[#387] Do not escape the "timeout" parameter. (#394)

* [#387] Do not escape the "timeout" parameter. When the "timeout" parameter is escaped and turned into a string, it does not pass validation checks. This PR ensures that timeout is passed through as-is, just like "request_timeout". There's no explicit test for this parameter, but I included it on some existing tests just to ensure that the code is exercised. Without the fix in place, one of the tests as modified will not pass. Signed-off-by: Charles Greer <[email protected]> * Unit test for timeout parameter Revert original test modifications. Signed-off-by: Charles Greer <[email protected]> * Extend test and linter Signed-off-by: Charles Greer <[email protected]> --------- Signed-off-by: Charles Greer <[email protected]>
2023-05-23 18:12:29 -04:00
parent 0395798d2b
commit fe3b5a8922
3 changed files with 33 additions and 5 deletions
@@ -165,16 +165,17 @@ def query_params(*opensearch_query_params):
            elif api_key is not None:
                headers["authorization"] = "ApiKey %s" % (_base64_auth_header(api_key),)

+            # don't escape ignore, request_timeout, or timeout
+            for p in ("ignore", "request_timeout", "timeout"):
+                if p in kwargs:
+                    params[p] = kwargs.pop(p)
+
            for p in opensearch_query_params + GLOBAL_PARAMS:
                if p in kwargs:
                    v = kwargs.pop(p)
                    if v is not None:
                        params[p] = _escape(v)

-            # don't treat ignore, request_timeout, and opaque_id as other params to avoid escaping
-            for p in ("ignore", "request_timeout"):
-                if p in kwargs:
-                    params[p] = kwargs.pop(p)
            return func(*args, params=params, headers=headers, **kwargs)

        return _wrapped