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!
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 $? # Prints 0
causes the exit code for the assignment to a to be returned which is 0.
EDIT:
Quoting from the manual:
If one of the expansions contained a command substitution, the exit
status of the command is the exit status of the last command
substitution performed.
Quoting from BASHFAQ/002:
How can I store the return value and/or output of a command in a
variable?
...
output=$(command)
status=$?
The assignment to output has no effect on command's exit status, which
is still in $?.
Note that this isn't the case when combined with local, as in local variable="$(command)". That form will exit successfully even if command failed.
Take this Bash script for example:
#!/bin/bash
function funWithLocalAndAssignmentTogether() {
local output="$(echo "Doing some stuff.";exit 1)"
local exitCode=$?
echo "output: $output"
echo "exitCode: $exitCode"
}
function funWithLocalAndAssignmentSeparate() {
local output
output="$(echo "Doing some stuff.";exit 1)"
local exitCode=$?
echo "output: $output"
echo "exitCode: $exitCode"
}
funWithLocalAndAssignmentTogether
funWithLocalAndAssignmentSeparate
Here is the output of this:
nick.parry#nparry-laptop1:~$ ./tmp.sh
output: Doing some stuff.
exitCode: 0
output: Doing some stuff.
exitCode: 1
This is because local is actually a builtin command, and a command like local variable="$(command)" calls local after substituting the output of command. So you get the exit status from local.
I came across the same problem yesterday (Aug 29 2018).
In addition to local mentioned in Nick P.'s answer and #sevko's comment in the accepted answer, declare in global scope also has the same behavior.
Here's my Bash code:
#!/bin/bash
func1()
{
ls file_not_existed
local local_ret1=$?
echo "local_ret1=$local_ret1"
local local_var2=$(ls file_not_existed)
local local_ret2=$?
echo "local_ret2=$local_ret2"
local local_var3
local_var3=$(ls file_not_existed)
local local_ret3=$?
echo "local_ret3=$local_ret3"
}
func1
ls file_not_existed
global_ret1=$?
echo "global_ret1=$global_ret1"
declare global_var2=$(ls file_not_existed)
global_ret2=$?
echo "global_ret2=$global_ret2"
declare global_var3
global_var3=$(ls file_not_existed)
global_ret3=$?
echo "global_ret3=$global_ret3"
The output:
$ ./declare_local_command_substitution.sh 2>/dev/null
local_ret1=2
local_ret2=0
local_ret3=2
global_ret1=2
global_ret2=0
global_ret3=2
Note the values of local_ret2 and global_ret2 in the output above. The exit codes are overwritten by local and declare.
My Bash version:
$ echo $BASH_VERSION
4.4.19(1)-release
(not an answer to original question but too long for comment)
Note that export A=$(false); echo $? outputs 0! Apparently the rules quoted in devnull's answer no longer apply. To add a bit of context to that quote (emphasis mine):
3.7.1 Simple Command Expansion
...
If there is a command name left after expansion, execution proceeds as described below. Otherwise, the command exits. If one of the expansions contained a command substitution, the exit status of the command is the exit status of the last command substitution performed. If there were no command substitutions, the command exits with a status of zero.
3.7.2 Command Search and Execution [ — this is the "below" case]
IIUC the manual describes var=foo as special case of var=foo command... syntax (pretty confusing!). The "exit status of the last command substitution" rule only applies to the no-command case.
While it's tempting to think of export var=foo as a "modified assignment syntax", it isn't — export is a builtin command (that just happens to take assignment-like args).
=> If you want to export a var AND capture command substitution status, do it in 2 stages:
A=$(false)
# ... check $?
export A
This way also works in set -e mode — exits immediately if the command substitution return non-0.
As others have said, the exit code of the command substitution is the exit code of the substituted command, so
FOO=$(false)
echo $?
---
1
However, unexpectedly, adding export to the beginning of that produces a different result:
export FOO=$(false)
echo $?
---
0
This is because, while the substituted command false fails, the export command succeeds, and that is the exit code returned by the statement.
