I have a bug that I don't know how to fix or even reproduce:
query = "SELECT id, name FROM names ORDER BY id"
results = database.execute(query)
where the class Database
contains:
def execute(self, query):
cursor = self.db.cursor()
try:
cursor.ex开发者_StackOverflow社区ecute(query)
return cursor.fetchall()
except:
import traceback
traceback.print_exc(file=debugFile)
return []
This is how I open the database connection:
self.db = MySQLdb.connect(
host=mysqlHost,
user=mysqlUser,
passwd=mysqlPasswd,
db=mysqlDB
)
This is the stacktrace of the error:
File "foo.py", line 169, in application results = config.db.execute(query)
File "Database.py", line 52, in execute
return cursor.fetchall()
File "/usr/lib/pymodules/python2.6/MySQLdb/cursors.py", line 340, in fetchall
self._check_executed()
File "/usr/lib/pymodules/python2.6/MySQLdb/cursors.py", line 70, in _check_executed
self.errorhandler(self, ProgrammingError, "execute() first")
File "/usr/lib/pymodules/python2.6/MySQLdb/connections.py", line 35, in defaulterrorhandler
raise errorclass, errorvalue
ProgrammingError: execute() first
Do you have any ideas of why this is happening and how can I fix it? I searched on the internet and I found out that the reason may be having 2 cursors, but I have only one.
try this in your traceback it's for debugging:
except ProgrammingError as ex:
if cursor:
print "\n".join(cursor.messages)
# You can show only the last error like this.
# print cursor.messages[-1]
else:
print "\n".join(self.db.messages)
# Same here you can also do.
# print self.db.messages[-1]
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