I am new to shell script.
I am trying to use one curl request like this in my shell script.
curl -X POST --header "Content-Type: application/json" --header "Accept: */*" "http://localhost:8080/api/rest/v1/category/p1/{id}?id=${a}&name=${b}&typecode=${c}
where $a or $b or $c may contains words seperated by spaces due to which curl request is getting failed.
You should also be able to get curl to do the work of encoding space and other special characters for you in the query string by using --data-urlencode, whilst adding -G to make it part of the url:
curl --header "Content-Type: application/json" --header "Accept: */*" \
"http://localhost:8080/api/rest/v1/category/p1/{id}" \
--data-urlencode id="${a}" \
--data-urlencode name="${b}" \
--data-urlencode typecode="${c}" \
-G -X POST
Test by setting a='a a' b='b b' c='a&b&c' and adding -v. You get the header:
POST /api/rest/v1/category/p1/id?id=a%20a&name=b%20b&typecode=a%26b%26c HTTP/1.1
By the way, {} should be encoded in urls so curl has removed them.
You can use:
curl -X POST --header 'Content-Type: application/json' --header 'Accept: */*' \
"http://localhost:8080/api/rest/v1/category/p1/{id}?id=${a// /%20}&name=${b// /%20}&typecode=${c// /%20}"
${a// /%20} (an similarly for $b and $c) will replace all space by %20 (encoded space) so that your rest API gets all parameter values correctly.