Related
I am running my scripts with:
#!/bin/bash -eu
Which aborts the script whenever a problem occurs, as wanted. But sometimes I expect one of the commands to eventually fail, and I would like to tell bash to ignore the fail condition. In make you can ignore the status of one command with the handy:
-command
Is there something similar in bash? The only thing that comes to mind is the ugly:
set +e
command
set -e
You could just do a no-op on the command failure or set an explicit true condition as
command || true
or for no-op as
command || :
Doing so forces the command-list (even a pipeline) to return an exit status of 0 on failure. See
true | false
echo $?
1
true | false || true
echo $?
0
true | false || :
echo $?
0
Just prepend a ! to the command so that its exit status does not make the script exit when running it with e:
! command
As seen in What's the meaning of a ! before a command in the shell?, having ! command negates the exit status of the given command and, used with set -e, prevents the shell to exit whatever the exit result is on that line.
From Bash Reference Manual → Pipelines:
Each command in a pipeline is executed in its own subshell. The exit status of a pipeline is the exit status of the last command in the pipeline (...). If the reserved word ‘!’ precedes the pipeline, the exit status is the logical negation of the exit status as described above. The shell waits for all commands in the pipeline to terminate before returning a value.
Then we have the info about 4.3.1 The set Builtin:
-e
Exit immediately if a pipeline (see Pipelines), which may consist of a single simple command (see Simple Commands), a list (see Lists), or a compound command (see Compound Commands) returns a non-zero status. The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test in an if statement, part of any command executed in a && or || list except the command following the final && or ||, any command in a pipeline but the last, or if the command’s return status is being inverted with !.
All together, and quoting my own answer:
When you have:
set -e
! command1
command2
What you are doing is to by-pass the set -e flag in the command1.
Why?
if command1 runs properly, it will return a zero status. ! will negate it, but set -e won't trigger an exit by the because it comes
from a return status inverted with !, as described above.
if command1 fails, it will return a non-zero status. ! will negate it, so the line will end up returning a zero status and the
script will continue normally.
Don't think there is
Could just write your own function though
#!/bin/bash -eu
-(){
set +e
"$#"
set -e
}
- command
echo got here
May want to use a function name since - is already used in bash.
As chepner pointed out it only works for simple commands, though, not pipelines or lists.
I seem to have encountered a very, very strange inconsistency in the way both dash and bash check for error conditions with the errexit option.
Using both dash and bash without the set -e/set -o errexit option, the following program:
foo()
{
echo pre
bar=$(fail)
echo post
}
foo
will print the following (with slightly different error strings for dash):
pre
./foo.sh: line 4: fail: command not found
post
With the errexit option, it will print the following:
pre
./foo.sh: line 4: fail: command not found
Surprisingly, however, if bar is local, the program will always echo both pre and post. More specifically, using both dash and bash with our without the errexit option, the following program:
foo()
{
echo pre
local bar=$(fail)
echo post
}
foo
will print the following:
pre
./foo.sh: line 4: fail: command not found
post
In other words, it seems that the return value of a command substitution that is assigned to a local variable is not checked by errexit, but it is if the variable is global.
I would have been inclined to think this was simply a corner case bug, if it didn't happen with both shells. Since dash is specifically designed to be POSIX conformant I wonder if this behavior is actually specified by the POSIX standard, though I have a hard time imagining how that would make sense.
dash(1) has this to say about errexit:
If not interactive, exit immediately if any untested command fails. The exit status of a command is considered to be explicitly tested if the command is used to control an if, elif, while, or until; or if the command is the left hand operand of an “&&” or “||” operator.
bash(1) is somewhat more verbose, but I have a hard time making sense of it:
Exit immediately if a pipeline (which may consist of a single simple command), a list, or a compound command (see SHELL GRAMMAR above), exits with a non-zero status. The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test following the if or elif reserved words, part of any command executed in a && or || list except the command following the final && or ||, any command in a pipeline but the last, or if the command's return value is being inverted with !. If a compound command other than a subshell returns a non-zero status because a command failed while -e was being ignored, the shell does not exit. A trap on ERR, if set, is executed before the shell exits. This option applies to the shell environment and each subshell environment separately (see COMMAND EXECUTION ENVIRONMENT above), and may cause subshells to exit before executing all the commands in the subshell.
