How do I read the output of a curl into a ruby variable?

Question

Say I have the following curl call :

"curl  -v --insecure -X POST -H 'Host: api.test.com' -H 'Tester #{@@test_name}' -d 'token=#{token_raw}' https://teststage.BLAHBLAH:token/terminate"

I'm trying to get the output of the curl call. Such as the 200 OK or 404 ERROR status.

So, if I do :

a = `curl  -v --insecure -X POST -H 'Host: api.test.com' -H 'Tester #{@@test_name}' -d 'token=#{token_raw}' https://teststage.BLAHBLAH:token/terminate`

I get nothing back in a

However, if I do

puts `curl  -v --insecure -X POST -H 'Host: api.test.com' -H 'Tester #{@@test_name}' -d 'token=#{token_raw}' https://teststage.BLAHBLAH:token/terminate`

Then I can see the output. How do I read it into the variable. I'd prefer an answer without any import like OPEN3, if possible.

Have you tried: `a = ``curl "blah.blah"`` - Does anyone know how to escape backticks? — BroiSatse, Mar 26 '14 at 23:50
As in the variable is also in backticks? Or just add a backtick to the variable? `a = `curl "blah.blah"`` ?? — gran_profaci, Mar 26 '14 at 23:52

Amiel Martin · Answer 1 · 2014-03-28T02:50:58.497

3

I'm not sure why

a = `curl  -v --insecure -X POST -H 'Host: api.test.com' -H 'Tester #{@@test_name}' -d 'token=#{token_raw}' https://teststage.BLAHBLAH:token/terminate`

isn't working for you.

Here's what I get with a simpler test:

RUBY_VERSION # => "1.9.3"

`curl --version`
# => "curl 7.30.0 (x86_64-apple-darwin13.0) libcurl/7.30.0 SecureTransport zlib/1.2.5\nProtocols: dict file ftp ftps gopher http https imap imaps ldap ldaps pop3 pop3s rtsp smtp smtps telnet tftp \nFeatures: AsynchDNS GSS-Negotiate IPv6 Largefile NTLM NTLM_WB SSL libz \n"

a = `curl http://echo.jsontest.com/hello/world`

a # => "{\"hello\": \"world\"}\n"

UPDATE

I missed a part of your question and didn't realize you were looking for the headers.

Try this:

a = `curl -I http://echo.jsontest.com/hello/world`
a.lines.first # => "HTTP/1.1 200 OK\r\n"

Hopefully that helps.

edited Mar 28 '14 at 02:50

answered Mar 27 '14 at 22:34

Amiel Martin

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So, my update is literally that if I get a 404 error, I get some random JSON back. If it works, blank. That's how I'm working with it. – gran_profaci Mar 27 '14 at 23:13
Technically no. But I guess I could live with it. Thanks. – gran_profaci Mar 27 '14 at 23:23
Oh, upon re-reading your question. I misunderstood part of it. I will edit. – Amiel Martin Mar 28 '14 at 02:49

coreyward · Answer 2 · 2014-03-27T00:12:55.583

You probably want to just use Ethon to interface with libcurl instead of calling out to the shell interface, especially if you're unwilling to use the tools Ruby gives you for interfacing with the shell and other processes more readily (you know, like Open3).

Note: the Ruby standard library (including Open3) is part of Ruby and distributed with it. It is in no sense an import, but if you really don't want to use the require method for some inane reason, IO is available without loading additional code and provides the low level interface that Open3 utilizes.

Sabuj Hassan · Answer 3 · 2014-03-27T22:22:05.987

1

Try this:

HTTP_RESP_CODE=$(curl -s -o out.html -w '%{http_code}' http://www.example.com)

echo $HTTP_RESP_CODE

This works with my ruby.

HTTP_RESP_CODE=`curl -s -o out.html -w '%{http_code}' http://www.example.com`
print HTTP_RESP_CODE

edited Mar 27 '14 at 22:22

answered Mar 27 '14 at 08:06

Sabuj Hassan

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t.ruby:1: `$(' is not allowed as a global variable name t.ruby:1: syntax error, unexpected end-of-input @@r = $(curl -v --insecure -X POST ... – gran_profaci Mar 27 '14 at 22:15
1

@gran_profaci no idea about it. You can try to put my example curl command inisde the \`...\` instead of `$(...)` – Sabuj Hassan Mar 27 '14 at 22:18
Same Error. This is a syntax problem apparently. – gran_profaci Mar 27 '14 at 22:19
1

@gran_profaci I have just checked the code with ruby(see updated answer) and it works. May be your problem is at elsewhere! – Sabuj Hassan Mar 27 '14 at 22:22

How do I read the output of a curl into a ruby variable?

3 Answers3