How do you avoid breaking a promise chain in JavaScript?
Chain .then calls directly: fetchUser(id).then(user => fetchPosts(user)).then(posts => ...). Do not store the promise in a variable and call .then multiple times on it (that runs them in parallel, not sequentially).
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More FAQs in Promise Chaining Pitfalls in JavaScript
Not returning from .then (undefined flows to next), breaking the chain (parallel not sequential), nesting .then (mini callback hell), forgetting .catch (unhandled rejections), sequential instead of parallel await (slow), and wrapping promises in unnecessary new Promise.
It defeats the purpose of promises (creating mini callback hell). Instead of nesting, chain flatly. Or use async/await, which is the cleanest way to sequence async operations.
The chain continues as fulfilled with the value returned by .catch. If .catch returns 'fallback', the next .then receives 'fallback'. If .catch throws or re-throws, the next .catch catches it.
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