Re: std.process spawnShell/pipeShell dont capture output of the shell
I'm sorry for the "necro-bumping", but I've not found the
solution for this yet, and i'm getting a different error message
at the execution moment:
"std.stdio.StdioException@/build/ldc/src/ldc/runtime/phobos/std/stdio.d(4066): Bad
file descriptor"
This is my testing code:
module dd_test;
import std.stdio, core.stdc.stdlib;
import std.process, std.string;
int main(string[] args) {
string command_find = "(find /usr/share/icons/ -name a*) ";
//string command_dd = "(dd if=/dev/urandom | pv -ptrbef -i 2
-s 2339876653 | dd of=/dev/null) 2>&1";
auto p = pipeShell(command_find, Redirect.all);
foreach(str; p.stdin.byLine){
writefln("IN: %s", str);
}
foreach(str; p.stdout.byLine){
writefln("OUT: %s", str);
}
foreach(str; p.stderr.byLine){
writefln("Error: %s", str);
}
return 0;
}
Re: std.process spawnShell/pipeShell dont capture output of the shell
On Thursday, 10 October 2013 at 01:24:03 UTC, Jesse Phillips wrote: On Wednesday, 9 October 2013 at 14:54:32 UTC, Colin Grogan wrote: is blocking. However, its not meant to be blocking is it not? That new /bin/bash process is meant to run in parallel to the main process? I'm not sure exactly the implementation. But if you're asking to run bash and then print its output, wouldn't it have to block because the output won't be complete until the program has finished running? And since bash will run until you exit the program will block for ever. From what I understood in the documentation, I didnt think it should block. I understood it's meant to run the child process in the background and you communicate with it via the stdin/stdout/stderr pipes. But, from experimentation I can see that running /bin/bash does appear to block. Ill have to do some more digging when I have the time...
std.process spawnShell/pipeShell dont capture output of the shell
Hi folks,
Is there anyway to make std.process.spawnShell or
std.process.pipeShell capture the output of the shell interface?
For example, the code:
import std.stdio;
import std.process;
void main(string[] args){
writefln(Executing: %s, args[1]);
auto processPipes = pipeShell(args[1], Redirect.all);
foreach(str; processPipes.stdout.byLine){
writefln(STDOUT: %s,str);
}
foreach(str; processPipes.stderr.byLine){
writefln(STDERR: %s,str);
}
}
is compiled to an executable called mess.
When executed with the following produces:
~/test$ ./mess ls
Executing: ls
STDOUT: mess
STDOUT: text.txt
Thats all fine, however, I'd expect it to print another ~/test$
at the end, as if its an interactive shell waiting for input.
Is that an incorrect assumption to make or am I doing something
wrong that is stopping that from happening?
Also, If I run /bin/bash through program, it hangs.
~/test$ ./mess /bin/bash
Executing: /bin/bash
Thanks,
Colin
Re: std.process spawnShell/pipeShell dont capture output of the shell
On Wednesday, 9 October 2013 at 11:22:26 UTC, Colin Grogan wrote: ~/test$ ./mess ls Executing: ls STDOUT: mess STDOUT: text.txt Thats all fine, however, I'd expect it to print another ~/test$ at the end, as if its an interactive shell waiting for input. It is extracting the output of the program, 99.9% of the time this is what everyone wants and including the shells behavior would be terrible. If you run bash, then it is going to be waiting for stdin so you'd need to pass it ls and exit to get the output and quit. I'm not sure this would get you want you wanted.
Re: std.process spawnShell/pipeShell dont capture output of the shell
On Wednesday, 9 October 2013 at 14:27:30 UTC, Jesse Phillips
wrote:
On Wednesday, 9 October 2013 at 11:22:26 UTC, Colin Grogan
wrote:
~/test$ ./mess ls
Executing: ls
STDOUT: mess
STDOUT: text.txt
Thats all fine, however, I'd expect it to print another
~/test$ at the end, as if its an interactive shell waiting
for input.
It is extracting the output of the program, 99.9% of the time
this is what everyone wants and including the shells behavior
would be terrible.
But its exactly what I want ;)
If you run bash, then it is going to be waiting for stdin so
you'd need to pass it ls and exit to get the output and
quit. I'm not sure this would get you want you wanted.
That would imply that the call to
pipeShell(/bin/bash, Redirect.all);
is blocking. However, its not meant to be blocking is it not?
That new /bin/bash process is meant to run in parallel to the
main process?
If I use the program to execute the following script:
#!/bin/bash
for i in {1..100}
do
for j in {1..100}
do
echo $i, $j
done
sleep 15
done
and monitor the output that is printed, I can see that it is
indeed sleeping for 15 seconds, so the pipeShell call is
definitely non-blocking.
I assumed it should behave the same way for /bin/bash. It is just
another process after all, surely pipeShell knows nothing more
about /bin/bash that it does about my silly script above?
Re: std.process spawnShell/pipeShell dont capture output of the shell
On Wednesday, 9 October 2013 at 14:54:32 UTC, Colin Grogan wrote: is blocking. However, its not meant to be blocking is it not? That new /bin/bash process is meant to run in parallel to the main process? I'm not sure exactly the implementation. But if you're asking to run bash and then print its output, wouldn't it have to block because the output won't be complete until the program has finished running? And since bash will run until you exit the program will block for ever.
