Python调用外部软件命令

Question

我有一个小问题.我有一个软件,它有一个带两个输入的命令.命令是:maf2hal inputfile outputfile.

我需要从Python脚本中调用此命令.Python脚本询问用户输入文件的路径和输出文件的路径,并将它们存储在两个变量中.问题是当我调用命令maf2hal给出两个变量名作为参数时,我得到的错误是找不到文件.

有没有解决的办法？这是我的代码:

folderfound = "n" # looping condition
while (folderfound == "n"):
    path = raw_input("Enter path of file to convert (with the extension) > ")
    if not os.path.exists(path):
        print "\tERROR! file not found. Maybe file doesn't exist or no extension was provided. Try again!\n"
    else:
        print "\tFile found\n"
        folderfound = "y"

folderfound = "y" # looping condition
while (folderfound == "y"):
    outName = raw_input("Enter path of output file to be created > ")
    if os.path.exists(outName):
        print "\tERROR! File already exists \n\tEither delete the existing file or enter a new file name\n\n"
    else:
        print "Creating output file....\n"
        outputName = outName + ".maf"
        print "Done\n"
        folderfound = "n"
hal_input = outputName #inputfilename, 1st argument
hal_output = outName + ".hal" #outputfilename, 2nd argument

call("maf2hal hal_input hal_output", shell=True)

Answer 1

这是错的:

call("maf2hal hal_input hal_output", shell=True)

它应该是:

call(["maf2hal", hal_input, hal_output])

否则,您将"hal_input"作为实际文件名,而不是使用该变量.

shell=True除非绝对必要,否则你不应该使用,在这种情况下,它不仅是不必要的,而且是毫无意义的低效率.只需直接调用可执行文件,如上所述.

对于奖励积分,请使用check_call()而不是call(),因为前者将实际检查返回值并在程序失败时引发异常.使用call()没有,所以错误可能会被忽视.