I give some supplementary additional explanation to my question.
On 2017/11/08 18:49, Rikimaru Honjo wrote:
Hello,
* I'm using Zuul v3 ver.2.5.3.dev1585.
I created a job which runs following two tasks for zuul v3.[1]
1. Run tests on remote host which is created by Nodepool.
(In this time, "hosts" value is "all".)
2. Run zuul_return on localhost where is zuul-executor.
And, submit a URL of logs specified as "log_url" attribute to gerrit.
(In this time, "hosts" value is "localhost".)
In this case, task 2 won't be run if task 1 is failed.
And, the below URL will be submitted to gerrit.
The pseudocode is this:
# task-1
- hosts: all
tasks:
- command: "Run tests"
register: result
ignore_errors: True
- command: "Process after testing"
# task-2
- hosts: localhost
tasks:
- zuul_return:
[...]
when: result.rc == 1
In this case, the "result" value of task-1 is not passed to task-2.
So I'm planning to save "result" value to a local file.
Are there any other solutions?
finger://<hostname of zuul-executor>/e8ce4c7f627a43689dd9ad4ecc43c881
How do you run "zuul_return" if the test task will be failed?
P.S.
If I specified "ignore_errors:True" in task 1, task2 was run even if
task1 was failed.
But, the submitted result was "SUCCESS" in this case.
[1]
I refered the following suggestion.
http://lists.openstack.org/pipermail/openstack-infra/2017-October/005646.html
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Rikimaru Honjo
E-mail:[email protected]
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