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[py] Avoid waiting indefinitely on a frozen chromedriver process (#14578
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Add a default timeout of 2 minutes two requests to the webdriver
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dbernhard-0x7CD authored Oct 18, 2024
1 parent 97c29e7 commit 7c9f000
Showing 1 changed file with 3 additions and 3 deletions.
6 changes: 3 additions & 3 deletions py/selenium/webdriver/remote/remote_connection.py
Original file line number Diff line number Diff line change
Expand Up @@ -305,7 +305,7 @@ def execute(self, command, params):
LOGGER.debug("%s %s %s", command_info[0], url, str(trimmed))
return self._request(command_info[0], url, body=data)

def _request(self, method, url, body=None):
def _request(self, method, url, body=None, timeout=120):
"""Send an HTTP request to the remote server.
:Args:
Expand All @@ -323,12 +323,12 @@ def _request(self, method, url, body=None):
body = None

if self.keep_alive:
response = self._conn.request(method, url, body=body, headers=headers)
response = self._conn.request(method, url, body=body, headers=headers, timeout=timeout)
statuscode = response.status
else:
conn = self._get_connection_manager()
with conn as http:
response = http.request(method, url, body=body, headers=headers)
response = http.request(method, url, body=body, headers=headers, timeout=timeout)
statuscode = response.status
data = response.data.decode("UTF-8")
LOGGER.debug("Remote response: status=%s | data=%s | headers=%s", response.status, data, response.headers)
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