CodeForPhilly / CodeForPhilly/stately

Initiate action should/could be specified in URL

Abierto
#17 0 comentarios 0 reacciones 0 asignados Ver en GitHub
discussion
Lenguaje dominante
Python
Estrellas
22
Forks
5
Métricas de merge de PR
Sin PR fusionados en 30 d

Descripción

The way the client is designed, it gets the workflow definition from `GET /api/travel-request/`. If there's an `id` property, it renders the `data` and the `events`. If there's multiple `state.actions`, it renders action buttons. If there's only 1 `state.actions`, it renders the form for it. If there's more than 1 `state.actions`, it waits until you select one of the action buttons, and renders that form. On submission, the form posts to `POST /api/travel-request//?token=xx`.

At the moment, this fails on the initiate action, I think because by design we expected clients to not include the action on the initiate `POST`. But it's actually pretty simple to do that since (a) we're providing the name of the action in the response to their `GET` request, and (b) they're already using part of that response for their `POST` request (the template).

My guess is you probably added extra logic to identify what the default action was. Perhaps this isn't necessary, and we can just expect clients to include the action name in every `POST` request.

Thoughts @mjumbewu ?

Guía de contribución

No hay ninguna guía de contribución indexada para este repositorio

Evaluación

Este issue todavía no se ha evaluado.

Recibe los nuevos issues en tu correo

Un resumen breve de issues de GitHub para principiantes.