Invoke-WebRequest, POST with parameters

Question

I'm attempting to POST to a uri, and send the parameter username=me

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

How do I pass the parameters using the method POST?

See [this answer](http://stackoverflow.com/a/13732610/1630171) to a similar question. — Ansgar Wiechers, Jun 26 '13 at 16:28
You can also consider [this Q/A or the referenced answer](https://stackoverflow.com/a/13732610/4575793). — Cadoiz, Sep 14 '21 at 07:51

score 373 · Accepted Answer · answered Jun 26 '13 at 21:40

373

Put your parameters in a hash table and pass them like this:

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

answered Jun 26 '13 at 21:40

Jellezilla

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For my future reference as much as anyone else's info, the hash table can also be passed, one-liner-style, directly to the -Body parameter. – cori Jun 29 '18 at 13:07
4

add $ProgressPreference = 'SilentlyContinue' to speed things up by factor of 10. – andrej Nov 30 '18 at 00:13
I would try this `non-json` `hash-table` solution first before going to the json version, see @rob. – Timo Mar 25 '21 at 20:27
Just some more content. Thanks to Timo, [link to rob's answer](https://stackoverflow.com/a/37323588/4575793). The oneliner like cori suggested would be `Invoke-WebRequest -Uri http://example.com/foobar -Method POST -Body @{username='me';moredata='qwerty'}` (possibly with `$ProgressPreference = 'SilentlyContinue'`). Pay attention that in comparison to `curl` you have no quotation marks `"` for the variable names and `=` instead of `:` and `;` instead of `,`. – Cadoiz Sep 14 '21 at 08:41
`-UseDefaultCredentials` to pass in the Windows authentication user – Volodymyr Kotylo Jan 05 '22 at 09:50

rob · Answer 2 · 2019-05-16T11:24:38.447

125

For some picky web services, the request needs to have the content type set to JSON and the body to be a JSON string. For example:

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

or the equivalent for XML, etc.

edited May 16 '19 at 11:24

answered May 19 '16 at 12:43

rob

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score 33 · Answer 3 · answered Sep 27 '18 at 22:26

33

This just works:

$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

answered Sep 27 '18 at 22:26

Francesco Mantovani

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Possibly stupid question, but how do I know the `connectapitoken`? Or is this optional? – Cadoiz Sep 14 '21 at 08:02
1

@Cadoiz, it's optional, as other Headers. Depends on the service you are consuming, if it cares about those values. – Volodymyr Kotylo Jan 04 '22 at 08:49

GorvGoyl · Answer 4 · 2022-02-12T17:49:38.440

16

Single command without ps variables when using JSON as body {lastName:"doe"} for POST api call:

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

See more: Power up your PowerShell

edited Feb 12 '22 at 17:49

answered Dec 19 '19 at 16:05

GorvGoyl

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Attention! In comparison to curl you have `=` instead of `:`. You're doing it correct in the code block, but maybe not above. `;` instead of `,` is correct and the quotation marks `"` for the variable names are alright and just not wanted by PowerShell. – Cadoiz Sep 14 '21 at 08:42

Invoke-WebRequest, POST with parameters

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