Question

I am new to python and was trying to read a specific column from a CSV file.I am editing the problem now.

Here is the CSV file for eg:

one two three four
1    3    5    7
2    3    5    7

I wrote a very simple Python program as :(thanks to Amrit below for the modified mode).

import csv
with open('Activity_PSR.csv','rU') as csvFile:
    reader=csv.reader(csvFile,delimiter=',')
    for row in reader:
         print row[6]
csvFile.close()

I am getting the specific column with Column header.I want the output of just the elements without the column header.

Once I get the output of this column ,I want to store it in a variable and use it in a CURL command to POST Data. CURL command equiv I beleive is urlib.

Here is the CURL POST Command:

curl -v -X POST -d ‘{“Organization_Id___ORACO__Account_Order":300100051155060,"}' slcaf961.us.oracle.com:10615/crmCommonApi/resources/11.1.10/__ORACO__Order_c
 -H "Content-Type:application/vnd.oracle.adf.resourceitem+json" --user “sales_representative”

And the urllib code I am planning to use:

import urllib2
data = '{“Organization_Id___ORACO__Account_Order":300100051155060"}'
url = 'slcad745.us.oracle.com:10618/crmCommonApi/resources/11.1.10/__ORACO__Order_c'
req = urllib2.Request(url, data, {'Content-Type': 'application/json'})
f = urllib2.urlopen(req)
for x in f:
    print(x)
f.close()

In the data field here Instead of passing 300100051155060 directly I want to use the value I got from the CSV above and pass that as the Account-Id.

How to pass the value I got from the CSV and pass it to the JSON payload as a parameter ? Need to store the value I got from CSV and store it in a Variable and pass it to the JSON payload/data as a parameter and execute the POST method. Request folks to help me on how to acheive the above scenario

Answer 1

This means that a row does not have a column on index 2. You need to create a guard:

for row in csv_f:
    # make sure row has value on column 2 - otherwise skip row
    if len(row) > 2:
        print row[2]

You can also use an exception, which in Python is considered cheap:

for row in csv_f:
    try:
        print row[2]
    except IndexError:
        print("skipping row %s" % str(row))

If you know all rows need to have a value there, make sure you're using the right delimiter using the delimiter argument to the reader's constructor.

Also, make sure you're closing the file after the loop:

for row in csv_f:
    # do stuff with 'row'
csv_f.close()

Answer 2

You should consider working with pandas. It has a function called read_csv that works very well. In your case the code to print out column three is

from pandas import *
csv_f=read_csv('test.csv', sep=r"\s*")
for a in csv_f['three']:
     print a

Of course, it has a bunch of functionalities that you should check out and see if you can make an use of.

Answer 3

请改用此代码：

import csv
with open('test.csv','r') as csvFile:
    reader=csv.reader(csvFile,delimiter=',')
    reader.next()   # if you don't want to print columns headers
    for row in reader:
         print row[2]
csvFile.close()

Reading specific Column Values from CSV and pass it to urllib as parameter in Python

3 个答案: