Related
Hi… Need a little help here…
I tried to emulate the DOS' dir command in Linux using bash script. Basically it's just a wrapped ls command with some parameters plus summary info. Here's the script:
#!/bin/bash
# default to current folder
if [ -z "$1" ]; then var=.;
else var="$1"; fi
# check file existence
if [ -a "$var" ]; then
# list contents with color, folder first
CMD="ls -lgG $var --color --group-directories-first"; $CMD;
# sum all files size
size=$(ls -lgGp "$var" | grep -v / | awk '{ sum += $3 }; END { print sum }')
if [ "$size" == "" ]; then size="0"; fi
# create summary
if [ -d "$var" ]; then
folder=$(find $var/* -maxdepth 0 -type d | wc -l)
file=$(find $var/* -maxdepth 0 -type f | wc -l)
echo "Found: $folder folders "
echo " $file files $size bytes"
fi
# error message
else
echo "dir: Error \"$var\": No such file or directory"
fi
The problem is when the argument contains an asterisk (*), the ls within the script acts differently compare to the direct ls command given at the prompt. Instead of return the whole files list, the script only returns the first file. See the video below to see the comparation in action. I don't know why it behaves like that.
Anyone knows how to fix it? Thank you.
Video: problem in action
UPDATE:
The problem has been solved. Thank you all for the answers. Now my script works as expected. See the video here: http://i.giphy.com/3o8dp1YLz4fIyCbOAU.gif
The asterisk * is expanded by the shell when it parses the command line. In other words, your script doesn't get a parameter containing an asterisk, it gets a list of files as arguments. Your script only works with $1, the first argument. It should work with "$#" instead.
This is because when you retrieve $1 you assume the shell does NOT expand *.
In fact, when * (or other glob) matches, it is expanded, and broken into segments by $IFS, and then passed as $1, $2, etc.
You're lucky if you simply retrieved the first file. When your first file's path contains spaces, you'll get an error because you only get the first segment before the space.
Seriously, read this and especially this. Really.
And please don't do things like
CMD=whatever you get from user input; $CMD;
You are begging for trouble. Don't execute arbitrary string from the user.
Both above answers already answered your question. So, i'm going a bit more verbose.
In your terminal is running the bash interpreter (probably). This is the program which parses your input line(s) and doing "things" based on your input.
When you enter some line the bash start doing the following workflow:
parsing and lexical analysis
expansion
brace expansion
tidle expansion
variable expansion
artithmetic and other substitutions
command substitution
word splitting
filename generation (globbing)
removing quotes
Only after all above the bash
will execute some external commands, like ls or dir.sh... etc.,
or will do so some "internal" actions for the known keywords and builtins like echo, for, if etc...
As you can see, the second last is the filename generation (globbing). So, in your case - if the test* matches some files, your bash expands the willcard characters (aka does the globbing).
So,
when you enter dir.sh test*,
and the test* matches some files
the bash does the expansion first
and after will execute the command dir.sh with already expanded filenames
e.g. the script get executed (in your case) as: dir.sh test.pas test.swift
BTW, it acts exactly with the same way for your ls example:
the bash expands the ls test* to ls test.pas test.swift
then executes the ls with the above two arguments
and the ls will print the result for the got two arguments.
with other words, the ls don't even see the test* argument - if it is possible - the bash expands the wilcard characters. (* and ?).
Now back to your script: add after the shebang the following line:
echo "the $0 got this arguments: $#"
and you will immediatelly see, the real argumemts how your script got executed.
also, in such cases is a good practice trying to execute the script in debug-mode, e.g.
bash -x dir.sh test*
and you will see, what the script does exactly.
Also, you can do the same for your current interpreter, e.g. just enter into the terminal
set -x
and try run the dir.sh test* = and you will see, how the bash will execute the dir.sh command. (to stop the debug mode, just enter set +x)
Everbody is giving you valuable advice which you should definitely should follow!
But here is the real answer to your question.
To pass unexpanded arguments to any executable you need to single quote them:
./your_script '*'
The best solution I have is to use the eval command, in this way:
#!/bin/bash
cmd="some command \"with_quetes_and_asterisk_in_it*\""
echo "$cmd"
eval $cmd
The eval command takes its arguments and evaluates them into the command as the shell does.
This solves my problem when I need to call a command with asterisk '*' in it from a script.
