Powershell如何从REST获取结果代码 [英] Powershell how to get the result code from REST

问题描述

我正在发送一个带有 ID/密码的 POST 请求，我需要取回一个响应令牌，我该如何获取并保存它以供以后在脚本中使用?

I'm sending a POST request with ID/password and I need to get back a respond token, how can I get it and save it for later use in the script?

$loginUrl = "https://some-ip"

$params = @{
 "username"="$username"
 "password"="$password"
}

Invoke-WebRequest -Uri $loginUrl -Method POST -Body ($params|ConvertTo-Json) -ContentType "application/json"

推荐答案

按照您的输入:

$url = "https://some-ip"

$params = @{
 "username" = $username
 "password" = $password
} | ConvertTo-Json

$apiReturn = Invoke-RestMethod -Uri $url -Method POST -Body $params -ContentType "application/json"

$apiReturn 然后可以用作响应.

此外，您可以使用Invoke-RestMethod的SessionVariable参数.

Furthermore, you can use the SessionVariable parameter of Invoke-RestMethod.

$apiReturn = Invoke-RestMethod -Uri $url -Method POST -Body $params -ContentType "application/json" -SessionVariable sessionToken
$sessionToken.Headers.Add('Authorization', $apiReturn)
$sessionToken.Headers.Add('Content-Type', 'application/json')

在这种情况下，您将响应令牌添加到授权"并将整个令牌转发到您后续的 API 调用.像这样你只需要添加 $sessionToken 和 Content-Type 例如已经提供.

In this scenario, you add the response token to 'Authorization' and forward the whole token to your subsequent API calls. Like this you only need to add $sessionToken and Content-Type for example is already provided.

Invoke-RestMethod -Method Post -Uri $url -WebSession $sessionToken 

如果需要，您可以在标题中添加更多参数.

You can add more parameters to your Header in case it is required.

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