Running Python With STDIN From Bash

Question

I have a bash code (Mybash1.sh) where the result I need to pass to another bash code (Mybash2.sh) that contain Python

Here are the codes. Mybash1.sh

#! /bin/bash
# Mybash1.sh
cut -f1,3 input_file.txt | sort | ./Mybash2.sh 

Mybash2.sh is this:

#! /bin/bash
#Mybash2.sh
python mycode.py foo.txt <("$@") > output.txt
# do something for output.txt

The problem I have is that "output.txt" in Mybash2.sh contain no result. Is there a correct way to execute Python in Mybash2.sh?

Note that mycode.py will work if I run it on an intermediate temporary file given from Mybash1.sh. But I wanted to avoid using that, since I will call Mybash2.sh in many instances within Mybash1.sh.

Snippet of mycode.py looks like this:

if __name__ == "__main__":
    import sys, os, fileinput
    progName = os.path.basename(sys.argv[0])
    if len(sys.argv) != 3:
        sys.exit('Usage: ' + progName + ' file1 file2')
    file1 = fileinput.input(sys.argv[1])
    file2 = fileinput.input(sys.argv[2])

    # do something for file1 and file2

    file1.close()
    file2.close()

Answer 1

in your myscript1.sh, you are passing a pipeline to myscript2.sh, therefore, its some sort of STDIN for myscript2.sh. You should read that STDIN from myscript2.sh, not taking in input arguments. eg myscript2.sh

#!/bin/bash
while read myinput
do
 echo "myinput is $myinput"
 # assuming you are passing each input line from the cut command into Python
 python mycode.py foo.txt $myinput > output.txt
done

Lastly, why all these dependencies? can't you do everything in Python, or shell??

Answer 2

In python, you want to have file2 = sys.stdin.

And then:

#! /bin/bash
#Mybash2.sh
python mycode.py foo.txt > output.txt

EDIT: I've just seen the fileinput docs and it seems that if you supply '-' to fileinput.input(), it will read stdin, so without any changes in your Python, this should work:

#! /bin/bash
#Mybash2.sh
python mycode.py foo.txt - > output.txt