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I have some url which has space in it's query param. I want to use this in curl, e.g.

curl -G "http://localhost:30001/data?zip=47401&utc_begin=2013-8-1 00:00:00&utc_end=2013-8-2 00:00:00&country_code=USA"

which gives out

Malformed Request-Line

As per my understanding o/p is due to the space present in query param.

Is there any away to encode the url automatically before providing it to curl command?

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2 Answers 2

curl supports url-encoding internally with --data-urlencode:

$ curl -G -v "http://localhost:30001/data" --data-urlencode "msg=hello world"

Trace headers

> GET /data?hello%20world HTTP/1.1
> User-Agent: curl/7.19.7 (x86_64-redhat-linux-gnu)
> Host: localhost
> Accept: */*
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 curl -G "$( echo "$URL" | sed 's/ /%20/' )"

Where $URL is the url you want to do the translations on.

There are also more than one type of translation (encoding) you can have in a URL, so you may want to do:

curl -G "$(perl -MURI::Escape -e 'print uri_escape shift, , q{^A-Za-z0-9\-._~/:}' -- "$URL")"


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Note that echo "$URL" | sed 's/ /%20/' won't do the right thing if there are % characters in the URL. Also, spaces are normally encoded as + (and + as %2b). I recommend the Perl solution, which is reliable. –  Gilles Aug 14 '13 at 22:02
sed 's/ /%20/g' if you have more than one space to translate... –  sebthebert Feb 6 at 14:45

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