I am facing a problem while reading input from command prompt in shell script. My script's name is status.ksh, and I have to take parameter from command prompt. This script accepts 2 parameter. 1st is "-e" and second is "server_name".
When I am running the script like this,
status.ksh -e server_name
echo $#
is giving output "server_name" only, where as expected output should be "-e server_name"
and echo $1 is giving output as NULL, where as expected output should be "-e".
Please guide me, how to read get the 1st parameter, which is "-e" .
Thanks & Regards
The problem was caused by -e. This is a flag for echo.
-e enable interpretation of backslash escapes
Most of the unix commands allow -- to be used to separate flags and the rest of the arguments, but echo doesn't support this, so you need another command:
printf "%s\n" "$1"
If you need complex command line argument parsing, definitely go with getopts as Joe suggested.
Have you read this reference? http://www.lehman.cuny.edu/cgi-bin/man-cgi?getopts+1
You shouldn't use $1, $2, $#, etc to parse options. There are builtins that can handle this for you.
Example 2 Processing Arguments for a Command with Options
The following fragment of a shell program processes the
arguments for a command that can take the options -a or -b.
It also processes the option -o, which requires an option-
argument:
while getopts abo: c
do
case $c in
a | b) FLAG=$c;;
o) OARG=$OPTARG;;
\?) echo $USAGE
exit 2;;
esac
done
shift `expr $OPTIND - 1`
More examples:
http://linux-training.be/files/books/html/fun/ch21s05.html
http://publib.boulder.ibm.com/infocenter/pseries/v5r3/index.jsp?topic=/com.ibm.aix.cmds/doc/aixcmds2/getopts.htm
http://www.livefirelabs.com/unix_tip_trick_shell_script/may_2003/05262003.htm
I'm new to shell programming. I intend to get directory name after zip file was extracted. The print statement of it is
$test.sh helloworld.zip
helloworld
Let's take a look at test.sh:
#! /bin/sh
length=echo `expr index "$1" .zip`
a=$1
echo $(a:0:length}
However I got the Bad substitution error from the compiler.
And when I mention about 'shell'.I just talking about shell for I don't know the difference between bash or the others.I just using Ubuntu 10.04 and using the terminal. (I am using bash.)
If your shell is a sufficiently recent version of bash, that parameter expansion notation should work.
In many other shells, it will not work, and a bad substitution error is the way the shell says 'You asked for a parameter substitution but it does not make sense to me'.
Also, given the script:
#! /bin/sh
length=echo `expr index "$1" .zip`
a=$1
echo $(a:0:length}
The second line exports variable length with value echo for the command that is generated by running expr index "$1" .zip. It does not assign to length. That should be just:
length=$(expr index "${1:?}" .zip)
where the ${1:?} notation generates an error if $1 is not set (if the script is invoked with no arguments).
The last line should be:
echo ${a:0:$length}
Note that if $1 holds filename.zip, the output of expr index $1 .zip is 2, because the letter i appears at index 2 in filename.zip. If the intention is to get the base name of the file without the .zip extension, then the classic way to do it is:
base=$(basename $1 .zip)
and the more modern way is:
base=${1%.zip}
There is a difference; if the name is /path/to/filename.zip, the classic output is filename and the modern one is /path/to/filename. You can get the classic output with:
base=${1%.zip}
base=${base##*/}
Or, in the classic version, you can get the path with:
base=$(dirname $1)/$(basename $1 .zip)`.)
If the file names can contain spaces, you need to think about using double quotes, especially in the invocations of basename and dirname.
Try running it with bash.
bash test.sh helloworld.zip
-likewise-
"try changing the first line to #!/bin/bash" as comment-answered by – #shellter
Try that in bash :
echo $1
len=$(wc -c <<< "$1")
a="${1}.zip"
echo ${a:0:$len}
Adapt it to fit your needs.
I want to loop through a path list that I have gotten from an echo $VARIABLE command.
For example:
echo $MANPATH will return
/usr/lib:/usr/sfw/lib:/usr/info
So that is three different paths, each separated by a colon. I want to loop though each of those paths. Is there a way to do that? Thanks.
Thanks for all the replies so far, it looks like I actually don't need a loop after all. I just need a way to take out the colon so I can run one ls command on those three paths.
You can set the Internal Field Separator:
( IFS=:
for p in $MANPATH; do
echo "$p"
done
)
I used a subshell so the change in IFS is not reflected in my current shell.
