Use kubectl to wait for Jobs to succeed
No need to poll: there's a subcommand for that.pull/37596/head
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@ -244,7 +244,13 @@ So, now run the Job:
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kubectl apply -f ./job.yaml
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```
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Now wait a bit, then check on the job.
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You can wait for the Job to succeed, with a timeout:
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```shell
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# The check for condition name is case insensitive
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kubectl wait --for=condition=complete --timeout=300s job/job-wq-1
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```
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Next, check on the Job:
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```shell
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kubectl describe jobs/job-wq-1
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@ -285,7 +291,9 @@ Events:
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14s 14s 1 {job } Normal SuccessfulCreate Created pod: job-wq-1-p17e0
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```
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All our pods succeeded. Yay.
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All the pods for that Job succeeded. Yay.
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@ -208,9 +208,18 @@ Events:
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FirstSeen LastSeen Count From SubobjectPath Type Reason Message
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--------- -------- ----- ---- ------------- -------- ------ -------
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33s 33s 1 {job-controller } Normal SuccessfulCreate Created pod: job-wq-2-lglf8
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```
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You can wait for the Job to succeed, with a timeout:
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```shell
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# The check for condition name is case insensitive
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kubectl wait --for=condition=complete --timeout=300s job/job-wq-2
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```
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```shell
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kubectl logs pods/job-wq-2-7r7b2
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```
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```
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Worker with sessionID: bbd72d0a-9e5c-4dd6-abf6-416cc267991f
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Initial queue state: empty=False
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Working on banana
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@ -107,7 +107,14 @@ When you create this Job, the control plane creates a series of Pods, one for ea
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Because `.spec.parallelism` is less than `.spec.completions`, the control plane waits for some of the first Pods to complete before starting more of them.
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Once you have created the Job, wait a moment then check on progress:
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You can wait for the Job to succeed, with a timeout:
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```shell
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# The check for condition name is case insensitive
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kubectl wait --for=condition=complete --timeout=300s job/indexed-job
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```
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Now, describe the Job and check that it was successful.
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```shell
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kubectl describe jobs/indexed-job
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