powershell invoke-restmethod multipart / form-data [英] powershell invoke-restmethod multipart/form-data

问题描述

我目前正在尝试使用REST API将文件上传到Web服务器。正如前面提到的，我使用PowerShell。与curl这没有问题。调用如下：

I'm currently trying to upload a file to a Webserver by using a REST API. And as mentioned I'm using PowerShell for this. With curl this is no problem. The call looks like this:

curl -H "Auth_token:"$AUTH_TOKEN -H "Content-Type:multipart/form-data" -X POST -F appInfo='{"name": "test","description": "test"}' -F uploadFile=@/test/test.test https://server/api/

但是对于使用Invoke-Restmethod命令将它导出到powershell，我完全无奈。就我的搜索，不可能使用Invoke-Restmethod。 https://www.snip2code.com/Snippet/396726/PowerShell -V3-Multipart-formdata-example
但是即使有Snipped我不够聪明，得到这个工作，因为我不想上传两个文件，而是一个文件和一些参数。

But I'm completely helpless when it comes to exporting this to powershell with a Invoke-Restmethod command. As far as I searched it is not possible to use the Invoke-Restmethod for this. https://www.snip2code.com/Snippet/396726/PowerShell-V3-Multipart-formdata-example But even with that Snipped I'm not smart enough to get this working since I don´t want to upload two files but instead one file and some arguments.

我会非常感谢，如果有人可以让我回到轨道上：o
谢谢！

I would be very thankful if someone could get me back on the track with this :o Thanks!

推荐答案

这应该是很直接。从此回答：

$Uri = 'https://server/api/';
$Headers = @{'Auth_token'=$AUTH_TOKEN};
$FileContent = [IO.File]::ReadAllText('C:\test\test.test');
$Fields = @{'appInfo'='{"name": "test","description": "test"}';'uploadFile'=$FileContent};

Invoke-RestMethod -Uri $Uri -ContentType 'multipart/form-data' -Method Post -Headers $Headers -Body $Fields;

您可能想使用 [IO.File] :: ReadAllBytes

You may want to use [IO.File]::ReadAllBytes() if the file isn't a text file.

如果您正在上传一个巨大的文件，这也可能无法正常工作。

This also may not work well if you're uploading a huge file.

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