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?
发布于 2013-06-27 05:40:01
将您的参数放入哈希表中,并像这样传递它们:
$postParams = @{username='me';moredata='qwerty'}
Invoke-WebRequest -Uri http://example.com/foobar -Method POST -Body $postParams
发布于 2016-05-19 20:43:39
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的等价物,等等。
发布于 2018-09-28 06:26:54
这就行了:
$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
https://stackoverflow.com/questions/17325293
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