Linux Error Code Variable
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Resources Polls Contact Me / Advertising Search This Blog Monday, March 24, 2008 How to check the exit status code When a command finishes execution, it returns an exit bash if exit code code. The exit code is not displayed on the screen by default. To bash script exit on error examine the exit code, you need to examine a special variable, "$?" Say, you are searching for a string in a
Exit Bash Shell
text file. $ grep x1y2z3 somefile.txt $ The standard output of the command returns null, which is a pretty good indication that the string cannot be found in the file. But what if
Bash Return Value From Function
you embed the grep command in a script? How can you tell if the string is found or not? Checking the exit code will tell you. Let's first try it out interactively. $ grep x1y2z3 somefile.txt $ echo $? 1 Note that in bash, the exit status is 0 if the command succeeded, and 1 if failed. For grep, 0 means that the string was found, and bash if exit code not 0 1 (or higher), otherwise. To check the exit status in a script, you may use the following pattern: somecommand argument1 argument2 RETVAL=$? [ $RETVAL -eq 0 ] && echo Success [ $RETVAL -ne 0 ] && echo Failure Posted by Peter Leung at 7:50 PM Email ThisBlogThis!Share to TwitterShare to FacebookShare to Pinterest 4 comments: Mad MAx said... $RETVAL=$?is not correct.when you assign a value to a variable, in bash, you can't prepend variable name with $.The correct form isRETVAL=$?regards, Mad Max. August 4, 2009 at 3:07 AM Peter Leung said... Thanks, Mad MAx.I made the correction.Peter August 4, 2009 at 8:50 PM Anonymous said... You can also just avoid the RETVAL altogether and use the "||" or "&&" operands which are called when the command on the left returns 1 or 0 respectively, e.g.# grep returns 1, e.g. no matchgrep foo /tmp/bar.txt || echo "text not found"# grep returns 0, e.g. matchgrep baz /tmp/bar.txt && echo "found it!" April 25, 2011 at 2:56 PM Wang said... Thanks for sharing, and thanks for the above comment. July 31, 2012 at 11:59 AM Post a Comment Newer Post Older Post Home Subscribe to: Post Comments (Atom) Followers Subscribe in a read
exit codes, exit codes are important and this article describes how to use them in your scripts and understand them in
Shell Script Exit
general. Written by Benjamin Cane on 2014-09-02 14:45:00| 4 min read Sponsored by exit code 0 Lately I've been working on a lot of automation and monitoring projects, a big part of these projects are write a shell script to calculate sum multiplication of three variables using expr keyword taking existing scripts and modifying them to be useful for automation and monitoring tools. One thing I have noticed is sometimes scripts use exit codes and sometimes they don't. It seems http://linuxcommando.blogspot.com/2008/03/how-to-check-exit-status-code.html like exit codes are easy for poeple to forget, but they are an incredibly important part of any script. Especially if that script is used for the command line. What are exit codes? On Unix and Linux systems, programs can pass a value to their parent process while terminating. This value is referred to as an exit code or exit status. On http://bencane.com/2014/09/02/understanding-exit-codes-and-how-to-use-them-in-bash-scripts/ POSIX systems the standard convention is for the program to pass 0 for successful executions and 1 or higher for failed executions. Why is this important? If you look at exit codes in the context of scripts written to be used for the command line the answer is very simple. Any script that is useful in some fashion will inevitably be either used in another script, or wrapped with a bash one liner. This becomes especially true if the script is used with automation tools like SaltStack or monitoring tools like Nagios, these programs will execute scripts and check the status code to determine whether that script was successful or not. On top of those reasons, exit codes exist within your scripts even if you don't define them. By not defining proper exit codes you could be falsely reporting successful executions which can cause issues depending on what the script does. What happens if I don't specify an exit code In Linux any script run from the command line has an exit code. With Bash scripts, if the exit code is not specified in the s
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the http://unix.stackexchange.com/questions/207957/assigning-exit-code-to-a-shell-local-variable company Business Learn more about hiring developers or posting ads with us Unix & Linux http://stackoverflow.com/questions/20157938/bash-exit-code-of-variable-assignment-to-command-substitution Questions Tags Users Badges Unanswered Ask Question _ Unix & Linux Stack Exchange is a question and answer site for users of Linux, FreeBSD and other Un*x-like operating systems. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the exit code top Assigning exit code to a shell local variable up vote 11 down vote favorite 2 #!/bin/bash function0() { local t1=$(exit 1) echo $t1 } function0 echo prints empty value. I expected: 1 Why doesn't t1 variable get assigned the exit command's return value - 1? bash shell variable command-substitution share|improve this question edited Jun 6 '15 at 18:43 Gilles 372k696761127 asked Jun 6 '15 at 15:06 Madhavan Kumar 1,1282927 add a comment| 3 Answers bash if exit 3 active oldest votes up vote 19 down vote accepted local t1=$(exit 1) tells the shell to: run exit 1 in a subshell; store its output (as in, the text it outputs to standard output) in a variable t1, local to the function. It's thus normal that t1 ends up being empty. ($() is known as command substitution.) The exit code is always assigned to $?, so you can do function0() { (exit 1) echo "$?" } to get the effect you're looking for. You can of course assign $? to another variable: function0() { (exit 1) local t1=$? echo "$t1" } share|improve this answer edited Jun 7 '15 at 16:09 answered Jun 6 '15 at 15:13 Stephen Kitt 42.5k56494 You know, you can always put the return into the pipe, too. `$(trap 'printf "::ERRNO:$?"' 0; # now do whatever however - that trap will ensure the last string written is the last return for the whole substitution context. –mikeserv Jun 7 '15 at 16:28 add a comment| up vote 6 down vote Exit code was stored in $? variable. Using Command Substitution only capture the output, you should use (...) to create subshell: #!/bin/bash func() { (exit 1) local t1=$? printf '%d\n' "$t1" } func share|improve this answer edited Jun 7 '15 at 15:33 answered Jun 6 '15 at 15:14 c
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up Bash: exit code of variable assignment to command substitution up vote 24 down vote favorite I am confused about what error code the command will return when executing a variable assignment plainly and with command substitution: a=$(false); echo $? It outputs 1, which let me think that variable assignment doesn't sweep or produce new error code upon the last one. But when I tried this: false; a=""; echo $? It outputs 0, obviously this is what a="" returns and it override 1 returned by false. I want to know why this happens, is there any particularity in variable assignment that differs from other normal commands? Or just be cause a=$(false) is considered to be a single command and only command substitution part make sense? -- UPDATE -- Thanks everyone, from the answers and comments I got the point "When you assign a variable using command substitution, the exit status is the status of the command." (by @Barmar), this explanation is excellently clear and easy to understand, but speak doesn't precise enough for programmers, I want to see the reference of this point from authorities such as TLDP or GNU man page, please help me find it out, thanks again! linux bash shell exit-code share|improve this question edited Jul 27 '15 at 10:21 Ciro Santilli 烏坎事件2016六四事件 法轮功 52.8k10226167 asked Nov 23 '13 at 2:52 Reorx 9651724 add a comment| 2 Answers 2 active oldest votes up vote 34 down vote accepted Upon executing a command as $(command) allows the output of the command to replace itself. When you say: a=$(false) # false fails; the output of false is stored in the variable a the output produced by the command false is stored in the variable a. Moreover, the exit code is the same as produced by the command. help false would tell: false: false Return an unsuccessful result. Exit Status: Always fails. On the other hand, saying: $ false # Exit code: 1 $ a="" # Exit code: 0 $ echo $? # Prints 0 causes the exit code for the assignment to a to be returned which is