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Multiple rows from MySQL query saved in an array structure

+1 vote

I have this Python code:

self.lock_tables("read", ['nets_permissions as n', 'devices_permissions as d'])
usrs = self.db.get("SELECT n.user_id FROM nets_permissions as n \
                    left join devices_permissions as d \
                    on n.user_id = d.user_id \
                    where d.user_id is null \
                    and n.network_id=%s and n.perm<>3", netid)

for usr in usrs:
    self.lock_tables("write", ['devices_permissions'])
    self.db.execute("INSERT devices_permissions SET \
                     user_id=%s, network_id=%s, device_id=%s, perm=%s",\
                     usr, netid, sensid, perm)

I first do a query to retrieve some user_id from two tables. I want save this user_id in one variable and after do a for loop to insert this records in another table...

This code doesn't work. I obtain this error:    
Exception: Multiple rows returned for Database.get() query

How can I retrieve this multiple rows and then process everyone of them at one time?

posted Mar 20, 2013 by Salil Agrawal

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you can simplify irrelevant code (e.g. don't need to see a very long select statement)

1 Answer

+1 vote

What database adapter module are you using... The above sure doesn'tlook like MyS QLdb... Using MySQLdb's db-api functions the above would probably look something like (ignoring that your code sample appears to part of some method in a class):

import MySQLdb as db

con = db.connection(specify DB, user, password, etc.)

cur = con.cursor()

rslt = cur.execute("""select n.user_id from nets_permissions as n left join devices_permissions as d on n.user_id = d.user_id where d.user_id is null and n.network_id = %s and n.perm3""", netid)

***** UHM! "where d.user_id is null" implies (to me) that the join won't find anything -- n.user_id = d.user_id suggests you want NULL ids on both sides, but by definition NULL is never equal to NULL!

dta = cur.fetchall()
con.commit() #end implicit transaction

for usr in dta:
cur.execute("""insert devices_permissions (user_id, network_id, device_id, perm) values (%s, %s, %s, %s)""", (usr, netid, sensid, perm) )

answer Mar 20, 2013 by anonymous
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The problem description:

There are set of processes running on my system say process_set_1. Build a process agent which runs an asynchronous socket listening to incoming requests from process_set_1. Each process sends an id. The agent stores these ids in a dictionary and sends response that ur id has been accepted. Now agent process receives some data from another process (which is some sort of sister process for the agent). This data contains id along with a command which is to be sent by the agent to the process_set_1 through HTTP client over AF_UNIX socket, since each process in process_set_1 has a HTTP listening CLI. The process agent sends an HTTP request by recognizing id stored in dictionary to the process_set_1. A service running in the process_set_1 routes this HTTP command to the respective process.

Now my problem is the HTTP request which to be sent must go through AF_UNIX socket. I got this solution.

class UnixStreamHTTPConnection(httplib.HTTPConnection):

    def __init__(self, path, host='localhost/rg_cli',port=None, strict=None,
        httplib.HTTPConnection.__init__(self, host, port=port, strict=strict,

    def connect(self):
        self.sock = socket.socket(socket.AF_UNIX, socket.SOCK_STREAM)

But this cannot work since normal socket will not work and thus i thought of asyncore module in python. To use asyncore module again i will have to subclass asyncore.dispatcher. This class also contains connect() method.

Another problem is I don't know how asyncore module works and thus not able to find a way to mix the work of 1) listening forever, accept the connection, store id and the sock_fd.
2) accept data from the process' agent sister process, retrieve the sock_fd by matching the id in the dictionary and send it through the AF_UNIX socket.

Please help since i have already spent 2 days digging it out. Sorry if could explain my problem very well.

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