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include a string variable that contains spaces into the 'subprocess.run()' [duplicate]

I'm working on a wrapper script that will exercise a vmware executable, allowing for the automation of virtual machine startup/shutdown/register/deregister actions. I'm trying to use subprocess to handle invoking the executable, but the spaces in the executables path and in parameters of the executable are not being handled correctly by subprocess. Below is a code fragment:
vmrun_cmd = r"c:/Program Files/VMware/VMware Server/vmware-cmd.bat"
def vm_start(target_vm):
list_arg = "start"
list_arg2 = "hard"
if vm_list(target_vm):
p = Popen([vmrun_cmd, target_vm, list_arg, list_arg2], stdout=PIPE).communicate()[0]
print p
else:
vm_register(target_vm)
vm_start(target_vm)
def vm_list2(target_vm):
list_arg = "-l"
p = Popen([vmrun_cmd, list_arg], stdout=PIPE).communicate()[0]
for line in p.split('\n'):
print line
If I call the vm_list2 function, I get the following output:
$ ./vmware_control.py --list
C:\Virtual Machines\QAW2K3Server\Windows Server 2003 Standard Edition.vmx
C:\Virtual Machines\ubunturouter\Ubuntu.vmx
C:\Virtual Machines\vacc\vacc.vmx
C:\Virtual Machines\EdgeAS-4.4.x\Other Linux 2.4.x kernel.vmx
C:\Virtual Machines\UbuntuServer1\Ubuntu.vmx
C:\Virtual Machines\Other Linux 2.4.x kernel\Other Linux 2.4.x kernel.vmx
C:\Virtual Machines\QAClient\Windows XP Professional.vmx
If I call the vm_start function, which requires a path-to-vm parameter, I get the following output:
$ ./vmware_control.py --start "C:\Virtual Machines\ubunturouter\Ubuntu.vmx"
'c:\Program' is not recognized as an internal or external command,
operable program or batch file.
Apparently, the presence of a second parameter with embedded spaces is altering the way that subprocess is interpreting the first parameter. Any suggestions on how to resolve this?
python2.5.2/cygwin/winxp

If you have spaces in the path, the easiest way I've found to get them interpreted properly is this.
subprocess.call('""' + path + '""')
I don't know why exactly it needs double double quotes, but that is what works.

I believe that list2cmdline(), which is doing the processing of your list args, splits any string arg on whitespace unless the string contains double quotes. So I would expect
vmrun_cmd = r'"c:/Program Files/VMware/VMware Server/vmware-cmd.bat"'
to be what you want.
You'll also likely want to surround the other arguments (like target_vm) in double quotes on the assumption that they, too, each represent a distinct arg to present to the command line. Something like
r'"%s"' % target_vm
(for example) should suit.
See the list2cmdline documentation

'c:\Program' is not recognized as an internal or external command, operable program or batch file.
To get this message, you are either:
Using shell=True:
vmrun_cmd = r"c:\Program Files\VMware\VMware Server\vmware-cmd.bat"
subprocess.Popen(vmrun_cmd, shell=True)
Changing vmrun_cmd on other part of your code
Getting this error from something inside vmware-cmd.bat
Things to try:
Open a python prompt, run the following command:
subprocess.Popen([r"c:\Program Files\VMware\VMware Server\vmware-cmd.bat"])
If that works, then quoting issues are out of the question. If not, you've isolated the problem.

In Python on MS Windows, the subprocess.Popen class uses the CreateProcess API to started the process. CreateProcess takes a string rather than something like an array of arguments. Python uses subprocess.list2cmdline to convert the list of args to a string for CreateProcess.
If I were you, I'd see what subprocess.list2cmdline(args) returns (where args is the first argument of Popen). It would be interesting to see if it is putting quotes around the first argument.
Of course, this explanation might not apply in a Cygwin environment.
Having said all this, I don't have MS Windows.

