Submit and vote on feature ideas.

Welcome to the new Parasoft forums! We hope you will enjoy the site and try out some of the new features, like sharing an idea you may have for one of our products or following a category.

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

Options
27farzana
27farzana 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

  • williammccusker
    williammccusker Posts: 644 ✭✭✭
    Options

    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_parasoft
    benken_parasoft Posts: 1,232 ✭✭✭
    Options

    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.