Aggarwal-Raghav commented on PR #6749:
URL: https://github.com/apache/hive/pull/6749#issuecomment-5704342870
> please check
>
> ```
> @Test
> public void testResourceNamedNamespacesIsNotTreatedAsPrefix() {
> assertEquals(Route.LOAD_NAMESPACE,
route("v1/namespaces/namespaces").first());
> assertEquals(Route.LOAD_TABLE,
route("v1/namespaces/db/tables/namespaces").first());
> assertEquals(Route.UPDATE_TABLE,
> Route.from(HTTPMethod.POST,
"v1/namespaces/db/tables/namespaces").first());
> assertEquals(Route.LOAD_TABLE,
> route("v1/catalogs/sales/namespaces/db/tables/namespaces").first());
> }
> ```
>
> I think an exact-length match (no prefix) should always win, and among
prefixed matches, the fewest prefix segments should win:
>
> ```
> public static Pair<Route, Map<String, String>> from(HTTPMethod method,
String path) {
> List<String> parts = SLASH.splitToList(path);
> Route best = null;
> for (Route candidate : Route.values()) {
> if (candidate.matches(method, parts)
> && (best == null || candidate.prefixLength(parts) <
best.prefixLength(parts))) {
> best = candidate;
> }
> }
> return best == null ? null : Pair.of(best, best.variables(parts));
> }
> ```
**Yes, this edge case was a miss from me** but 1 question in this approach,
if I do `/v1/config` then it will search for all the Route enum values i.e. all
25 routes LIST_NAMESPACES, CREATE_NAMESPACE etc. but as the number of routes is
just 25 so there shouldn't be any performance concern but this cost will be for
all REST HTTP calls.
**there might be a better way,**
how about we pre-sort (descending order) the routes based on
`requiredLength` at startup. `prefixLength = request_path_size -
requiredLength` **_The greater the requiredLength the lower the
prefixLenght_**. So the first one will be best match. WDYT?
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