python - if statement return wrong value

Question

I used the followng code to trigger IF statement based on the value of row[0] my problem is always IF statement give me wrong output even if the value is greater than 0 its keep printing NO FAILURE .

conn = psycopg2.connect(conn_string)
        cursor = conn.cursor()
        cursor.execute("select count(*) as result from events_log where match_event_timestamp  > (NOW() - INTERVAL '147 hours');")
        row = cursor.fetchone()
        cursor.execute("SELECT COUNT(*) AS result FROM events_log  WHERE status = 'FAILURE' AND match_event_timestamp  > (NOW() - INTERVAL '24 hours');")
        row += cursor.fetchone()
        logger.info("Number of Events in last 24hr : %s ; Number of Event FAILURE  in last 24h: %s" % (row[0],row[1]))
        output = { 'api_key':'jhfsdf', 'data': { 'item' : [
                        {'text': 'Today ', 'value': row[0]},
                        {'text': 'Current Month ', 'value': row[1]}
                    ] } }

        if row[0] > '0':
                print('FAILURE')
        else:
                print("NO FAILURE")

what did i missed in IF statement here ?

i dont know what row[0] contains... but what happens if you convert them to int before evaluating the condition of the if/else ? like if int(row[0]) > 0: — dot.Py
– dot.Py, Commented May 29, 2016 at 14:00

Ani Menon · Accepted Answer · 2016-05-29 14:04:57Z

1

'0' is a string use 0. Like this:

if( row[0] > 0 ):
   ...

Otherwise you get this:

TypeError: unorderable types: int() > str()

which means you may not compare an int to a string.

answered May 29, 2016 at 14:04

Ani Menon

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