try running:

    su -c &quot;Your command right here&quot; -s /bin/sh username

This will run the command as username given that you have permissions to sudo as that user.

Call `visudo` and add this:

    user1 ALL=(user2) NOPASSWD: /home/user2/bin/test.sh

The **command paths must be absolute**! Then call `sudo -u user2 /home/user2/bin/test.sh` from a `user1` shell. Done.


    `su -c &quot;Your command right here&quot; -s /bin/sh username`

The above command is correct, but on Red Hat if selinux is enforcing it will not allow cron to execute scripts as another user. example; 
`execl: couldn&#39;t exec /bin/sh
execl: Permission denied`

I had to install setroubleshoot and setools and run the following to allow it:

```
yum install setroubleshoot setools
sealert -a /var/log/audit/audit.log
grep crond /var/log/audit/audit.log | audit2allow -M mypol
semodule -i mypol.p
```

I&#39;m fairly new to Javascript development so this might be a real newbie question.

I&#39;ve got a [tag:sencha-touch] application riddled with `console.log();` for debugging purposes.

I&#39;ve got [tag:chirpy] doing all of my build time combining. It outputs a `app.debug.js` for debugging as well as a `app.min.js` for production

Now I could go through all of my code files looking for `console.log();` and delete it manually when I&#39;m ready to go to production, but I&#39;m wondering if there&#39;s a way to override the method.

Basically, whenever the `console.log();` method is called, DO NOTHING.

That way, I can put the override code file in my production config, and NOT in my debug config.

Is this possible?

Override console.log(); for production

Is there an easy way to set conditional breakpoints in Visual Studio?

If I want to hit a breakpoint only when the value of a variable becomes something, how can I do it?


How to set conditional breakpoints in Visual Studio?

I have script.sh that must be run as user2.  However, this script can only be run under user1 in my application.

I would like the following command to run:

    su user2 -C script.sh

but be able to run without password.

I also want this to be very restrictive, as in user1 can only run script.sh under user2 and nothing else.

I&#39;ve tried doing this with sudoers file and just got endlessly confused after hours of trying.

If somebody can provide an explicit example of how this can be accomplished (instead of something generic like use sudoers), it would be greatly appreciated.


How to run script as another user without password?

I have script.sh that must be run as user2. However, this script can only be run under user1 in my application.
I would like the following command to run:
<pre><code class="hljs">su user2 -C script.sh
</code></pre>
but be able to run without password.
I also want this to be very restrictive, as in user1 can only run script.sh under user2 and nothing else.
I've tried doing this with sudoers file and just got endlessly confused after hours of trying.
If somebody can provide an explicit example of how this can be accomplished (instead of something generic like use sudoers), it would be greatly appreciated.

I know this is not so much a programming question but it is relevant.

I work on a fairly &lt;a href=&quot;http://chromium.org&quot;&gt;large cross platform project&lt;/a&gt;. On Windows I use VC++ 2008. On Linux I use gcc. There are around 40k files in the project. Windows is 10x to 40x slower than Linux at compiling and linking the same project. How can I fix that?

A single change incremental build 20 seconds on Linux and &gt; 3 mins on Windows. Why? I can even install the &#39;gold&#39; linker in Linux and get that time down to 7 seconds.

Similarly git is 10x to 40x faster on Linux than Windows. 

In the git case it&#39;s possible git is not using Windows in the optimal way but VC++? You&#39;d think Microsoft would want to make their own developers as productive as possible and faster compilation would go a long way toward that. Maybe they are trying to encourage developers into C#?

As simple test, find a folder with lots of subfolders and do a simple

    dir /s &gt; c:\list.txt

on Windows. Do it twice and time the second run so it runs from the cache. Copy the files to Linux and do the equivalent 2 runs and time the second run.

    ls -R &gt; /tmp/list.txt

I have 2 workstations with the exact same specs. HP Z600s with 12gig of ram, 8 cores at 3.0ghz. On a folder with ~400k files Windows takes 40seconds, Linux takes &lt; 1 second.

Is there a registry setting I can set to speed up Windows? What gives?

---

A few slightly relevant links, relevant to compile times, not necessarily i/o.

* Apparently [there&#39;s an issue in Windows 10 (not in Windows 7) that closing a process holds a global lock](https://randomascii.wordpress.com/2017/07/09/24-core-cpu-and-i-cant-move-my-mouse/). When compiling with multiple cores and therefore multiple processes this issue hits.

