CODEWITHSUNDEEP
pythoncsv
user1453786
user1453786

Reputation: 225

Open URL stored in a csv file

I'm almost an absolute beginner in Python, but I am asked to manage some difficult task. I have read many tutorials and found some very useful tips on this website, but I think that this question was not asked until now, or at least in the way I tried it in the search engine.

I have managed to write some url in a csv file. Now I would like to write a script able to open this file, to open the urls, and write their content in a dictionary. But I have failed : my script can print these addresses, but cannot process the file.

Interestingly, my script dit not send the same error message each time. Here the last : req.timeout = timeout AttributeError: 'list' object has no attribute 'timeout'

So I think my script faces several problems : 1- is my method to open url the right one ? 2 - and what is wrong in the way I build the dictionnary ?

Here is my attempt below. Thanks in advance to those who would help me !

import csv
import urllib

dict = {}

test = csv.reader(open("read.csv","rb"))

for z in test:  
    sock = urllib.urlopen(z)
    source = sock.read()
    dict[z] = source
    sock.close()
print dict

Upvotes: 0

Views: 1083

Answers (1)

Christian Witts
Christian Witts

Reputation: 11585

First thing, don't shadow built-ins. Rename your dictionary to something else as dict is used to create new dictionaries.

Secondly, the csv reader creates a list per line that would contain all the columns. Either reference the column explicitly by urllib.urlopen(z[0]) # First column in the line or open the file with a normal open() and iterate through it.

Apart from that, it works for me.

Upvotes: 1

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