python - TypeError: expected string or buffer on sql query

Question

I have an sql query that's returning the following value -

{'jobrun_deps': 'T=1 \x01ID=56494759 \x01DID=583887 \x01O=N \x01M=N \x01J=76732 \x01R=0 \x01P=1 \x01S=101 \x01WR=N \x01T=1 \x01ID=56494760 \x01DID=418400 \x01O=N \x01M=N \x01J=48064 \x01R=14780471 \x01P=1 \x01S=101 \x01WR=N \x01T=1 \x01ID=56494761 \x01DID=583889 \x01O=N \x01M=N \x01J=76733 \x01R=0 \x01P=1 \x01S=101 \x01WR=N \x01'}

So when I try to do this -

results = (re.findall(r'ID=(\d+)', my_query))

I get this error -

Traceback (most recent call last):
  File "D:\Scripts\OPS\TIDAL\dependency_check.py", line 25, in <module>
    results = (re.findall(r'ID=(\d+)', my_query))
  File "D:\Python27\lib\re.py", line 177, in findall
    return _compile(pattern, flags).findall(string)
TypeError: expected string or buffer

This is because the re.findall is expecting it to not have the column name in the results. how do I either change my re.findall to use the output I'm providing or change the sql query to give a string like it expects.

'T=1 \x01ID=56494759 \x01DID=583887 \x01O=N \x01M=N \x01J=76732 \x01R=0 \x01P=1 \x01S=101 \x01WR=N \x01T=1 \x01ID=56494760 \x01DID=418400 \x01O=N \x01M=N \x01J=48064 \x01R=14780471 \x01P=1 \x01S=101 \x01WR=N \x01T=1 \x01ID=56494761 \x01DID=583889 \x01O=N \x01M=N \x01J=76733 \x01R=0 \x01P=1 \x01S=101 \x01WR=N \x01'

Kasravnd · Accepted Answer · 2014-12-04 21:39:51Z

3

Its because of that my_query is a dictionary and you can not pass it to re.findall you need to pass the value of the dictionary :

results = (re.findall(r'ID=(\d+)', my_query['jobrun_deps']))

answered Dec 4, 2014 at 21:39

Kasravnd

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1 Comment

whoisearth Over a year ago

Thanks! knew it was going to be something simple.

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