How to pass file location through 'Create Soap client' REST API call?

27farzana27farzana Posts: 2

I am trying to create a scenario to create a soap client through below REST API call,
https://localhost:port/soavirt/api/v6/tools/soapClients
Ideally we can pass request payload in soap client through text, file in Request tab through Parasoft UI (view: literal xml)
When i try to do the same through above API call, I couldnt see part in that API post payload to select and pass an xml request through file. I tried adding below part in payload,
"request" : {
"inputMode": "literal",
"literal":{
"file":{
"location" : {
"id": "Request.txt",
"external": "Request.txt"
}
},
"type" : "file"
}
}
After executing, soap client gets created but The file location is not populating in that input file field.. could anyone please help in this?

Answers

  • williammccuskerwilliammccusker Posts: 433 ✭✭

    Hi,

    Are you trying to configure the request message to be read from a file using the REST API? The REST API currently doesn't support configuring the request message to be read from a file. The REST API unfortunately does not cover all the options available in the UI so if this is a feature you need please contact support and let them know about your use case.

    Sorry for the inconvenience!

  • benken_parasoftbenken_parasoft Posts: 908 ✭✭✭

    According to the schema type definitions in the API docs, "file" does not appear to be available at all. You are showing it as a child of "literal" but that contradicts the schema. So, you can set "type" to "file" but can not specify a file location. This appears to be something unimplemented/unsupported as confirmed by the previous comment.

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