CODEWITHSUNDEEP
bashcurl
newbie
newbie

Reputation: 607

curl command not accepting input from file

I'm trying to run a curl command such as:

curl -Sks -XPOST https://localhost:9999/some/local/service -d 'some text arguments'

The above works fine. However, when I place

'some text arguments'

in a file and call the command as:

curl -Sks -XPOST https://localhost:9999/some/local/service -d `cat file_with_args`

I get many exceptions from the service. Does anyone know why this is?

Thanks!

Upvotes: 0

Views: 602

Answers (1)

ruakh
ruakh

Reputation: 183201

If the file contains 'some text arguments', then your command is equivalent to this:

curl -Sks -XPOST https://localhost:9999/some/local/service -d \'some text arguments\'

— that is, it's passing 'some, text, and arguments' as three separate arguments to curl.

Instead, you should put just some text arguments in the file (no single-quotes), and run this command:

curl -Sks -XPOST https://localhost:9999/some/local/service -d "`cat file_with_args`"

(wrapping the `cat file_with_args` part in double-quotes so that the resulting some text arguments doesn't get split into separate arguments).

Incidentally, I recommend writing $(...) rather than `...`, because it's more robust in general (though in your specific command it doesn't make a difference).

Upvotes: 3

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