One problem is that if the command is surrounded with quotes and doesn't have spaces, that could also confuse the shell.
So I do this:
if ' ' in raw_cmd:
fmt = '"%s"'
else:
fmt = '%s'
cmd = fmt % raw_cmd

That was quite a hard problem for the last three ours....nothing stated so far did work, neither using r"" or Popen with a list and so on. What did work in the end was a combination of format string and r"". So my solution is this:
subprocess.Popen("{0} -f {1}".format(pathToExe, r'"%s"' % pathToVideoFileOrDir))
where both variables pathToExe and pathToVideoFileOrDir have whitespaces in their path. Using \" within the formatted string did not work and resulted in the same error that the first path is not detected any longer correctly.

Possibly stupid suggestion, but perhaps try the following, to remove subprocess + spaces from the equation:
import os
from subprocess Popen, PIPE
os.chdir(
os.path.join("C:", "Program Files", "VMware", "VMware Server")
)
p = Popen(
["vmware-cmd.bat", target_vm, list_arg, list_arg2],
stdout=PIPE
).communicate()[0]
It might also be worth trying..
p = Popen(
[os.path.join("C:", "Program Files", "VMware", "VMware Server", "vmware-cmd.bat"), ...

You probably don't want to use Pipe
If the output of the subprogram is greater than 64KB it is likely your process will crash.
http://thraxil.org/users/anders/posts/2008/03/13/Subprocess-Hanging-PIPE-is-your-enemy/
Subprocess.Popen has a keyword argument shell, making it as if the shell has been parsing your arguments, setting shell=True should do what you want.

Why are you using r""? I believe that if you remove the "r" from the beginning, it will be treated as a standard string which may contain spaces. Python should then properly quote the string when sending it to the shell.

Here's what I don't like
vmrun_cmd = r"c:/Program Files/VMware/VMware Server/vmware-cmd.bat"
You've got spaces in the name of the command itself -- which is baffling your shell. Hence the "'c:\Program' is not recognized as an internal or external command,
operable program or batch file."
Option 1 -- put your .BAT file somewhere else. Indeed, put all your VMWare somewhere else. Here's the rule: Do Not Use "Program Files" Directory For Anything. It's just wrong.
Option 2 -- quote the vmrun_cmd value
vmrun_cmd = r'"c:/Program Files/VMware/VMware Server/vmware-cmd.bat"'

Subprocess with a variable that contains a whitespace (path)

Just started Python again and now I'm already stuck on the following...
I'm trying to use subprocess.Popen with a variable with a whitespace in it (a Windows path).
Doing a print on the variable the variable seems to work fine. But when using the variable in subprocess.Popen, the variable is cut off by the first whitespace.
Below a part of the script (the variable 'image_file' contains the Windows Path(s) with spaces)
def start_phone(image_file):
cmd = tar_exe + " -tf "+ image_file
print (cmd)
subprocess.Popen(cmd, shell=True)
How can I use subprocess with a variable with whitespaces (path) in it?

You can either put double quotes around each argument with potential white spaces in it:
cmd = f'"{tar_exe}" -tf "{image_file}"'
subprocess.Popen(cmd, shell=True)
or don't use shell=True and instead put arguments in a list:
subprocess.Popen([tar_exe, '-tf', image_file])

If you take a look at the subprocess documentation you'll see that arguments must be provided to subprocess commands in the form of a list, so your code sample should look like
def start_phone(image_file):
subprocess.Popen([tar_exe, "-tf", image_file])

Using a glob to generate arguments with subprocess.run()

I want to use metaflac (https://linux.die.net/man/1/metaflac) command from within a python script.
from subprocess import run
flac_files = "/home/fricadelle/Artist - Album (2008)/*.flac"
run(['metaflac', '--add-replay-gain', flac_files])
I get
The FLAC file could not be opened. Most likely the file does not exist
or is not readable.
if I add shell = True to the run function I'd get:
ERROR: you must specify at least one FLAC file;
metaflac cannot be used as a pipe
So what do I do wrong? Thanks!
PS: of course the command works fine in a shell:
metaflac --add-replay-gain /home/fricadelle/Artist\ -\ Album \(2008\)/*.flac

