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How to pass database in base url like 'base_url = 'http://localhost:8069/odoo_test' 
'

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There is no Rest Api from Odoo by default?

编写者

Yeah its a custom api that

编写者 最佳答案

I tried following your steps, like http://localhost:8069/web?db=dbname/api/verify but still getting this error in response.

{'error': {'message': 'Odoo Server Error', 'data': {'name': 'werkzeug.exceptions.BadRequest', 'message': "400 Bad Request: <function Home.web_client at 0x7ff4e0fc3d90>, /web: Function declared as capable of handling request of type 'http' but called with a request of type 'json'", 'debug': 'Traceback (most recent call last):\n File "/usr/lib/python3/dist-packages/odoo/http.py", line 654, in _handle_exception\n return super(JsonRequest, self)._handle_exception(exception)\n File "/usr/lib/python3/dist-packages/odoo/http.py", line 312, in _handle_exception\n raise pycompat.reraise(type(exception), exception, sys.exc_info()[2])\n File "/usr/lib/python3/dist-packages/odoo/tools/pycompat.py", line 87, in reraise\n raise value\n File "/usr/lib/python3/dist-packages/odoo/http.py", line 696, in dispatch\n result = self._call_function(**self.params)\n File "/usr/lib/python3/dist-packages/odoo/http.py", line 320, in _call_function\n raise werkzeug.exceptions.BadRequest(msg % params)\nwerkzeug.exceptions.BadRequest: 400 Bad Request: <function Home.web_client at 0x7ff4e0fc3d90>, /web: Function declared as capable of handling request of type \'http\' but called with a request of type \'json\'\n', 'exception_type': 'internal_error', 'arguments': []}, 'code': 200}, 'jsonrpc': '2.0', 'id': None}

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最佳答案

Hi Keerthi,


I guess you need to pass the db name along with the url. Try this


http://localhost:8069/web?db=dbname
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