im trying to hit gitlab release api to update my release note description field from CHANGELOG.md file
these are what ive tried:
#!/bin/sh
releaseNote=$(cat CHANGELOG.md)
curl --header 'Content-Type: application/json' --request PUT --data '{"description": "'"${releaseNote}"'"}' --header "PRIVATE-TOKEN: mytokenhere" "https://gitlab.com/api/v4/projects/1234/releases/0.1"
#!/bin/sh
releaseNote=$(cat CHANGELOG.md)
curl --header 'Content-Type: application/json' --request PUT --data '{\"description\": \"${releaseNote}\" }' --header "PRIVATE-TOKEN: mytokenhere" "https://gitlab.com/api/v4/projects/1234/releases/0.1"
when i try to hard-code description field like this, it works
#!/bin/sh
curl --header 'Content-Type: application/json' --request PUT --data '{"description": "foo"}' --header "PRIVATE-TOKEN: mytokenhere" "https://gitlab.com/api/v4/projects/1234/releases/0.1"
and here is whats inside my CHANGELOG.md file
# Changelog
All notable changes to this project will be documented in this file. See [standard-version](https://github.com/conventional-changelog/standard-version) for commit guidelines.
## 1.0.0 (2022-07-22)
### Features
* abc
* def
any suggestion guys?
as per #LéaGris suggestion above, finally it works.. here is my code snippet
#!/bin/sh
# format CHANGELOG.md to proper JSON format first
releaseNote=$( jq -sR '{"description": .}' CHANGELOG.md )
# hit the API
curl --header 'Content-Type: application/json' --request PUT --data "$releaseNote" --header "PRIVATE-TOKEN: mytokenhere" "https://gitlab.com/api/v4/projects/1234/releases/0.1"
Related
I am novice and first timer to bash. Trying to run bash under command line to invoke an API, by passing token and payload received from two different APIs and are set as parameters. Below is my command. I am trying to add this bash script to a task in AzureBatch service job.
It has 3 curl requests,
First one (Line#1 in the code snippet below)- gets payload by
calling an API. ---- This is working fine, I am able to verify the
payload using the echo statement following the first curl command.
Second one(Line#3 in the code snippet below) - gets token by
calling the token provider ----- This is working fine as well,
verified using the echo statement.
Third one (Line#5 in the code snippet below),This is the problematic command. I am trying to pass the token and payload received from the above two commands and the curl is not able to resolve them.
both token and payload or not resolving to their values..
My Bash COmmand
/bin/bash -c
"payload=$(curl --location --request GET 'http://url/OutreachData')
&& echo -e \"The value of payload is: "'$payload'"\"
&& token=$(curl --location --request POST 'https://login.microsoftonline.com/<<tenantId>>/oauth2/v2.0/token' --header 'Content-Type: application/x-www-form-urlencoded' --data-urlencode 'client_id=<<clientId>>' --data-urlencode 'scope=api://<<applicationId>>/.default' --data-urlencode 'client_secret=<<clientSecret>>' --data-urlencode 'grant_type=client_credentials' --data-urlencode 'Audience=api://<<applicationId>>'|jq -j '.access_token')
&& echo -e \"value of token is "'$token'"\n\"
&& result=$(curl --location --request POST 'https://url/api/<<Resource>>' --header 'accept: */*' --header 'Content-Type: application/json' --header 'Authorization: Bearer '"'$token'" --data-raw "'$payload'")
&& echo -e \"Result is "'$result'"\""
This is how the third Curl is resolving to, payload and token are not getting replaced as we can see in the authorization header and data-raw elements
++ curl --location --request POST https://url/api/ --header 'accept: /' --header 'Content-Type: application/json' --header 'Authorization: Bearer ' --data-raw ''''''''
There should be no need to explicitly run this with bash -c unless you are in a very constrained environment where you simply cannot run Bash by any other means.
The immediate problem is that code like
bash -c "echo "'$token'" && true"
ends up with $token being single-quoted in the shell which you run bash -c from. But the blazingly obvious fix is to not have this complex quoting in the first place.
payload=$(curl --location --request GET 'http://url/OutreachData')
echo "The value of payload is: '$payload'"
token=$(curl --location --request POST \
'https://login.microsoftonline.com/<<tenantId>>/oauth2/v2.0/token' \
--header 'Content-Type: application/x-www-form-urlencoded' \
--data-urlencode 'client_id=<<clientId>>' \
--data-urlencode 'scope=api://<<applicationId>>/.default' \
--data-urlencode 'client_secret=<<clientSecret>>' \
--data-urlencode 'grant_type=client_credentials' \
--data-urlencode 'Audience=api://<<applicationId>>' |
jq -j '.access_token')
echo "value of token is "'$token'"
result=$(curl --location --request POST \
'https://url/api/<<Resource>>' --header 'accept: */*' \
--header 'Content-Type: application/json' \
--header 'Authorization: Bearer '"$token" \
--data-raw "$payload")
echo "Result is "'$result'"
If your current shell is not Bash and you need these commands to be run in Bash, a simpler workaround is to put the script in a here document, which drastically simplifies the quoting needs (or if this is in an interactive session, just run bash and run these commands at the interactive Bash prompt, then exit when you no longer want to be in Bash).