Unless you specify shell=True (and as a first approximation, you should never specify shell=True), the arguments you provide are passed as is, with no shell expansions, word-splitting or dequoting. So the filename you pass as an argument is precisely /home/fricadelle/Artist - Album (2008)/*.flac, which is not the name of any file. (That's why you don't need to add backslashes before the spaces and parentheses. If you specified shell=True -- and I repeat, you really should avoid that -- then you would need to include backslashes so that the shell doesn't split the name into several different words.)
When you type
flac_files = "/home/fricadelle/Artist - Album (2008)/*.flac unquoted in a shell, the shell will try to expand that to a list of all the files whose names match then pattern, and will then pass that list as separate arguments. Since subprocess.run doesn't do this, you will have to do it yourself, which you would normally do with glob.glob. For example,
from subprocess import run
from glob import glob
flac_files = "/home/fricadelle/Artist - Album (2008)/*.flac"
run(['metaflac', '--add-replay-gain'] + glob(flac_files))
Note: unlike the shell, glob.glob will return an empty list if the pattern matches no files. You really should check for this error rather than invoke metaflac with no filename options.

See the answer here for a better explanation.
Globbing doesn't work the way you're expecting it to here, you need to specify shell=True, but then you'll need to drop the list.
run('metaflac --add-replay-gain ' + flac_files, shell=True)
Should do the trick.

PYTHON: How to pass asterisk (*) as command line argument

I'm trying to make a python script that search words in files.
If I pass txt it will only look in files with .txt extension, but I want to pass * as argument to search in every files.
if sys.argv[4] == "*"
Don't work and if I try
print sys.argv[4]
It print the name of the script
find.py
But not the same way as
print sys.argv[0]
As it will return
./find.py
So, someone already had this problem and, of course, solved it ?

Your shell attaches meaning to * as well. You need to escape it when calling your script to prevent the shell from expanding it:
python find.py \*
sys.argv[0] is the exact name passed used to run the script. That can be a relative path (./find.py, ../bin/find.py) or an absolute path, depending on how it was invoked. Use os.path.abspath() to normalize it.

How to use subprocess when multiple arguments contain spaces?

I'm working on a wrapper script that will exercise a vmware executable, allowing for the automation of virtual machine startup/shutdown/register/deregister actions. I'm trying to use subprocess to handle invoking the executable, but the spaces in the executables path and in parameters of the executable are not being handled correctly by subprocess. Below is a code fragment:
vmrun_cmd = r"c:/Program Files/VMware/VMware Server/vmware-cmd.bat"
def vm_start(target_vm):
list_arg = "start"
list_arg2 = "hard"
if vm_list(target_vm):
p = Popen([vmrun_cmd, target_vm, list_arg, list_arg2], stdout=PIPE).communicate()[0]
print p
else:
vm_register(target_vm)
vm_start(target_vm)
def vm_list2(target_vm):
list_arg = "-l"
p = Popen([vmrun_cmd, list_arg], stdout=PIPE).communicate()[0]
for line in p.split('\n'):
print line
If I call the vm_list2 function, I get the following output:
$ ./vmware_control.py --list
C:\Virtual Machines\QAW2K3Server\Windows Server 2003 Standard Edition.vmx
C:\Virtual Machines\ubunturouter\Ubuntu.vmx
C:\Virtual Machines\vacc\vacc.vmx
C:\Virtual Machines\EdgeAS-4.4.x\Other Linux 2.4.x kernel.vmx
C:\Virtual Machines\UbuntuServer1\Ubuntu.vmx
C:\Virtual Machines\Other Linux 2.4.x kernel\Other Linux 2.4.x kernel.vmx
C:\Virtual Machines\QAClient\Windows XP Professional.vmx
If I call the vm_start function, which requires a path-to-vm parameter, I get the following output:
$ ./vmware_control.py --start "C:\Virtual Machines\ubunturouter\Ubuntu.vmx"
'c:\Program' is not recognized as an internal or external command,
operable program or batch file.
Apparently, the presence of a second parameter with embedded spaces is altering the way that subprocess is interpreting the first parameter. Any suggestions on how to resolve this?
python2.5.2/cygwin/winxp

If you have spaces in the path, the easiest way I've found to get them interpreted properly is this.
subprocess.call('""' + path + '""')
I don't know why exactly it needs double double quotes, but that is what works.

