Converting a curl command to Powershell

Question

I can use curl to call a REST api

curl  -H 'Authorization: Bearer <token>' \
      -H 'Accept: application/json' \
      -H 'Content-Type: application/json' \
      -X POST \
      -d '<json>' \
      https://api.dnsimple.com/v2/1010/zones/example.com/records

and the json needs to be in the format of:

{
  "name": "",
  "type": "MX",
  "content": "mxa.example.com",
}

How can I call this API using Invoke-WebRequest? I'm trying the following (of course I'm using variables here) When I call this I'm getting an error 400

$uri =  [string]::Format("https://api.dnsimple.com/v2/{0}/zones/{1}/records",$account,$domain)
$headers = @{}
$headers["Authorization"] = [string]::Format("Bearer {0}", $token)
$headers["Accept"] = "application/json"
$headers["Content-Type"] = "application/json"

$json = @{}
$json["name"] = $subdomain
$json["type"] = "A"
$json["content"] = $ip

Invoke-WebRequest -Uri $uri -Body $json -Headers $headers -Method Post

Answer 1

Problem seems to be that you're passing a raw hashtable to the -Body parameter, rather than an actual valid JSON string. Use ConvertTo-Json for this. Also, no need to use [string]::Format() explicitly, you can use the -f operator instead!

$uri = 'https://api.dnsimple.com/v2/{0}/zones/{1}/records' -f $account,$domain
$headers = @{
  Authorization = 'Bearer {0}' -f $token
  Accept        = 'application/json'
  Content-Type  = 'application/json'
}
$json = @{
  name    = "$subdomain"
  type    = "A"
  content = "$ip"
} |ConvertTo-Json
Invoke-WebRequest -Uri $uri -Body $json -Headers $headers -Method Post

Answer 2

I'd use Invoke-RestMethod personally...

See

Saves re-inventing the wheel.