Making a system call that returns the stdout output as a string

Question

Perl and PHP do this with backticks. For example,

$output = `ls`;

Returns a directory listing. A similar function, system(\"foo\")

Answer 1

Haskell:

import Control.Exception
import System.IO
import System.Process
main = bracket (runInteractiveCommand "ls") close $ \(_, hOut, _, _) -> do
    output <- hGetContents hOut
    putStr output
  where close (hIn, hOut, hErr, pid) =
          mapM_ hClose [hIn, hOut, hErr] >> waitForProcess pid

With MissingH installed:

import System.Cmd.Utils
main = do
    (pid, output) <- pipeFrom "ls" []
    putStr output
    forceSuccess pid

This is an easy operation in "glue" languages like Perl and Ruby, but Haskell isn't.

Answer 2

J:

output=:2!:0'ls'

Answer 3

Icon/Unicon:

stream := open("ls", "p")
while line := read(stream) do { 
    # stuff
}

The docs call this a pipe. One of the good things is that it makes the output look like you're just reading a file. It also means you can write to the app's stdin, if you must.

Answer 4

Years ago I wrote a plugin for jEdit that interfaced to a native application. This is what I used to get the streams off the running executable. Only thing left to do is while((String s = stdout.readLine())!=null){...}:

/* File:    IOControl.java
 *
 * created: 10 July 2003
 * author:  dsm
 */
package org.jpop.io;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintStream;

/**
 *  Controls the I/O for a process. When using the std[in|out|err] streams, they must all be put on
 *  different threads to avoid blocking!
 *
 * @author     dsm
 * @version    1.5
 */
public class IOControl extends Object {
    private Process process;
    private BufferedReader stdout;
    private BufferedReader stderr;
    private PrintStream stdin;

    /**
     *  Constructor for the IOControl object
     *
     * @param  process  The process to control I/O for
     */
    public IOControl(Process process) {
        this.process = process;
        this.stdin = new PrintStream(process.getOutputStream());
        this.stdout = new BufferedReader(new InputStreamReader(process.getInputStream()));
        this.stderr = new BufferedReader(new InputStreamReader(process.getErrorStream()));
    }

    /**
     *  Gets the stdin attribute of the IOControl object
     *
     * @return    The stdin value
     */
    public PrintStream getStdin() {
        return this.stdin;
    }

    /**
     *  Gets the stdout attribute of the IOControl object
     *
     * @return    The stdout value
     */
    public BufferedReader getStdout() {
        return this.stdout;
    }

    /**
     *  Gets the stderr attribute of the IOControl object
     *
     * @return    The stderr value
     */
    public BufferedReader getStderr() {
        return this.stderr;
    }

    /**
     *  Gets the process attribute of the IOControl object. To monitor the process (as opposed to
     *  just letting it run by itself) its necessary to create a thread like this: <pre>
     *. IOControl ioc;
     *.
     *. new Thread(){
     *.     public void run(){
     *.         while(true){    // only necessary if you want the process to respawn
     *.             try{
     *.                 ioc = new IOControl(Runtime.getRuntime().exec("procname"));
     *.                 // add some code to handle the IO streams
     *.                 ioc.getProcess().waitFor();
     *.             }catch(InterruptedException ie){
     *.                 // deal with exception
     *.             }catch(IOException ioe){
     *.                 // deal with exception
     *.             }
     *.
     *.             // a break condition can be included here to terminate the loop
     *.         }               // only necessary if you want the process to respawn
     *.     }
     *. }.start();
     *  </pre>
     *
     * @return    The process value
     */
    public Process getProcess() {
        return this.process;
    }
}

Answer 5

Why there is still no c# guy here :)

This is how to do it in C#. The built-in way.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Diagnostics;

namespace TestConsole
{
    class Program
    {
        static void Main(string[] args)
        {
            Process p = new Process();

            p.StartInfo.UseShellExecute = false;
            p.StartInfo.CreateNoWindow = true;
            p.StartInfo.RedirectStandardOutput = true;
            p.StartInfo.RedirectStandardError = true;
            p.StartInfo.FileName = "cmd";
            p.StartInfo.Arguments = "/c dir";
            p.Start();

            string res = p.StandardOutput.ReadToEnd();
            Console.WriteLine(res);
        }

    }
}

Answer 6

C# 3.0, less verbose than this one:

using System;
using System.Diagnostics;

class Program
{
    static void Main()
    {
        var info = new ProcessStartInfo("cmd", "/c dir") { UseShellExecute = false, RedirectStandardOutput = true };
        Console.WriteLine(Process.Start(info).StandardOutput.ReadToEnd());
    }
}

Caveat: Production code should properly dispose the Process object...