* [The `/analyse` option can adversely affect perf because it loads a web browser](https://randomascii.wordpress.com/2014/03/31/you-got-your-web-browser-in-my-compiler/). (Not relevant here but good to know)


How do I get Windows to go as fast as Linux for compiling C++?

Is there any easy way to have a system-wide mutex in Python on Linux? By &quot;system-wide&quot;, I mean the mutex will be used by a group of Python _processes_; this is in contrast to a traditional mutex, which is used by a group of _threads_ within the same process.

EDIT: I&#39;m not sure Python&#39;s [`multiprocessing`](https://docs.python.org/2/library/multiprocessing.html) package is what I need. For example, I can execute the following in two different interpreters:

    from multiprocessing import Lock
    L = Lock()
    L.acquire()

When I execute these commands simultaneously in two separate interpreters, I want one of them to hang. Instead, neither hangs; it appears they aren&#39;t acquiring the same mutex.


System-wide mutex in Python on Linux

The Linux specific `backtrace()` and `backtrace_symbols()` allows you to produce a call trace of the program. However, it only prints function addresses, not their names for my program. How can I make them print the function names as well ? I&#39;ve tried compiling the program with `-g` as well as `-ggdb`. The test case below just prints this:

&lt;pre&gt;

    BACKTRACE ------------
    ./a.out() [0x8048616]
    ./a.out() [0x8048623]
    /lib/libc.so.6(__libc_start_main+0xf3) [0x4a937413]
    ./a.out() [0x8048421]
    ----------------------
    &lt;/pre&gt;
I&#39;d want the first 2 items to also show the function names, `foo` and `main`
    
Code:

    #include &lt;execinfo.h&gt;
    #include &lt;string.h&gt;
    #include &lt;errno.h&gt;
    #include &lt;unistd.h&gt;
    #include &lt;stdlib.h&gt;
    
    static void full_write(int fd, const char *buf, size_t len)
    {
            while (len &gt; 0) {
                    ssize_t ret = write(fd, buf, len);
    
                    if ((ret == -1) &amp;&amp; (errno != EINTR))
                            break;
    
                    buf += (size_t) ret;
                    len -= (size_t) ret;
            }
    }
    
    void print_backtrace(void)
    {
            static const char start[] = &quot;BACKTRACE ------------\n&quot;;
            static const char end[] = &quot;----------------------\n&quot;;
    
            void *bt[1024];
            int bt_size;
            char **bt_syms;
            int i;
    
            bt_size = backtrace(bt, 1024);
            bt_syms = backtrace_symbols(bt, bt_size);
            full_write(STDERR_FILENO, start, strlen(start));
            for (i = 1; i &lt; bt_size; i++) {
                    size_t len = strlen(bt_syms[i]);
                    full_write(STDERR_FILENO, bt_syms[i], len);
                    full_write(STDERR_FILENO, &quot;\n&quot;, 1);
            }
            full_write(STDERR_FILENO, end, strlen(end));
        free(bt_syms);
    }
    void foo()
    {
        print_backtrace();
    }
    
    int main()
    {
        foo();
        return 0;
    }



How to make backtrace()/backtrace_symbols() print the function names?

I have an application that runs on an embedded Linux device and every now and then changes are made to the software and occasionally also to the root file system or even the installed kernel.

In the current update system the contents of the old application directory are simply deleted and the new files are copied over it. When changes to the root file system have been made the new files are delivered as part of the update and simply copied over the old ones.

Now, there are several problems with the current approach and I am looking for ways to improve the situation:

* The root file system of the target that is used to create file system images is not versioned (I don&#39;t think we even have the original rootfs).
* The rootfs files that go into the update are manually selected (instead of a diff)
* The update continually grows and that becomes a pita. There is now a split between update/upgrade where the upgrade contains larger rootfs changes.
* I have the impression that the consistency checks in an update are rather fragile if at all implemented.

Requirements are:

* The application update package should not be too large and it must also be able to change the root file system in the case modifications have been made.
* An upgrade can be much larger and only contains the stuff that goes into the root file system (like new libraries, kernel, etc.). An update can require an upgrade to have been installed.&lt;br&gt;
Could the upgrade contain the whole root file system and simply do a `dd` on the flash drive of the target?
* Creating the update/upgrade packages should be as automatic as possible.


I absolutely need some way to do versioning of the root file system. This has to be done in a way, that I can compute some sort of `diff` from it which can be used to update the rootfs of the target device. 

I already looked into Subversion since we use that for our source code but that is inappropriate for the Linux root file system (file permissions, special files, etc.).

I&#39;ve now created some shell scripts that can give me something similar to an `svn diff` but I would really like to know if there already exists a working and tested solution for this.

