关于无状态判定及修改后REST图片示例状态的技术咨询
Great question—let's break this down clearly, since state management is one of the trickier bits about getting REST right.
问题一:修改后的架构是否仍为有状态?
Short answer: Yes, it’s still stateful—here’s why.
The original example you’re referencing was stateful because the server kept track of a "current image" value for each user; every GET /next relied on that stored state to know what to return next.
Your tweak changes how you update that state (using an explicit PUT /currentImage?value=X instead of letting GET update it implicitly), but it doesn’t get rid of the server’s need to store user-specific state. Think about it: to handle a GET /nextImage?userId request, the server still has to look up the saved "current image" value for that userId to figure out what the next image should be. Even though this GET doesn’t modify the state, the server is still holding onto client-specific data between requests. That’s the key sign of a stateful system in REST terms.
问题二:什么样的设计才是无状态的?(以及为什么)
I think what you’re getting at here is: What would make this scenario stateless, and why does that align with REST’s rules?
REST’s stateless constraint boils down to this: every request must contain all the info the server needs to process it—no server-side storage of client session state allowed.
Instead of having the server track which image a user is on, the client would send the current image’s identifier directly in the request. For example:
GET /images/{currentImageId}/next– The server uses the providedcurrentImageIdto look up the next image, no need to store anything about the user.- Or even simpler: the client just requests the next image by its known ID directly:
GET /images/456right after viewing/images/455.
This design is stateless because:
- No client-specific state sticks around on the server between requests.
- The server doesn’t need to remember any prior interactions to handle the current request.
- Scaling becomes way easier (any server instance can handle any request, no need to sync session state across machines).
内容的提问来源于stack exchange,提问作者Woodgnome

