On Monday, October 28, 2013 6:53:09 PM UTC+11, Alex Kocharin wrote:
>
>
> var fs = require('fs')
>   , childProcess = require('child_process')
>   , stream = fs.createWriteStream('/tmp/output.log')
>   , child = childProcess.spawn('bash', ['-c', 'echo STDOUT; echo STDERR 
> >&2; echo DONE;'])
>
> child.stderr.pipe(stream)
> child.stdout.pipe(stream)
>
>
Thanks for the suggestion, although unfortunately I don't think that would 
work for my real scenario. I'm spawning long running tasks, which may 
outlive the process that spawned them (which is why I'm directing the 
output into a file). Presumably these streams would all stop working / die 
when my parent process ends.
 

> On Sunday, October 27, 2013 5:19:50 AM UTC+4, Tim Cuthbertson wrote:
>>
>> I want to spawn a child process and have *all* output go to the same file.
>>
>> I tried this:
>>
>>     var fs = require('fs'), childProcess = require('child_process');
>>     var output = fs.openSync('/tmp/output.log', 'w');
>>     childProcess.spawn('bash', ['-c', 'echo STOUT; echo STDERR >&2; echo 
>> DONE;'], {stdio: ['ignore', output, output]});
>>
>> Which presumably provides the same FD for both stdout and stderr in the 
>> child process. Looking at the file produced, I get:
>>
>>     STOUT
>>     DONE
>>
>> stderr is apparently being ignored. Looking at the child_process docs, 
>> there's an example that opens *two* versions of the same file (in append 
>> mode), and uses this for stdout / stderr. Since I want to truncate the file 
>> (not append), I tried:
>>
>>     var fs = require('fs'), childProcess = require('child_process');
>>     var output = fs.openSync('/tmp/output.log', 'w');
>>     var output2 = fs.openSync('/tmp/output.log', 'a');
>>     childProcess.spawn('bash', ['-c', 'echo STOUT; echo STDERR >&2; echo 
>> DONE;'], {stdio: ['ignore', output, output2]});
>>
>> But it looks like the write stream is not playing nice and is overwriting 
>> the other stream's output, as I get:
>>
>>     STOUT
>>     DONE
>>     R
>>
>> So I guess I have to explicitly truncate the file first, then open two 
>> append streams to it?
>>
>>     fs.closeSync(fs.openSync('/tmp/output.log', 'w'));
>>     var output = fs.openSync('/tmp/output.log', 'a');
>>     var output2 = fs.openSync('/tmp/output.log', 'a');
>>     childProcess.spawn('bash', ['-c', 'echo STOUT; echo STDERR >&2; echo 
>> DONE;'], {stdio: ['ignore', output, output2]});
>>
>> Which finally gives me the desired output:
>>
>>     STOUT
>>     STDERR
>>     DONE
>>
>> For kicks, I also tried reusing a single append stream:
>>
>>     fs.closeSync(fs.openSync('/tmp/output.log', 'w'));
>>     var output = fs.openSync('/tmp/output.log', 'a');
>>     childProcess.spawn('bash', ['-c', 'echo STOUT; echo STDERR >&2; echo 
>> DONE;'], {stdio: ['ignore', output, output]});
>>
>> But that gave me the same results as my initial attempt (no stderr at 
>> all).
>>
>>
>> I guess I've discovered *how* to do this (my third attempt), but that 
>> seems pretty ugly. Can anyone suggest a better way? In python, we can 
>> explicitly redirect the child's stderr to stdout:
>>
>>     output = open('/tmp/output.log', 'w');
>>     subprocess.Popen(['bash', '-c', 'echo STOUT; echo STDERR >&2; echo 
>> DONE;'], stderr = subprocess.STDOUT)
>>
>> Is there anything similar in nodejs, or am I stuck with explicitly 
>> truncating the file, followed by opening two append-mode descriptors?
>>
>> Cheers,
>>  - Tim.
>>
>

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