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I think that it is handled in the code here, but not sure when it is called. When I debugged it, I could not trace the

super()

call.

hey there, not sure I'm following. You have an existing VPC, and you're using

awsx.ec2.Vpc.fromExistingIds

and you passed in the subnet ids, and now you want to grab the subnets from the

const vpc = awsx.ec2.Vpc.fromExistingIds

right? is there a reason you wouldn't just get the subnets from the

publicSubnetIds

you already passed in?

Hi! My goal there is to grab the associated route table IDs and create a route in them to do VPC peering. Since

awsx.ec2.Subnet

has a

routeTable

attribute, it was an easy task. I was wondering if I could do the same with the

awsx.ec2.Vpc

object created using

awsx.ec2.Vpc.fromExistingIds

i don't believe it can be done with

fromExistingIds

no, because that particular method only populates the vpc itself, not the res of the resources inside the component (I think)

Yeah, it is working like that. However, when I looked in the source code of

awsx.ec2.Vpc

, I found some populating code. Not sure when it gets called

heya, I am having the same issues, it seems you have to set it manually

const publicSubnets = await aws.ec2.getSubnetIds({
      vpcId: vpcId,
      filters: [{ name: 'tag:type', values: ['public'] }],
    });
    const privateSubnets = await aws.ec2.getSubnetIds({
      vpcId: vpcId,
      filters: [{ name: 'tag:type', values: ['private'] }],
    });

awsx.ec2.Vpc.fromExistingIds(name, {
      vpcId: vpcId,
      publicSubnetIds: publicSubnetIds,
      privateSubnetIds: privateSubnetIds,
    });

that looks correct to me

@miniature-advantage-31279 Thanks for the solution, for a quick testing, I exported the

publicSubnetIds

and

privateSubnetIds

from other stack and using

awsx.ec2.Vpc.fromExistingIds

. Unfortunately, that seemed not populating

routeTables

properties of

awsx.ec2.Subnet

instances. Ultimately, I ended up exporting

routeTableIds

itself from another stack 😅

yeah that also alternative that I did

Ah, okay 👍