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Cannot call ubuntu 'ulimit' from python subprocess without using shell option


Cannot call ubuntu 'ulimit' from python subprocess without using shell option

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Cannot call ubuntu 'ulimit' from python subprocess without using shell option
Tag : python , By : Frank
Date : January 12 2021, 07:00 PM

To fix this issue ulimit is a wrapper around a system call to limit the resources of the current process. Because it acts on the current process, it needs to be called on the current process or it has no effect.
For this reason, the shell implements it as a built-in, so there is no such binary.

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Python: ulimit and nice for subprocess.call / subprocess.Popen?


Tag : python , By : MK.
Date : March 29 2020, 07:55 AM
should help you out You can set limits for subprocesses with the ulimit and nice shell commands like this:
import subprocess
subprocess.Popen('ulimit -t 60; nice -n 15 cpuhog', shell=True)

How to do this Python subprocess call without using shell=True?


Tag : python , By : Mare Astra
Date : March 29 2020, 07:55 AM
Hope this helps I believe that other guy is spot on (haven't tried it myself though). However if you ever find yourself having similar issues again shlex.split(s) might be helpful. It takes the string 's' and splits it "using shell-like syntax".
In [3]: shlex.split(s)
Out[3]: ['bash', '-c', 'shred -n 5 -uz /tmp/{*.txt,*.pdf,*.doc}']

python subprocess call ubuntu 14.04


Tag : java , By : Eran Yahav
Date : March 29 2020, 07:55 AM
I wish this help you Just as @MarkB mentioned in his comment this is not really a Python error message, but instead is Python simply explaining what command it ran to execute a Java program, and its return value, and that something went wrong during its execution.
This message shows that the Java program returned with a value of 1. Usually, a program exiting with a non-zero status means that something went wrong. You will need to look up what the value of 1 for this application means (if anything).

Subprocess.Popen vs .call: What is the correct way to call a C-executable from shell script using python where all 6 job


Tag : python-3.x , By : Frank
Date : March 29 2020, 07:55 AM
I wish did fix the issue. You tried to simulate multiprocessing with subprocess.Popen() which does not work like you want: the output is blocked after a while unless you consume it, for instance with communicate() (but this is blocking) or by reading the output, but with 6 concurrent handles in a loop, you are bound to get deadlocks.
The best way is run the subprocess.call lines in separate threads.
import threading,time

lock=threading.Lock()
def func1(a,b,c):
    lock.acquire()
    print(a,b,c)
    lock.release()
    time.sleep(10)

tl=[]
t = threading.Thread(target=func1,args=[1,2,3])
t.start()
tl.append(t)
t=threading.Thread(target=func1,args=[4,5,6])
t.start()
tl.append(t)

# wait for all threads to complete (if you want to wait, else
# you can skip this loop)

for t in tl:
    t.join()
import threading,time,subprocess

lock=threading.Lock()

def func1(runrescompare,rescompare_dir):
    resrun_proc = subprocess.check_output(runrescompare, shell=True, cwd=rescompare_dir, stderr=subprocess.PIPE, universal_newlines=True)
    lock.acquire()
    print(resrun_proc)
    lock.release()

tl=[]
t=threading.Thread(target=func1,args=["ls","C:/"])
t.start()
tl.append(t)
t=threading.Thread(target=func1,args=["ls","D:/"])
t.start()
tl.append(t)

# wait for all threads to complete (if you want to wait, else
# you can skip this loop)

for t in tl:
    t.join()

Inexplicable shell command un-escaping behavior in Python's os.system and subprocess.check_output on Ubuntu 18.04


Tag : python , By : Liy
Date : March 29 2020, 07:55 AM
With these it helps While I can agree that the behavior is strange, it is not inexplicable. There is a reason for the behavior, which has nothing to do with Python or subprocess. Exactly the same behavior is seen in a C program, using the system call to the OS (Linux) as with your Python program.
The reason has to do with your shell, though, but not exactly with bash. The reason is rather that when calling os.system() or the subprocess.Popen() family (including subprocess.check_output()) with shell=True. The documentation states that "On POSIX with shell=True, the shell defaults to /bin/sh." Thus, the shell that invokes your echo command is not bash even if that is your default shell and the shell from which you are running your script/starting Python.
$ ll /bin/sh
lrwxrwxrwx 1 root root 4 sep 23 12:53 /bin/sh -> dash*
#!/usr/bin/env python3
import sys
import subprocess
import shlex

def get_command(n):
    return "echo 'Should be {} backslahes: {}'".format(n, "\\"*n)

print("")
print("Using subprocess.check_output:")
print("")
for n in range(1, 5):

    # Direct invocation:
    cmd = get_command(n)
    sys.stdout.write(subprocess.check_output(shlex.split(cmd)).decode())

    # Controlling invocation shell:
    bash_cmd = ['/bin/bash', '-c'] + [cmd]
    sys.stdout.write(subprocess.check_output(bash_cmd).decode())

    # Using shell=True but point to /bin/echo
    echo_cmd = '/bin/' + cmd
    sys.stdout.write(subprocess.check_output(echo_cmd, shell=True).decode())
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