CODEWITHSUNDEEP
bashshellcurl
user2219963
user2219963

Reputation: 1840

How to specify command parameters in one variable?

In the test script so many times I use the command "curl". In order to optimize the code, I want options of "curl" to carry out in a global variable.

I read the terms of use of "curl", and it says that to pass a parameter that contains spaces must be to frame it in single quotes.

But it is not working. 

$ curl_options="-i -L -k -S --connect-timeout 30 --user-agent 'Opera/9.80 (Windows NT 6.1; WOW64) Presto/2.12.388 Version/12.14'"
$ curl $curl_options "http://google.com"

Output Result: 

curl: (6) Couldn't resolve host'' Opera ' 
curl: (6) Couldn't resolve host '(Windows' 
curl: (6) Couldn't resolve host 'NT' 
curl: (6) Couldn't resolve host '6 .1; ' 
curl: (6) Couldn't resolve host 'WOW64)' 
curl: (6) Couldn't resolve host 'Presto' 
curl: (6) Couldn't resolve host 'Version'

Upvotes: 0

Views: 357

Answers (1)

chepner
chepner

Reputation: 530843

In bash, you should use an array. This way, you don't need to worry whether a space in the string is part of an option, or separating two options:

curl_options=( ... )
curl_options+=( "--user-agent" "Opera/9.80 (Windows NT 6.1; WOW64) Presto/2.12.388 Version/12.14")

curl "${curl_options[@]}" "http://google.com"

If you can't use arrays (e.g., they aren't available in the shell you are using), you'll have to fall back to using eval:

$ curl_options="-i -L -k -S --connect-timeout 30 --user-agent 'Opera/9.80 (Windows NT 6.1; WOW64) Presto/2.12.388 Version/12.14'"
$ eval "curl $curl_options http://google.com"

This isn't ideal, since you need to be very careful about how you set the value of curl_options, because eval has no knowledge about what the value represents. The shell merely interpolates the value into the string passed to eval, and eval executes it. Typos can have unintended consequences.

Upvotes: 3

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