neo4j.exceptions.CypherSyntaxError: {code: Neo.ClientError.Statement.SyntaxError} {message: The old parameter syntax `{param}` is no longer supported. Please use `$param` instead (line 1, column 22 (offset: 21))

[lanjianchun]

Traceback (most recent call last): File "/usr/local/bin/aclpwn", line 8, in sys.exit(main()) File "/usr/local/lib/python3.8/dist-packages/aclpwn/init.py", line 108, in main fromres = session.run(q % args.from_type.capitalize(), name=from_object) File "/usr/local/lib/python3.8/dist-packages/neo4j/work/simple.py", line 217, in run self._autoResult._run(query, parameters, self._config.database, self._config.default_access_mode, self._bookmarks, kwparameters) File "/usr/local/lib/python3.8/dist-packages/neo4j/work/result.py", line 101, in _run self._attach() File "/usr/local/lib/python3.8/dist-packages/neo4j/work/result.py", line 202, in _attach self._connection.fetch_message() File "/usr/local/lib/python3.8/dist-packages/neo4j/io/_bolt4x0.py", line 353, in fetch_message response.on_failure(summary_metadata or {}) File "/usr/local/lib/python3.8/dist-packages/neo4j/io/_bolt4x0.py", line 552, in on_failure raise Neo4jError.hydrate(metadata) neo4j.exceptions.CypherSyntaxError: {code: Neo.ClientError.Statement.SyntaxError} {message: The old parameter syntax {param} is no longer supported. Please use $param instead (line 1, column 22 (offset: 21)) "MATCH (n:User {name: {name}}) RETURN n"

seems that neo4j need to update! how to fix it!

[tiyeuse]

Hello, I'm facing the same issue with a fresh install of aclpwn. It can be great to also modify the requirements.txt and change this:

neo4j-driver==1.7.0
neo4j==1.7.0
requests

[eriakort]

Hi, having similar issues right now. A fresh install of aclpwn. When I change the parameter as mentioned from {name} to $name it does skip this file without issues but shows something else from another file as follows:

Traceback (most recent call last): File "/usr/local/bin/aclpwn", line 8, in sys.exit(main()) File "/usr/local/lib/python3.9/dist-packages/aclpwn/init.py", line 120, in main paths = pathfinding.dijkstra_find(fromid, toid, args.database) File "/usr/local/lib/python3.9/dist-packages/aclpwn/pathfinding.py", line 34, in dijkstra_find data = resp.json() File "/usr/local/lib/python3.9/dist-packages/requests/models.py", line 900, in json return complexjson.loads(self.text, **kwargs) File "/usr/lib/python3/dist-packages/simplejson/init.py", line 525, in loads return _default_decoder.decode(s) File "/usr/lib/python3/dist-packages/simplejson/decoder.py", line 370, in decode obj, end = self.raw_decode(s) File "/usr/lib/python3/dist-packages/simplejson/decoder.py", line 400, in raw_decode return self.scan_once(s, idx=_w(s, idx).end()) simplejson.errors.JSONDecodeError: Expecting value: line 1 column 1 (char 0)

I honestly don't know how to fix this. Any solutions out there? Thanks in advance.