If a compound command or shell function executes in a context where -e is being ignored, none of the commands executed within the compound command or function body will be affected by the -e setting, even if -e is set and a command returns a failure status. If a compound command or shell function sets -e while executing in a context where -e is ignored, that setting will not have any effect until the compound command or the command containing the function call completes.
TL;DR The exit status of local "hides" the exit status of any command substitutions appearing in one of its arguments.
The exit status of a variable assignment is poorly documented (or at least, I couldn't find any specifics in a quick skim of the various man pages and the POSIX spec). As far as I can tell, the exit status is taken as the exit status of the last command substitution that occurs in the value of the assignment, or 0 if there are no command substitutions. Non-final command substitutions appear to be included in the list of "tested" situations, as an assignment like
foo=$(false)$(true)
does not exit with errexit set.
local, however, is a command itself whose exit status is normally 0, independent of any command substitutions that occur in its arguments. That is, while
foo=$(false)
has an exit status of 1,
local foo=$(false)
will have an exit status of 0, with any command substitutions in an argument seemingly considered "tested" for the purposes of errexit.
Try this:
#!/bin/bash
set -e
foo()
{
echo pre
local bar
bar=$(fail)
echo post
}
foo
exit
!! OR !!
#!/bin/bash
foo()
{
set -e
echo pre
local bar
bar=$(fail)
echo post
}
foo
exit
OUTPUT:
$ ./errexit_function
pre
./errexit_function: line 8: fail: command not found
$ echo $?
127
As far as I can tell this is a work around for a bug in bash, but try this,
#!/bin/bash
set -e
foo()
{
echo true || return_value=$?
echo the command returned a value of ${return_value:-0}
$(fail) || return_value=$?
echo the command returned a value of ${return_value:-0}
echo post
}
foo
exit
I'm currently working on some init scripts that should both use set -e and confirm to the Linux Standard Base core specification. Now those two don't really work together:
Since due to the set -e the first command that fails causes the scripts to exit with the return value of the failed command I cannot set the exit status of the script to something LSB conformant for commands that fail with LSB incompatible return values. I could unset -e before each such command, but that's quite a hassle and in that case I'd actually rather not use set -e at all?
I guess another way to achieve the same result as with unset -e would be to do something like
returns_1() { return 1; }
...
cmd_that_fails_with_non_LSB_return_values || returns_1
but that seems quite bulky as well and I'd again have to check each and every command's possible return values.
Is there a way to set the error code returned by the script when it is terminated due to set -e to a fixed value so it would return 1 (i.e. LSB general/unspecified error) no mather what return value the failed command had? Should I not bother with LSB conformant return codes and/or set -e? (this will probably turn into a discussion about the merrits of set -e anyway judging from the amount of search results you get for that)
Small code snippet to illustrate the problem:
#!/bin/bash
# init script for service foo
set -e
start() {
echo "bar"
cmd_fails_with_return_code_3 # script exits with return code 3, not LSB conformant
echo "baz"
}
...
case "$1" in
start)
start
;;
...
esac
From the man page, it appears you can set a trap on ERR:
A trap on ERR, if set, is executed before the shell exits.
I haven't tried this but this would mean something like this might help you:
trap "exit 1" ERR
Be sure to read the man page for other useful options such as -E to inherit the ERR trap in subshells.
You basically have it nailed, although the returns function is superfluous.
cmd_that_fails_with_non_LSB_return_values || exit 1
More typically, you would actually somehow handle the error, if only just to report what happened.
die () {
echo "$0: $#" >&2
exit 1
}
:
cmd_that_fails_with_non_LSB_return_values ||
die "Long command with underscores failed; aborting"
Philosophically, I suppose the purpose of set -e is mainly to "gently remind" (that is, force) you to handle all possible error conditions.
When I use exit command in a shell script, the script will terminate the terminal (the prompt). Is there any way to terminate a script and then staying in the terminal?
My script run.sh is expected to execute by directly being sourced, or sourced from another script.
EDIT:
To be more specific, there are two scripts run2.sh as
...
. run.sh
echo "place A"
...
and run.sh as
...
exit
...
when I run it by . run2.sh, and if it hit exit codeline in run.sh, I want it to stop to the terminal and stay there. But using exit, the whole terminal gets closed.
