powershell - Invoke-WebRequest, POST 带参数

Question

我正在尝试 POST 到 uri，并发送参数username=me

Invoke-WebRequest -Uri http://example.com/foobar -Method POST

如何使用 POST 方法传递参数？

Answer 1

将您的参数放在哈希表中并像这样传递它们：

$postParams = @{username='me';moredata='qwerty'}
Invoke-WebRequest -Uri http://example.com/foobar -Method POST -Body $postParams

Answer 2

对于一些挑剔的 Web 服务，请求需要将内容类型设置为 JSON，并将正文设置为 JSON 字符串。例如：

Invoke-WebRequest -UseBasicParsing http://example.com/service -ContentType "application/json" -Method POST -Body "{ 'ItemID':3661515, 'Name':'test'}"

或 XML 等的等价物。

Answer 3

这很有效：

$body = @{
 "UserSessionId"="12345678"
 "OptionalEmail"="MyEmail@gmail.com"
} | ConvertTo-Json

$header = @{
 "Accept"="application/json"
 "connectapitoken"="97fe6ab5b1a640909551e36a071ce9ed"
 "Content-Type"="application/json"
} 

Invoke-RestMethod -Uri "http://MyServer/WSVistaWebClient/RESTService.svc/member/search" -Method 'Post' -Body $body -Headers $header | ConvertTo-HTML

Answer 4

用作 POST api 调用JSON的主体时没有 ps 变量的单个命令：{lastName:"doe"}

Invoke-WebRequest -Headers @{"Authorization" = "Bearer N-1234ulmMGhsDsCAEAzmo1tChSsq323sIkk4Zq9"} `
                  -Method POST `
                  -Body (@{"lastName"="doe";}|ConvertTo-Json) `
                  -Uri https://api.dummy.com/getUsers `
                  -ContentType application/json

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