Getting subprocess.Popen stdout when running by cron

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I want to get a service status and if it's not up, to send the status (stdout) in email.

This script is scheduled to run every hour by cron.

When running manually, the following works fine:

def is_service_running(name):
with open(os.devnull, 'wb') as hide_output:
 proc = subprocess.Popen(['service', name, 'status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)
 output = proc.stdout.read()
 exit_code = proc.wait()
 return exit_code == 0, output

But when running by cron. output is empty.

How can I capture stdout when running by cron?
Thank you

asked Jun 14 at 15:14

SagiLow

1537

Is it because /usr/sbin is in your interactive environment but not cronÃ¢Â€Â™s?
â€“Â Jeff Schaller
Jun 14 at 17:58

Hi @JeffSchaller, no idea.. that's why i'm asking :)
â€“Â SagiLow
Jun 15 at 9:14

I think you should see if that popen call is successful. Maybe also use the full path to the service command.
â€“Â Jeff Schaller
Jun 15 at 9:21

@JeffSchaller You were close, see the answer
â€“Â SagiLow
Jun 15 at 9:22

add a commentÂ |Â

up vote
0
down vote

favorite

I want to get a service status and if it's not up, to send the status (stdout) in email.

This script is scheduled to run every hour by cron.

When running manually, the following works fine:

def is_service_running(name):
with open(os.devnull, 'wb') as hide_output:
 proc = subprocess.Popen(['service', name, 'status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)
 output = proc.stdout.read()
 exit_code = proc.wait()
 return exit_code == 0, output

But when running by cron. output is empty.

How can I capture stdout when running by cron?
Thank you

asked Jun 14 at 15:14

SagiLow

1537

Is it because /usr/sbin is in your interactive environment but not cronÃ¢Â€Â™s?
â€“Â Jeff Schaller
Jun 14 at 17:58

Hi @JeffSchaller, no idea.. that's why i'm asking :)
â€“Â SagiLow
Jun 15 at 9:14

I think you should see if that popen call is successful. Maybe also use the full path to the service command.
â€“Â Jeff Schaller
Jun 15 at 9:21

@JeffSchaller You were close, see the answer
â€“Â SagiLow
Jun 15 at 9:22

add a commentÂ |Â

up vote
0
down vote

favorite

I want to get a service status and if it's not up, to send the status (stdout) in email.

This script is scheduled to run every hour by cron.

When running manually, the following works fine:

def is_service_running(name):
with open(os.devnull, 'wb') as hide_output:
 proc = subprocess.Popen(['service', name, 'status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)
 output = proc.stdout.read()
 exit_code = proc.wait()
 return exit_code == 0, output

But when running by cron. output is empty.

How can I capture stdout when running by cron?
Thank you

asked Jun 14 at 15:14

SagiLow

1537

I want to get a service status and if it's not up, to send the status (stdout) in email.

This script is scheduled to run every hour by cron.

When running manually, the following works fine:

def is_service_running(name):
with open(os.devnull, 'wb') as hide_output:
 proc = subprocess.Popen(['service', name, 'status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)
 output = proc.stdout.read()
 exit_code = proc.wait()
 return exit_code == 0, output

But when running by cron. output is empty.

How can I capture stdout when running by cron?
Thank you

asked Jun 14 at 15:14

SagiLow

1537

asked Jun 14 at 15:14

SagiLow

1537

asked Jun 14 at 15:14

SagiLow

1537

asked Jun 14 at 15:14

SagiLow

1537

Is it because /usr/sbin is in your interactive environment but not cronÃ¢Â€Â™s?
â€“Â Jeff Schaller
Jun 14 at 17:58

Hi @JeffSchaller, no idea.. that's why i'm asking :)
â€“Â SagiLow
Jun 15 at 9:14

I think you should see if that popen call is successful. Maybe also use the full path to the service command.
â€“Â Jeff Schaller
Jun 15 at 9:21

@JeffSchaller You were close, see the answer
â€“Â SagiLow
Jun 15 at 9:22

add a commentÂ |Â

Is it because /usr/sbin is in your interactive environment but not cronÃ¢Â€Â™s?
â€“Â Jeff Schaller
Jun 14 at 17:58

Hi @JeffSchaller, no idea.. that's why i'm asking :)
â€“Â SagiLow
Jun 15 at 9:14

I think you should see if that popen call is successful. Maybe also use the full path to the service command.
â€“Â Jeff Schaller
Jun 15 at 9:21

@JeffSchaller You were close, see the answer
â€“Â SagiLow
Jun 15 at 9:22

Is it because /usr/sbin is in your interactive environment but not cronÃ¢Â€Â™s?
â€“Â Jeff Schaller
Jun 14 at 17:58

Hi @JeffSchaller, no idea.. that's why i'm asking :)
â€“Â SagiLow
Jun 15 at 9:14

I think you should see if that popen call is successful. Maybe also use the full path to the service command.
â€“Â Jeff Schaller
Jun 15 at 9:21

@JeffSchaller You were close, see the answer
â€“Â SagiLow
Jun 15 at 9:22

add a commentÂ |Â

1 Answer
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up vote
0
down vote

The problem wasn't cron but shell=True.

Apparently, when using shell=True, popen expects single string and not a list.

So when I updated my call to:

proc = subprocess.Popen(['service ' + name + ' status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)

everything worked.

answered Jun 15 at 9:22

SagiLow

1537

add a commentÂ |Â

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1 Answer
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active

oldest

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1 Answer
1

active

oldest

votes

up vote
0
down vote

The problem wasn't cron but shell=True.

Apparently, when using shell=True, popen expects single string and not a list.

So when I updated my call to:

proc = subprocess.Popen(['service ' + name + ' status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)

everything worked.

answered Jun 15 at 9:22

SagiLow

1537

add a commentÂ |Â

up vote
0
down vote

The problem wasn't cron but shell=True.

Apparently, when using shell=True, popen expects single string and not a list.

So when I updated my call to:

proc = subprocess.Popen(['service ' + name + ' status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)

everything worked.

answered Jun 15 at 9:22

SagiLow

1537

add a commentÂ |Â

up vote
0
down vote

The problem wasn't cron but shell=True.

Apparently, when using shell=True, popen expects single string and not a list.

So when I updated my call to:

proc = subprocess.Popen(['service ' + name + ' status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)

everything worked.

answered Jun 15 at 9:22

SagiLow

1537

The problem wasn't cron but shell=True.

Apparently, when using shell=True, popen expects single string and not a list.

So when I updated my call to:

proc = subprocess.Popen(['service ' + name + ' status'], stdout=subprocess.PIPE, stderr=subprocess.PIPE, shell=True)

everything worked.

answered Jun 15 at 9:22

SagiLow

1537

answered Jun 15 at 9:22

SagiLow

1537

answered Jun 15 at 9:22

SagiLow

1537

answered Jun 15 at 9:22

SagiLow

1537

add a commentÂ |Â

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