PS: I have tried to use return, but echo codeline will still gets executed....
The "problem" really is that you're sourcing and not executing the script. When you source a file, its contents will be executed in the current shell, instead of spawning a subshell. So everything, including exit, will affect the current shell.
Instead of using exit, you will want to use return.
Yes; you can use return instead of exit. Its main purpose is to return from a shell function, but if you use it within a source-d script, it returns from that script.
As §4.1 "Bourne Shell Builtins" of the Bash Reference Manual puts it:
return [n]
Cause a shell function to exit with the return value n.
If n is not supplied, the return value is the exit status of the
last command executed in the function.
This may also be used to terminate execution of a script being executed
with the . (or source) builtin, returning either n or
the exit status of the last command executed within the script as the exit
status of the script.
Any command associated with the RETURN trap is executed
before execution resumes after the function or script.
The return status is non-zero if return is used outside a function
and not during the execution of a script by . or source.
You can add an extra exit command after the return statement/command so that it works for both, executing the script from the command line and sourcing from the terminal.
Example exit code in the script:
if [ $# -lt 2 ]; then
echo "Needs at least two arguments"
return 1 2>/dev/null
exit 1
fi
The line with the exit command will not be called when you source the script after the return command.
When you execute the script, return command gives an error. So, we suppress the error message by forwarding it to /dev/null.
Instead of running the script using . run2.sh, you can run it using sh run2.sh or bash run2.sh
A new sub-shell will be started, to run the script then, it will be closed at the end of the script leaving the other shell opened.
Actually, I think you might be confused by how you should run a script.
If you use sh to run a script, say, sh ./run2.sh, even if the embedded script ends with exit, your terminal window will still remain.
However if you use . or source, your terminal window will exit/close as well when subscript ends.
for more detail, please refer to What is the difference between using sh and source?
This is just like you put a run function inside your script run2.sh.
You use exit code inside run while source your run2.sh file in the bash tty.
If the give the run function its power to exit your script and give the run2.sh
its power to exit the terminator.
Then of cuz the run function has power to exit your teminator.
#! /bin/sh
# use . run2.sh
run()
{
echo "this is run"
#return 0
exit 0
}
echo "this is begin"
run
echo "this is end"
Anyway, I approve with Kaz it's a design problem.
I had the same problem and from the answers above and from what I understood what worked for me ultimately was:
Have a shebang line that invokes the intended script, for example,
#!/bin/bash uses bash to execute the script
I have scripts with both kinds of shebang's. Because of this, using sh or . was not reliable, as it lead to a mis-execution (like when the script bails out having run incompletely)
The answer therefore, was
Make sure the script has a shebang, so that there is no doubt about its intended handler.
chmod the .sh file so that it can be executed. (chmod +x file.sh)
Invoke it directly without any sh or .
(./myscript.sh)
Hope this helps someone with similar question or problem.
To write a script that is secure to be run as either a shell script or sourced as an rc file, the script can check and compare $0 and $BASH_SOURCE and determine if exit can be safely used.
Here is a short code snippet for that
[ "X$(basename $0)" = "X$(basename $BASH_SOURCE)" ] && \
echo "***** executing $name_src as a shell script *****" || \
echo "..... sourcing $name_src ....."
I think that this happens because you are running it on source mode
with the dot
. myscript.sh
You should run that in a subshell:
/full/path/to/script/myscript.sh
'source' http://ss64.com/bash/source.html
It's correct that sourced vs. executed scripts use return vs. exit to keep the same session open, as others have noted.
Here's a related tip, if you ever want a script that should keep the session open, regardless of whether or not it's sourced.
The following example can be run directly like foo.sh or sourced like . foo.sh/source foo.sh. Either way it will keep the session open after "exiting". The $# string is passed so that the function has access to the outer script's arguments.
#!/bin/sh
foo(){
read -p "Would you like to XYZ? (Y/N): " response;
[ $response != 'y' ] && return 1;
echo "XYZ complete (args $#).";
return 0;
echo "This line will never execute.";
}
foo "$#";
Terminal result:
$ foo.sh
$ Would you like to XYZ? (Y/N): n
$ . foo.sh
$ Would you like to XYZ? (Y/N): n
$ |
(terminal window stays open and accepts additional input)
This can be useful for quickly testing script changes in a single terminal while keeping a bunch of scrap code underneath the main exit/return while you work. It could also make code more portable in a sense (if you have tons of scripts that may or may not be called in different ways), though it's much less clunky to just use return and exit where appropriate.