How can I set the environment variable in request header for curl
curl --location --request POST 'test.example.com' \
--header 'Content-Type: application/json' \
--header 'access_token: "${SOME_ENV_TOKEN}"' \
--header 'Authorization: Basic Og=='
Not working.
Use --header "access_token: \"${SOME_ENV_TOKEN}\""
The ' quotes disable variable expansion in sh.
I am using AWX curl for update template in bash script. Somehow this curl command is not able to run
curl --insecure -v -X PATCH https://somedomain.corp.com/api/v2/job_templates/\"$test_2_template\"/ -H 'Authorization: Basic keys==' -H 'Content-Type: application/json' -d '{"extra_vars":"{\"module_name\": \"$module_name\", \"env\": \"$appenv\", \"host_group_name\": \"$host_group_name\", \"rolename\": \"$rolename\", \"app_artifact_url\": \"$app_artifact_url\"}"}'
Looks like there is some issue in passing variables to curl command via bash
However if i run this from my terminal , it's working fine.
curl --insecure -v -X PATCH https://somedomain.com/api/v2/job_templates/279/ -H 'Authorization: Basic keys==' -H 'Content-Type: application/json' -d '{"extra_vars":"{\"module_name\": \"myservice\", \"env\": \"test\", \"host_group_name\": \"env2\", \"rolename\": \"myservice\", \"artifact_version\": \"2.1.0-SNAPSHOT\"}"}'
Please suggest some solution.
You can't substitute variables inside a single quoted string.
For terrible long commands like this, breaking them up into pieces is essential for readability and maintainability:
data=$(printf \
'{"extra_vars": "{\"module_name\": \"%s\", \"env\": \"%s\", \"host_group_name\": \"%s\", \"rolename\": \"%s\", \"app_artifact_url\": \"%s\"}"}' \
"$module_name" "$appenv" "$host_group_name" "$rolename" "$app_artifact_url"
)
curl_args=(
--insecure
-v
-X PATCH
-H 'Authorization: Basic keys=='
-H 'Content-Type: application/json'
-d "$data"
)
url="https://somedomain.example.com/api/v2/job_templates/$test_2_template/"
curl "${curl_args[#]}" "$url"
I am getting authentication failure error while using curl command in bash shell which password has ! char.
Below is curl command i am trying.
curl -D- -u 'some_user:somename#2019!' -X POST --data #data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
I have also tried escaping ! char using \, as below, but no luck.
curl -D- -u 'some_user:somename#2019''\!' -X POST --data #data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
can any one suggest.
You can try with entity code for ! = %21 (and # = %40). So:
curl -D- -u 'some_user:somename%402019%21' -X POST --data #data.txt -H 'Content-Type: application/json' https://help.myjira.com/rest/api/2/issue
I have a bunch of these to test my RESTful API
$CURL \
-v \
-H 'Content-Type: application/json' \
-H 'Accept: application/json' \
-X POST \
-d '{"user":{"email":"user#example.com","password":"secret"}}' \
$URL/$PATH/sessions
I kinda want to shorten it to something like
CURLOPTS="-v -H 'Content-Type: application/json' -H 'Accept: application/json'"
$CURL \
$CURLOPTS \
-X POST \
-d '{"user":{"email":"user#example.com","password":"secret"}}' \
$URL/$PATH/sessions
but the options don't seem to be passed in. Any clues?
Short answer: see BashFAQ #50:'m trying to put a command in a variable, but the complex cases always fail!.
Long answer: Putting commands (or parts of commands) into variables and then getting them back out intact is complicated. The reason your script doesn't work is because of the order in which the shell parses the command line: it parses (and removes) quotes and escapes, then replaces variable values. By the time $CURLOPTS gets replaced, it's too late for the quotes to have their intended effect; instead, they're passed to curl as part of the arguments, which confuses curl greatly.
The solution: store the options in an array rather than a plain string:
CURLOPTS=(-v -H 'Content-Type: application/json' -H 'Accept: application/json')
$CURL \
"${CURLOPTS[#]}" \
-X POST \
-d '{"user":{"email":"user#example.com","password":"secret"}}' \
"$URL/$PATH/sessions"
You can use an array and trigger word splitting
$ set -x
$ CURLOPTS=(-v -H 'Content-Type: application/json' -H 'Accept: application/json')
$ : curl "${CURLOPTS[#]}"
+ : curl -v -H 'Content-Type: application/json' -H 'Accept: application/json'