I'm trying to do something like the following to print 3 and 4:
input.txt
1
2
3
4
program1.py
import subprocess
inputfile = open('input.txt', 'r')
inputfile.readline()
inputfile.readline()
subprocess.call('python program2.py',stdin=inputfile)
inputfile.close()
program2.py
while True:
print raw_input()
It will print nothing. But if I remove the readlines() it will print 1 through 4 just fine.
How can I use a file starting at a certain line for a stdin for a subprocess?
You can use Popen() and .communicate() instead of call() . I am not completely sure why call() does not work in your case, but the following would work -
program1.py
import subprocess
import sys
inputfile = open('input.txt', 'r')
inputfile.readline()
inputfile.readline()
p = subprocess.call(['python', 'program2.py'],stdin=subprocess.PIPE,stdout=sys.stdout)
p.communicate(inputfile.read())
inputfile.close()
.decode() is needed as communicate() expects byte string, not normal string. I also redirected the stdout for the process to the stdout of your script, so that the results are printed. Also, a better way to write program2.py is -
import sys
for l in sys.stdin:
print l
This does not cause program2.py to go into infinite loop.
Example/Demo -
I have two files, a.py and b.py -
a.py -
import subprocess
import sys
inputfile = open('input.txt', 'r')
inputfile.readline()
inputfile.readline()
p = subprocess.Popen(['/usr/bin/python', 'b.py'],stdin=subprocess.PIPE,stdout=sys.stdout)
p.communicate(inputfile.read())
inputfile.close()
b.py -
import sys
for l in sys.stdin:
print l
Result of running a.py -
3
4
