8000 fix: warn when unregistering non existing module by kiaking · Pull Request #1786 · vuejs/vuex · GitHub
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fix: warn when unregistering non existing module #1786

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Merged
merged 2 commits into from
Jun 29, 2020
Merged

fix: warn when unregistering non existing module #1786

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054cee8
fix: warn when unregistering non existing module
kiaking Jun 29, 2020
4fae622
refactor: store child as a variable and reuse it
kiaking Jun 29, 2020
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fix: warn when unregistering non existing module
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@kiaking
kiaking committed Jun 29, 2020
commit 054cee89e4e0c02f135b629de35f75af79149ba9
15 changes: 14 additions & 1 deletion src/module/module-collection.js
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Original file line number Diff line number Diff line change
Expand Up @@ -49,7 +49,20 @@ export default class ModuleCollection {
unregister (path) {
const parent = this.get(path.slice(0, -1))
const key = path[path.length - 1]
if (!parent.getChild(key).runtime) return

if (!parent.getChild(key)) {
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Can we store the return value of getChild in a variable so that we reuse it with the next if condition?

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Oh yeah good point! Thanks. I've done it, and rebased (to the latest dev) and pushed 👍

if (__DEV__) {
console.warn(
`[vuex] trying to unregister module '${key}', which is ` +
`not registered`
)
}
return
}

if (!parent.getChild(key).runtime) {
return
}

parent.removeChild(key)
}
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8 changes: 8 additions & 0 deletions test/unit/module/module-collection.spec.js
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Expand Up @@ -92,4 +92,12 @@ describe('ModuleCollection', () => {
collection.unregister(['a'])
expect(collection.get(['a']).state.value).toBe(true)
})

it('warns when unregistering non existing module', () => {
const spy = jest.spyOn(console, 'warn').mockImplementation()

const collection = new ModuleCollection({})
collection.unregister(['a'])
expect(spy).toHaveBeenCalled()
})
})
0