Using such `diff`&#39;s I guess an Upgrade would then simply become a package that contains incremental updates based on a known root file system state.

What are your thoughts and ideas on this? How would you implement such a system? I prefer a simple solution that can be implemented in not too much time.



Implementing an update/upgrade system for embedded Linux devices

I would like to know how to view special characters while using &#39;less&#39; command.  For instance I want to see the non-printable characters with a special notation.  For instance in &#39;vi&#39; editor I use &quot;set list on&quot; to see the line termination characters represented by dollar &#39;$&#39; character.  Similarly I would want to do this using &#39;less&#39; command.

I referred Unix less manual, but to no avail.

Show special characters in Unix while using &#39;less&#39; Command

I have multiple files in a directory, example: `linux_file1.mp4`, `linux_file2.mp4` and so on. How do I move these files, using shell, so that the names are `file1.mp4`, `file2.mp4` and so on. I have about 30 files that I want to move to the new name.

Rename multiple files in shell

Is there a way to view a bash function&#39;s definition in bash?

For example, say I defined the function `foobar`

    function foobar {
        echo &quot;I&#39;m foobar&quot;
    }

Is there any way to later get the code that `foobar` runs?

    $ # non-working pseudocode
    $ echo $foobar
    echo &quot;I&#39;m foobar&quot;

Can bash show a function&#39;s definition?

    cat a.txt | xargs -I % echo %

In the example above, xargs takes `echo %` as the command argument. But in some cases, I need multiple commands to process the argument instead of one. For example:

    cat a.txt | xargs -I % {command1; command2; ... }

But xargs doesn&#39;t accept this form. One solution I know is that I can define a function to wrap the commands, but I want to avoid that because it is complex. Is there a better solution?


Running multiple commands with xargs

I&#39;m looking for exception handling mechanism in shell script. Is there any try,catch equivalent mechanism in shell script ?

Exception handling in shell scripting?

On the `diff` man-page I&#39;ve found these exit values:

        0     No differences were found. 
        1     Differences were found.
       &gt;1     An error occurred.

Are there different exit values above 1 for different errors?


What are the error exit values for diff?

When I use any command with sudo the environment variables are not there. For example after setting HTTP_PROXY the command `wget` works fine without `sudo`. However if I type `sudo wget` it says it can&#39;t bypass the proxy setting.

How to keep environment variables when using sudo

So, I&#39;m attempting to simply transfer folder of files from my local computer to a server via ssh and scp.  After sudoing I&#39;m using the command as follows:

    scp -r C:/desktop/myfolder/deployments/ user@host:/path/to/whereyouwant/thefile

I get the error:

&gt; ssh: C: Name or service not known

I&#39;m guessing its my syntax for `c:/desktop` etc.  Any ideas?  

BTW I&#39;m using putty + Windows 7.



  

Copying a local file from Windows to a remote server using scp

So I have a directory in /var/www (called cake) and I need to allow www-data to write to it, but I also want to write to it (without having to use sudo). I&#39;m afraid to change the permissions to 777 in case some other user on my machine (or a hacker) attempts to modify files in that directory. How do I only allow access for myself and Apache&#39;s www-data?

www-data permissions?

I want to write a shell script to automate a series of commands. The problem is some commands MUST be run as superuser and some commands MUST NOT be run as superuser. What I have done so far is something like this:

    #!/bin/bash
    
    command1
    sudo command2
    command3
    sudo command4

The problem is, this means somebody has to wait until command1 finishes before they are prompted for a password, then, if command3 takes long enough, they will then have to wait for command3 to finish. It would be nice if the person could get up and walk away, then come back an hour later and be done. For example, the following script has this problem:

    #!/bin/bash
    
    sleep 310
    sudo echo &quot;Hi, I&#39;m root&quot;
    sleep 310
    sudo echo &quot;I&#39;m still root?&quot;

How can I make it so that the user can just enter their password once, at the very start, and then walk away?

*Update:*

Thanks for the responses. I&#39;m running on Mac OS X Lion and ran Stephen P&#39;s script and got different results: (I also added $HOME)

    pair@abbey scratch$ ./test2.sh
    uid is 501
    user is pair
    username is 
    home directory is /Users/pair
    pair@abbey scratch$ sudo ./test2.sh 
    Password:
    uid is 0
    user is root
    username is root
    home directory is /Users/pair


How to write a shell script that runs some commands as superuser and some commands not as superuser, without having to babysit it?

