Powershell + Curl + Zimbra SOAP API

${my_xml} = @"
<?xml version=\"1.0\" encoding=\"UTF-8\"?>
<soapenv:Envelope xmlns:soapenv=\"http://schemas.xmlsoap.org/soap/envelope/\">
  <soapenv:Body>
   <GetFolderRequest xmlns=\"urn:zimbraMail\">
    <folder>
       <path>Folder Name</path>
    </folder>
   </GetFolderRequest>
  </soapenv:Body>
</soapenv:Envelope>
"@

${my_curl} = "c:\curl.exe"
${cookie} = "c:\cookie.txt"

${zimbra_soap_url} = "https://zimbra:7071/service/admin/soap"
${curl_getfolder_args} = "-b", "${cookie}",
            "--header", "Content-Type: text/xml;charset=UTF-8",
            "--silent",
            "--data-raw", "${my_xml}",
            "--url", "${zimbra_soap_url}"

[xml]${my_response} = & ${my_curl} ${curl_getfolder_args}
${my_response}.Envelope.Body.GetFolderResponse.folder.id

从手册上看，我相信这些就是你要找的机器人

-F/--form <name=content> (HTTP) This lets curl emulate a filled-in form in which a user has pressed the submit button. This causes curl to POST data using the Content-Type multipart/form-data according to RFC2388. This enables uploading of binary files etc. To force the 'content' part to be a file, prefix the file name with an @ sign. Example, to send your password file to the server, where 'password' is the name of the form-field to which /etc/passwd will be the input: curl -F password=@/etc/passwd www.mypasswords.com

在你的例子中，这就像 curl -F file=@/some/file/on/your/local/disk http://localhost:8080

您可以使用该命令:

curl -X POST --header 'Content-Type: multipart/form-data' --header 'Accept: application/json' --header 'Authorization: <<Removed>>' -F file=@"/home/xxx/Desktop/customers.json"  'API_SERVER_URL' -k 

如果你在Windows上使用curl:

curl -H "Content-Type: application/xml" -d "<?xml version="""1.0""" encoding="""UTF-8""" standalone="""yes"""?><message><sender>Me</sender><content>Hello!</content></message>" http://localhost:8080/webapp/rest/hello

Powershell + Curl + Zimbra SOAP API

${my_xml} = @"
<?xml version=\"1.0\" encoding=\"UTF-8\"?>
<soapenv:Envelope xmlns:soapenv=\"http://schemas.xmlsoap.org/soap/envelope/\">
  <soapenv:Body>
   <GetFolderRequest xmlns=\"urn:zimbraMail\">
    <folder>
       <path>Folder Name</path>
    </folder>
   </GetFolderRequest>
  </soapenv:Body>
</soapenv:Envelope>
"@

${my_curl} = "c:\curl.exe"
${cookie} = "c:\cookie.txt"

${zimbra_soap_url} = "https://zimbra:7071/service/admin/soap"
${curl_getfolder_args} = "-b", "${cookie}",
            "--header", "Content-Type: text/xml;charset=UTF-8",
            "--silent",
            "--data-raw", "${my_xml}",
            "--url", "${zimbra_soap_url}"

[xml]${my_response} = & ${my_curl} ${curl_getfolder_args}
${my_response}.Envelope.Body.GetFolderResponse.folder.id

你可以使用选项——data with file。

将xml内容写入名为soap_get.xml的文件，并使用curl命令发送请求:

curl -X POST——header "Content-Type:text/xml;charset=UTF-8"——data @soap_get.xml