从外壳脚本向Python传递参数

Question

I have run through some similar questions but none seem to answer my doubt, I am trying to pass the argument to python script from my bash script. I don't see errors nor I get the required output. What am I doing wrong here?

python:

import os
import glob
import unicodecsv as csv
import pandas as pd
import codecs
import sys




OUTPUT_PATH=sys.argv[2] 


def createFolder(directory):
  # print('createFolder')
  try:
        if not os.path.exists(directory):
            # print('createFolder')
            os.makedirs(directory)
            # print(directory)
  except OSError:
        print ('Error: Creating directory. ' +  directory)
createFolder(OUTPUT_PATH + 'csv20_out/')

bash script:

INPUT_PATH='home/pg/Public/test_data/test0002_2files_to_1flow/csv64_in/'
OUTPUT_PATH='home/pg/Public/test_data/test0002_2files_to_1flow/'

cd /
cd home/pg/Documents/LMS/kanjiforbeginner/convertToYAML/


cd converts_csv642csv20
python csvFile.py $INPUT_PATH $OUTPUT_PATH

Should create a folder csv20_out

when I print OUTPUT_PATH, I get 'home/pg/Public/test_data/test0002_2files_to_1flow/'

Answer 1

The contents of sys.argv look like this:

['csvFile.py', 'inputPath', 'outputPath']

To get the outputPath you need to access sys.argv[2] in your case.

Answer 2

By convention, the first argument is the name of the program itself—that is, the first thing specified on the command line ( whatever_name.py ).

So argv[0] is the program's name, argv[1] is your input path, and argv[2] is your output path.

Answer 3

So, as the other commenters have pointed out, you need to use argv[2] instead of argv[1] This is because in bash argv[0] is the python script being executed and argv[1] in your case becomes '$INPUT_PATH'

But, that won't solve your problem. In your script, you need to remove the ' surrounding $INPUT_PATH and $OUTPUT_PATH so that the last line becomes,
python csvFile.py $INPUT_PATH $OUTPUT_PATH

'$INPUT_PATH' sends the string $INPUT_PATH to your python script, whereas,

$INPUT_PATH sends 'home/pg/Public/test_data/test0002_2files_to_1flow/csv64_in/'

or whatever value you choose to assign to $INPUT_PATH

And lastly, you need to change, the values of $INPUT_PATH and $OUTPUT_PATH
INPUT_PATH='home/pg/Public/test_data/test0002_2files_to_1flow/csv64_in/' OUTPUT_PATH='home/pg/Public/test_data/test0002_2files_to_1flow/'

needs to become

INPUT_PATH='/home/pg/Public/test_data/test0002_2files_to_1flow/csv64_in/' OUTPUT_PATH='/home/pg/Public/test_data/test0002_2files_to_1flow/'

The / at the beginning gives the full path to the directory or file, if you didn't use the / then it becomes a relative path.

In your bash script, you're navigating folders using cd as a result, the folder

'home/pg/Public/test_data/test0002_2files_to_1flow/csv_out'

will get created under the directory,

home/pg/Documents/LMS/kanjiforbeginner/convertToYAML/

Adding the / at the beginning to $INPUT_PATH and $OUTPUT_PATH as shown before would ensure that your code:

• searches for input in the right directory
• creates the folder in the right directory

edit:

As Kim has suggested below, the last line in the script should be:

python csvFile.py "$INPUT_PATH" "$OUTPUT_PATH"

This is, in case $INPUT_PATH or $OUTPUT_PATH has spaces or special characters.