How to get the most recently launched EC2 instance with AWS CLI?

Question

I'm currently using the following CLI command to get the instance PublicIPAddress and LaunchTime for a given instance Name tag, 'myInstanceName':

aws ec2 describe-instances --filters 'Name=tag:Name,Values=myInstanceName' \ 
    --region us-east-1 \
    --query 'Reservations[*].Instances[*].{PublicIpAddress: PublicIpAddress, LaunchTime: LaunchTime}'

This results in the following:

[
    {
        "LaunchTime": "2019-01-25T11:49:06.000Z",
        "PublicIpAddress": "11.111.111.11"
    }
]

This is fine, but if there are two instances with the same name I will get two results in my result JSON. I need to find a way to get the most recent instance for a given name.

Solution Update

This question is quite specific to EC2 instances. The issue can be resolved using two different methods, answered below:
Parsing Result with jq
Using JMESPath

Please see this related question for more general sorting by date with JMESPath, and for further reading.

Possible duplicate of [Sort by date with jmespath](https://stackoverflow.com/questions/42414156/sort-by-date-with-jmespath) — bwest, Mar 21 '19 at 17:47

score 12 · Answer 1 · answered Mar 21 '19 at 21:35

Here's a method for finding the latest-launched instance, and displaying data about it:

aws ec2 describe-instances  --query 'sort_by(Reservations[].Instances[], &LaunchTime)[:-1].[InstanceId,PublicIpAddress,LaunchTime]'

sort_by(Reservations[].Instances[], &LaunchTime)[:-1] will return the last instance launched. The fields are then retrieved from those instances.

To understand this type of fun, see:

jarmod · Accepted Answer · 2019-03-21T22:32:21.743

Try using the jq utility. It's a command-line JSON parser. If you're not familiar with it then I'd recommend the jq playground for experimentation.

First flatten the awcli results, as follows:

aws ec2 describe-instances  \
    --query 'Reservations[].Instances[].{ip: PublicIpAddress, tm: LaunchTime}' \
    --filters 'Name=tag:Name,Values= myInstanceName'

Note that I've aliased LaunchTime to tm for brevity. That will result in (unsorted) output like this:

[
  {
    "ip": "54.4.5.6",
    "tm": "2019-01-04T19:54:11.000Z"
  },
  {
    "ip": "52.1.2.3",
    "tm": "2019-03-04T20:04:00.000Z"
  }
]

Next, pipe this result into jq and sort by descending tm (the alias for LaunchTime), as follows:

jq 'sort_by(.tm) | reverse'

That will result in output like this:

[
  {
    "ip": "52.1.2.3",
    "tm": "2019-03-04T20:04:00.000Z"
  },
  {
    "ip": "54.4.5.6",
    "tm": "2019-01-04T19:54:11.000Z"
  }
]

Finally, use jq to filter out all but the first result, as follows:

jq 'sort_by(.tm) | reverse | .[0]'

This will yield one result, the most recently launched instance:

{
  "ip": "52.1.2.3",
  "tm": "2019-03-04T20:04:00.000Z"
}

Putting it all together, the final command is:

aws ec2 describe-instances  \
    --query 'Reservations[].Instances[].{ip: PublicIpAddress, tm: LaunchTime}' \
    --filters 'Name=tag:Name,Values= myInstanceName' | \
    jq 'sort_by(.tm) | reverse | .[0]'

Thanks. This is very useful, I was not aware of the sort_by function. Great answer, I appreciate it! — x3nr0s, Mar 21 '19 at 19:20

How to get the most recently launched EC2 instance with AWS CLI?

2 Answers2

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