a 'o' it seems..
>
>
> Strangely :0-$ gets the whole link:
> echo !!:0-$:q
> echo 'curl' 'https://www.google.com/search?
> q=test&ie=utf-8&oe=utf-8&client=firefox-b-ab'
> curl
> https://www.google.com/search?q=test&ie=utf-8&oe=utf-8&client=firefox-b-ab
>
> Why is that?
>
>
>

-- 
Pier Paolo Grassi
email: pierpaolog@xxxxxxxxx
linkedin: https://www.linkedin.com/in/pier-paolo-grassi-19300217
founder: https://www.meetup.com/it-IT/Machine-Learning-TO


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