I have a script which runs another script via SSH on a remote server using sudo.  However, when I type the password, it shows up on the terminal. (Otherwise it works fine)

    ssh user@server &quot;sudo script&quot;

What&#39;s the proper way to do this so I can type the password for sudo over SSH without the password appearing as I type?

proper way to sudo over ssh

There are already [some existing questions][1] asked here about  running commands as another user. However, the question and answers focus on *a single command* instead of a long group of commands.

For example, consider the following script:

    #!/bin/bash
    set -e

    root_command -p param1  # run as root

    # these commands must be run as another user
    command1 -p &#39;parameter with &quot;quotes&quot; inline&#39;
    command2 -p &#39;parameter with &quot;quotes&quot; inline&#39;
    command3 -p &#39;parameter with &quot;quotes&quot; inline&#39;

There are a couple of important points to note here:

 - The final three commands must be run as another user using `su` or `sudo`. In the example there were three commands, but suppose that there were many more...

 - The commands themselves make use of single and double quotes.

The second point above prevents the use of the following syntax:

    su somebody -c &quot;command&quot;

...since the commands themselves contain quotes.

What is the proper way to &quot;group&quot; the commands and run them under another user account?

  [1]: https://stackoverflow.com/questions/10220019/how-to-write-a-shell-script-that-runs-some-commands-as-superuser-and-some-comman







How can I execute a group of commands as another user in Bash?

I have taken over a Ubuntu 14.04 server. It has a user called &quot;deployer&quot; (used with capistrano), and as such, it needs sudo privileges. With this setup, I can log into the server and do stuff like:

    workstation&gt; ssh deployer@myserver
    myserver&gt;  sudo apt-get install git
    myserver&gt; exit
    workstation&gt;

I am trying to figure out how to use Ansible (version 2.0.2.0 and python 2.7.3) to create a user called &quot;deployer&quot; and be able to log into the server with that id and then so sudo-ish things like &quot;apt-get install&quot;. My playbook looks like this:

    ---
    - hosts: example
      become: yes
      tasks:
      - name: Update apt cache
        apt:
          update_cache: yes
          cache_valid_time: 3600

      - group: name=sudo state=present

      - name: Add deployer user and add it to sudo
        user: name=deployer
              state=present
              createhome=yes
        become: yes
        become_method: &quot;sudo&quot;

      - name: Set up authorized keys for the deployer user
        authorized_key: user=deployer key=&quot;{{item}}&quot;
        with_file:
          - /home/jaygodse/.ssh/id_rsa.pub



After running this playbook, I am able to ssh into the machine as &quot;deployer&quot;, (e.g. ssh deployer@myserver) but if I run a sudo command, it always asks me for my sudo password. 

I understand that the &quot;deployer&quot; user ultimately has to find its way into the visudo users file, but I cannot figure out which magical Ansible incantations to invoke so that I can ssh into the machine as  deployer and then run a sudo command (e.g. sudo apt-get install git&quot;) without being prompted for a sudo  password. 

I have searched high and low, and I can&#39;t seem to find an Ansible playbook fragment which puts the user &quot;deployer&quot; into the sudo group without requiring a password.  How is this done?

Ansible: create a user with sudo privileges

Can not create folder during image building with non root user added to sudoers group. Here is my Dockerfile:

    FROM ubuntu:16.04
    
    RUN apt-get update &amp;&amp; \
        apt-get -y install sudo
    
    RUN adduser --disabled-password --gecos &#39;&#39; newuser \
        &amp;&amp; adduser newuser sudo \
        &amp;&amp; echo &#39;%sudo ALL=(ALL:ALL) ALL&#39; &gt;&gt; /etc/sudoers
    
    USER newuser
    
    RUN mkdir -p /newfolder
    WORKDIR /newfolder


I get error: `mkdir: cannot create directory &#39;/newfolder&#39;: Permission denied`

Cannot create directory. Permission denied inside docker container

After logging into ssh, I got this message:

&gt;&#39;Username&#39; is not in the sudoers file. This incident will be reported.

How can I resolve this? I&#39;m connecting ssh to my virtual private server.

Content Type	Original Author	Original Content on Stackoverflow
Question	user788171	View Question on Stackoverflow
Solution 1 - Linux	Jeronimo Robles	View Answer on Stackoverflow
Solution 2 - Linux	pyroscope	View Answer on Stackoverflow
Solution 3 - Linux	C Jaskoski	View Answer on Stackoverflow

How to run script as another user without password?

Linux Problem Overview

Linux Solutions

Solution 1 - Linux

Solution 2 - Linux

Solution 3 - Linux

How to set conditional breakpoints in Visual Studio?

Override console.log(); for production

Attributions