how to call a function (mutation) from another function (mutation)?

question in the title

4 Replies

snippet for editing

export const remove = mutation({
    args: { bmId: v.id("bookmarks") },
    handler: async (ctx, { bmId }) => {
        
    },
});

export const remove = mutation({
    args: { bmId: v.id("bookmarks") },
    handler: async (ctx, { bmId }) => {
        
    },
});

erquhart•16mo ago

3 approaches I use: 1. Ideally, see if there's a way to avoid this situation, often by splitting reusable logic out into separate functions, so you can do everything you need to in one mutation. There are a lot of approaches for this covered in Convex Stack and their example repos on GitHub. 2. If you must call a mutation from another mutation, use the scheduler arg to schedule it (you can schedule with a 0 wait in this case). 3. Last option is using an action instead of the initial mutation. This is not ideal as actions don't have the same transactional/atomicity guarantees as mutations and won't be retried in case of failure.

ashuvssut (ashu)OP•16mo ago

thanks! Btw I am really enjoying convex Its awesome Perfect DX

erquhart•16mo ago

Nice!! I'm totally not a convex team member, just a fellow avid user lol. But agree, and same!

how to call a function (mutation) from another function (mutation)?

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