Question 295389 in Azure Question posted in
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Link static web app's preview deployment to Azure Function's non-default deployment slot using az cli

Using this command:

az staticwebapp functions link --function-resource-id "/subscriptions/[sub-id]/resourceGroups/[resource-group-name]/providers/Microsoft.Web/sites/[func-name]/slots/[slot-name]" -n [swa-name] -g [swa-rg-name] --environment-name "[env-name]" --force

I get error:

ERROR: --function-resource-id must specify a function resource ID. To get resource ID, use the following commmand, inserting the function group/name as needed: 
az functionapp show --resource-group "[FUNCTION_RESOURCE_GROUP]" --name "[FUNCTION_NAME]" --query id

How do I specify the Azure Function deployment slot?

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2

Answers


  1. 0

    You should be able to use the generic az staticwebapp backends link command. It should accept any valid resourceId

    az staticwebapp backends link `
  --backend-resource-id <function app slot resource id>`
  --name <static website name>`
  --resource-group <resource group name>
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  2. 0

    In the resource id place you have to keep resource id of function app as below:

    $x=az functionapp show --resource-group "v-rbojja" --name "rith8976" --query id

    Then you can use below commad:

    az staticwebapp functions link --function-resource-id $x -n "rith98webapp" -g "v-rbojja" --environment-name "rith8976-rithwik" --force

    You can also use backend command as @Thomas mentioned :

    az staticwebapp backends link -n rith98webapp -g "rg" --backend-resource-id $x --backend-region "East US"

    enter image description here

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