loop through results looking for a string using python

Question

i have a csv and i'd like to find every occurrence of 'http://'. using the code below i get nothing. at this point, i'd like to know why i'm not getting any results at all (not even the 'hi' is bring printed). i'm a noob and it's just not making sense to me. explanations are welcome, links are fine but only if they help me understand...

import csv

with open('a_bunch_of_urls_and_other_stuff.csv', 'rb') as f:
    reader = csv.reader(f, delimiter=',')

    # remove results that dont have 'http://'     
    for result in reader:
        # print result      # this prints everything from the cvs
        # print result[2]   # this prints out the column with urls

        if result[2] == 'http://':
           print 'hi'

It looks like you want to use if "http://" in result[2]. E.g. use search_term in result[2] not search_term == result[2]. You might also want to consider using a regexp if you think in the future you might also care about https:// or other protocols too. — Tom Dalton
– Tom Dalton, Commented Apr 16, 2015 at 22:25
Ya gotta give us something to go on! What does print result[2] look like for some of the rows? Should the test be if result[2].lower().startswith('http://'):? — tdelaney
– tdelaney, Commented Apr 16, 2015 at 22:28
basically, i want to "check" to see if there's a 'http://' in the row. if so, then print...if not then don't print. — Silas
– Silas, Commented Apr 16, 2015 at 22:48

Saicharan S M · Accepted Answer · 2015-04-16 23:01:15Z

1

I am assuming you trying to check if result[2] contains 'http://'.
You can try:

if "http://" in result[2]:
    print result[2]

answered Apr 16, 2015 at 23:01

Saicharan S M

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1 Comment

Saicharan S M Over a year ago

if result[2] == 'http://': checks if the contents of result[2] is exactly 'http://'. Where as my answer checks if result[2] has at least one sub-string with the value 'http://'.

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