我试图POST到一个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?
我试图POST到一个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?
当前回答
对于一些挑剔的web服务,请求需要将内容类型设置为JSON,并将主体设置为JSON字符串。例如:
Invoke-WebRequest -UseBasicParsing http://example.com/service -ContentType "application/json" -Method POST -Body "{ 'ItemID':3661515, 'Name':'test'}"
或者等价的XML等等。
其他回答
把你的参数放到哈希表中,然后像这样传递:
$postParams = @{username='me';moredata='qwerty'}
Invoke-WebRequest -Uri http://example.com/foobar -Method POST -Body $postParams
对于一些挑剔的web服务,请求需要将内容类型设置为JSON,并将主体设置为JSON字符串。例如:
Invoke-WebRequest -UseBasicParsing http://example.com/service -ContentType "application/json" -Method POST -Body "{ 'ItemID':3661515, 'Name':'test'}"
或者等价的XML等等。
这是可行的:
$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
当使用JSON作为body {lastName:"doe"}进行POST api调用时,单个命令不带ps变量:
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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