獲取`perl file.pl <some_file.txt` shell命令的輸出

Question

output = cStringIO.StringIO() 
output.write('string') # this emulates some_file.txt 

p = subprocess.Popen(['perl', 'file.pl'], shell=True, stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.STDOUT) 
line, _ = p.communicate(output.getvalue()) 
print line 
output.close() 

這打印Can't open perl script "file.pl": No such file or directory。 python腳本和perl文件在同一個目錄下！

我想line是的perl file.pl < some_file.txt

輸出如何做到這一點？

Answer 1

我想行是perl file.pl < some_file.txt

#!/usr/bin/env python 
from subprocess import check_output 

with open('some_file.txt', 'rb', 0) as file: 
    line = check_output(['perl', 'file.pl'], stdin=file) 

輸出就可以了Can't open perl script "file.pl": No such file or directory，通過完整路徑file.pl例如，如果是在同一個目錄中的Python腳本，那麼你可以使用get_script_dir() function來獲取路徑。