CODEWITHSUNDEEP
jsonyamlaws-codebuildpyyaml
user12076139
user12076139

Reputation:

YAML_FILE_ERROR: did not find expected key

I tried to run Buildspec using aws codeBuild and trying to generate process.json file on-fly using jq command. But it gives an error while executing and build goes failed..

build: 
              commands:
                - cp $CODEBUILD_SRC_DIR/qe/performance/* apache-jmeter-5.2/bin/
                - cd apache-jmeter-5.2/bin/
                - DATE=`date "+%Y%m%d-%H-%M-%S"`
                - aws s3 cp $DATE-Report s3://$JMeterScanResultBucket/${ProjectName}/$DATE --recursive
                - jq -n --arg appname "$appname" '{apps: [ {project: wsg, issuetype: "Test Execution", summary: "Test Execution for junit Execution"}]}' > process.json

however, I have received following error: Line 20 goes to above "jq" command

DOWNLOAD_SOURCE 
Failed
YAML_FILE_ERROR: did not find expected key at line 20

Upvotes: 0

Views: 11680

Answers (1)

tinita
tinita

Reputation: 4336

A colon plus a space (or newline) in YAML means it's a key-value pair in a mapping:

key: value

Your jq command contains several colons followed by spaces.

Since you want a single string, you must quote it.

There are several ways to do that in YAML.

Single or double quoting wouldn't be ideal here because the string contains both quote types.

A folded block scalar is probably the best solution here. Newlines will be folded together as spaces.

- >
  jq -n --arg appname "$appname"
  '{apps: [ {project: wsg, issuetype: "Test Execution",
  summary: "Test Execution for junit Execution"}]}'
  > process.json

An alternative would be the literal block scalar, where you have to escape the linebreak like in a shell script:

- |
  jq -n --arg appname "$appname" \
  '{apps: [ {project: wsg, issuetype: "Test Execution", \
  summary: "Test Execution for junit Execution"}]}' \
  > process.json

Upvotes: 6

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