Also make sure to return with expected return value. Else if you use exit when you will encounter an exit it will exit from your base shell since source does not create another process (instance).
Improved the answer of Tzunghsing, with more clear results and error re-direction, for silent usage:
#!/usr/bin/env bash
echo -e "Testing..."
if [ "X$(basename $0 2>/dev/null)" = "X$(basename $BASH_SOURCE)" ]; then
echo "***** You are Executing $0 in a sub-shell."
exit 0
else
echo "..... You are Sourcing $BASH_SOURCE in this terminal shell."
return 0
fi
echo "This should never be seen!"
Or if you want to put this into a silent function:
function sExit() {
# Safe Exit from script, not closing shell.
[ "X$(basename $0 2>/dev/null)" = "X$(basename $BASH_SOURCE)" ] && exit 0 || return 0
}
...
# ..it have to be called with an error check, like this:
sExit && return 0
echo "This should never be seen!"
Please note that:
if you have enabled errexit in your script (set -e) and you return N with N != 0, your entire script will exit instantly. To see all your shell settings, use, set -o.
when used in a function, the 1st return 0 is exiting the function, and the 2nd return 0 is exiting the script.
if your terminal emulator doesn't have -hold you can sanitize a sourced script and hold the terminal with:
#!/bin/sh
sed "s/exit/return/g" script >/tmp/script
. /tmp/script
read
otherwise you can use $TERM -hold -e script
If a command succeeded successfully, the return value will be 0. We can check its return value afterwards.
Is there a “goto” statement in bash?
Here is some dirty workaround using trap which jumps only backwards.
#!/bin/bash
set -eu
trap 'echo "E: failed with exitcode $?" 1>&2' ERR
my_function () {
if git rev-parse --is-inside-work-tree > /dev/null 2>&1; then
echo "this is run"
return 0
else
echo "fatal: not a git repository (or any of the parent directories): .git"
goto trap 2> /dev/null
fi
}
my_function
echo "Command succeeded" # If my_function failed this line is not printed
Related:
https://stackoverflow.com/a/19091823/2402577
How to use $? and test to check function?
I couldn't find solution so for those who want to leave the nested script without leaving terminal window:
# this is just script which goes to directory if path satisfies regex
wpr(){
leave=false
pwd=$(pwd)
if [[ "$pwd" =~ ddev.*web ]]; then
# echo "your in wordpress instalation"
wpDir=$(echo "$pwd" | grep -o '.*\/web')
cd $wpDir
return
fi
echo 'please be in wordpress directory'
# to leave from outside the scope
leave=true
return
}
wpt(){
# nested function which returns $leave variable
wpr
# interupts the script if $leave is true
if $leave; then
return;
fi
echo 'here is the rest of the script, executes if leave is not defined'
}
I have no idea whether this is useful for you or not, but in zsh, you can exit a script, but only to the prompt if there is one, by using parameter expansion on a variable that does not exist, as follows.
${missing_variable_ejector:?}
Though this does create an error message in your script, you can prevent it with something like the following.
{ ${missing_variable_ejector:?} } 2>/dev/null
1) exit 0 will come out of the script if it is successful.
2) exit 1 will come out of the script if it is a failure.
You can try these above two based on ur req.
I am trying to do a homework assignment and it is very confusing. I am not sure if the professor's example is in Perl or bash, since it has no header. Basically, I just need help with the meat of the problem: capturing the input and outputting it. Here is the assignment:
In the session, provide a command prompt that includes the working directory, e.g.,
$./logger/home/it244/it244/hw8$
Accept user’s commands, execute them, and display the output on the screen.
During the session, create a temporary file “PID.cmd” (PID is the process ID) to store the command history in the following format (index: command):
1: ls
2: ls -l
If the script is aborted by CTRL+C (signal 2), output a message “aborted by ctrl+c”.