I believe that list2cmdline(), which is doing the processing of your list args, splits any string arg on whitespace unless the string contains double quotes. So I would expect
vmrun_cmd = r'"c:/Program Files/VMware/VMware Server/vmware-cmd.bat"'
to be what you want.
You'll also likely want to surround the other arguments (like target_vm) in double quotes on the assumption that they, too, each represent a distinct arg to present to the command line. Something like
r'"%s"' % target_vm
(for example) should suit.
See the list2cmdline documentation

'c:\Program' is not recognized as an internal or external command, operable program or batch file.
To get this message, you are either:
Using shell=True:
vmrun_cmd = r"c:\Program Files\VMware\VMware Server\vmware-cmd.bat"
subprocess.Popen(vmrun_cmd, shell=True)
Changing vmrun_cmd on other part of your code
Getting this error from something inside vmware-cmd.bat
Things to try:
Open a python prompt, run the following command:
subprocess.Popen([r"c:\Program Files\VMware\VMware Server\vmware-cmd.bat"])
If that works, then quoting issues are out of the question. If not, you've isolated the problem.

In Python on MS Windows, the subprocess.Popen class uses the CreateProcess API to started the process. CreateProcess takes a string rather than something like an array of arguments. Python uses subprocess.list2cmdline to convert the list of args to a string for CreateProcess.
If I were you, I'd see what subprocess.list2cmdline(args) returns (where args is the first argument of Popen). It would be interesting to see if it is putting quotes around the first argument.
Of course, this explanation might not apply in a Cygwin environment.
Having said all this, I don't have MS Windows.

One problem is that if the command is surrounded with quotes and doesn't have spaces, that could also confuse the shell.
So I do this:
if ' ' in raw_cmd:
fmt = '"%s"'
else:
fmt = '%s'
cmd = fmt % raw_cmd

That was quite a hard problem for the last three ours....nothing stated so far did work, neither using r"" or Popen with a list and so on. What did work in the end was a combination of format string and r"". So my solution is this:
subprocess.Popen("{0} -f {1}".format(pathToExe, r'"%s"' % pathToVideoFileOrDir))
where both variables pathToExe and pathToVideoFileOrDir have whitespaces in their path. Using \" within the formatted string did not work and resulted in the same error that the first path is not detected any longer correctly.

Possibly stupid suggestion, but perhaps try the following, to remove subprocess + spaces from the equation:
import os
from subprocess Popen, PIPE
os.chdir(
os.path.join("C:", "Program Files", "VMware", "VMware Server")
)
p = Popen(
["vmware-cmd.bat", target_vm, list_arg, list_arg2],
stdout=PIPE
).communicate()[0]
It might also be worth trying..
p = Popen(
[os.path.join("C:", "Program Files", "VMware", "VMware Server", "vmware-cmd.bat"), ...

You probably don't want to use Pipe
If the output of the subprogram is greater than 64KB it is likely your process will crash.
http://thraxil.org/users/anders/posts/2008/03/13/Subprocess-Hanging-PIPE-is-your-enemy/
Subprocess.Popen has a keyword argument shell, making it as if the shell has been parsing your arguments, setting shell=True should do what you want.

Why are you using r""? I believe that if you remove the "r" from the beginning, it will be treated as a standard string which may contain spaces. Python should then properly quote the string when sending it to the shell.

Here's what I don't like
vmrun_cmd = r"c:/Program Files/VMware/VMware Server/vmware-cmd.bat"
You've got spaces in the name of the command itself -- which is baffling your shell. Hence the "'c:\Program' is not recognized as an internal or external command,
operable program or batch file."
Option 1 -- put your .BAT file somewhere else. Indeed, put all your VMWare somewhere else. Here's the rule: Do Not Use "Program Files" Directory For Anything. It's just wrong.
Option 2 -- quote the vmrun_cmd value
vmrun_cmd = r'"c:/Program Files/VMware/VMware Server/vmware-cmd.bat"'