When you quit the logging session (either by “exit” or CTRL+C),
a. Delete the temporary file
b. Print out the total number of the commands in the session and the numbers of successful/failed commands (according to the exit status).
Here is my code so far (which did not go well, I would not try to run it):
#!/bin/sh
trap 'exit 1' 2
trap 'ctrl-c' 2
echo $(pwd)
while true
do
read -p command
echo "$command:" $command >> PID.cmd
done
Currently when I run this script I get
command read: 10: arg count
What is causing that?
======UPDATE=========
Ok I made some progress not quite working all the way it doesnt like my bashtrap or incremental index
#!/bin/sh
index=0
trap bashtrap INT
bashtrap(){
echo "CTRL+C aborting bash script"
}
echo "starting to log"
while :
do
read -p "command:" inputline
if [ $inputline="exit" ]
then
echo "Aborting with Exit"
break
else
echo "$index: $inputline" > output
$inputline 2>&1 | tee output
(( index++ ))
fi
done
This can be achieved in bash or perl or others.
Some hints to get you started in bash :
question 1 : command prompt /logger/home/it244/it244/hw8
1) make sure of the prompt format in the user .bashrc setup file: see PS1 data for debian-like distros.
2) cd into that directory within you bash script.
question 2 : run the user command
1) get the user input
read -p "command : " input_cmd
2) run the user command to STDOUT
bash -c "$input_cmd"
3) Track the user input command exit code
echo $?
Should exit with "0" if everything worked fine (you can also find exit codes in the command man pages).
3) Track the command PID if the exit code is Ok
echo $$ >> /tmp/pid_Ok
But take care the question is to keep the user command input, not the PID itself as shown here.
4) trap on exit
see man trap as you misunderstood the use of this : you may create a function called on the catched exit or CTRL/C signals.
5) increment the index in your while loop (on the exit code condition)
index=0
while ...
do
...
((index++))
done
I guess you have enough to start your home work.
Since the example posted used sh, I'll use that in my reply. You need to break down each requirement into its specific lines of supporting code. For example, in order to "provide a command prompt that includes the working directory" you need to actually print the current working directory as the prompt string for the read command, not by setting the $PS variable. This leads to a read command that looks like:
read -p "`pwd -P`\$ " _command
(I use leading underscores for private variables - just a matter of style.)
Similarly, the requirement to do several things on either a trap or a normal exit suggests a function should be created which could then either be called by the trap or to exit the loop based on user input. If you wanted to pretty-print the exit message, you might also wrap it in echo commands and it might look like this:
_cleanup() {
rm -f $_LOG
echo
echo $0 ended with $_success successful commands and $_fail unsuccessful commands.
echo
exit 0
}
So after analyzing each of the requirements, you'd need a few counters and a little bit of glue code such as a while loop to wrap them in. The result might look like this:
#/usr/bin/sh
# Define a function to call on exit
_cleanup() {
# Remove the log file as per specification #5a
rm -f $_LOG
# Display success/fail counts as per specification #5b
echo
echo $0 ended with $_success successful commands and $_fail unsuccessful commands.
echo
exit 0
}
# Where are we? Get absolute path of $0
_abs_path=$( cd -P -- "$(dirname -- "$(command -v -- "$0")")" && pwd -P )
# Set the log file name based on the path & PID
# Keep this constant so the log file doesn't wander
# around with the user if they enter a cd command
_LOG=${_abs_path}/$$.cmd
# Print ctrl+c msg per specification #4
# Then run the cleanup function
trap "echo aborted by ctrl+c;_cleanup" 2
# Initialize counters
_line=0
_fail=0
_success=0
while true
do
# Count lines to support required logging format per specification #3
((_line++))
# Set prompt per specification #1 and read command
read -p "`pwd -P`\$ " _command
# Echo command to log file as per specification #3
echo "$_line: $_command" >>$_LOG
# Arrange to exit on user input with value 'exit' as per specification #5
if [[ "$_command" == "exit" ]]
then
_cleanup
fi
# Execute whatever command was entered as per specification #2
eval $_command
# Capture the success/fail counts to support specification #5b
_status=$?
if [ $_status -eq 0 ]
then
((_success++))
else
((_fail++))
fi
done