The canonical way to do this, in Bash, is to use the read builtin appropriately:
IFS=: read -r -d '' -a path_array < <(printf '%s:\0' "$MANPATH")
This is the only robust solution: will do exactly what you want: split the string on the delimiter : and be safe with respect to spaces, newlines, and glob characters like *, [ ], etc. (unlike the other answers: they are all broken).
After this command, you'll have an array path_array, and you can loop on it:
for p in "${path_array[#]}"; do
printf '%s\n' "$p"
done
You can use Bash's pattern substitution parameter expansion to populate your loop variable. For example:
MANPATH=/usr/lib:/usr/sfw/lib:/usr/info
# Replace colons with spaces to create list.
for path in ${MANPATH//:/ }; do
echo "$path"
done
Note: Don't enclose the substitution expansion in quotes. You want the expanded values from MANPATH to be interpreted by the for-loop as separate words, rather than as a single string.
In this way you can safely go through the $PATH with a single loop, while $IFS will remain the same inside or outside the loop.
while IFS=: read -d: -r path; do # `$IFS` is only set for the `read` command
echo $path
done <<< "${PATH:+"${PATH}:"}" # append an extra ':' if `$PATH` is set
You can check the value of $IFS,
IFS='xxxxxxxx'
while IFS=: read -d: -r path; do
echo "${IFS}${path}"
done <<< "${PATH:+"${PATH}:"}"
and the output will be something like this.
xxxxxxxx/usr/local/bin
xxxxxxxx/usr/bin
xxxxxxxx/bin
Reference to another question on StackExchange.
for p in $(echo $MANPATH | tr ":" " ") ;do
echo $p
done
IFS=:
arr=(${MANPATH})
for path in "${arr[#]}" ; do # <- quotes required
echo $path
done
... it does take care of spaces :o) but also adds empty elements if you have something like:
:/usr/bin::/usr/lib:
... then index 0,2 will be empty (''), cannot say why index 4 isnt set at all
This can also be solved with Python, on the command line:
python -c "import os,sys;[os.system(' '.join(sys.argv[1:]).format(p)) for p in os.getenv('PATH').split(':')]" echo {}
Or as an alias:
alias foreachpath="python -c \"import os,sys;[os.system(' '.join(sys.argv[1:]).format(p)) for p in os.getenv('PATH').split(':')]\""
With example usage:
foreachpath echo {}
The advantage to this approach is that {} will be replaced by each path in succession. This can be used to construct all sorts of commands, for instance to list the size of all files and directories in the directories in $PATH. including directories with spaces in the name:
foreachpath 'for e in "{}"/*; do du -h "$e"; done'
Here is an example that shortens the length of the $PATH variable by creating symlinks to every file and directory in the $PATH in $HOME/.allbin. This is not useful for everyday usage, but may be useful if you get the too many arguments error message in a docker container, because bitbake uses the full $PATH as part of the command line...
mkdir -p "$HOME/.allbin"
python -c "import os,sys;[os.system(' '.join(sys.argv[1:]).format(p)) for p in os.getenv('PATH').split(':')]" 'for e in "{}"/*; do ln -sf "$e" "$HOME/.allbin/$(basename $e)"; done'
export PATH="$HOME/.allbin"
This should also, in theory, speed up regular shell usage and shell scripts, since there are fewer paths to search for every command that is executed. It is pretty hacky, though, so I don't recommend that anyone shorten their $PATH this way.
The foreachpath alias might come in handy, though.
Combining ideas from:
https://stackoverflow.com/a/29949759 - gniourf_gniourf
https://stackoverflow.com/a/31017384 - Yi H.
code:
PATHVAR='foo:bar baz:spam:eggs:' # demo path with space and empty
printf '%s:\0' "$PATHVAR" | while IFS=: read -d: -r p; do
echo $p
done | cat -n
output:
1 foo
2 bar baz
3 spam
4 eggs
5
You can use Bash's for X in ${} notation to accomplish this:
for p in ${PATH//:/$'\n'} ; do
echo $p;
done
OP's update wants to ls the resulting folders, and has pointed out that ls only requires a space-separated list.
ls $(echo $PATH | tr ':' ' ') is nice and simple and should fit the bill nicely.
After reading the Bash man pages and with respect to this post, I am still having trouble understanding what exactly the eval command does and which would be its typical uses.
For example, if we do:
$ set -- one two three # Sets $1 $2 $3
$ echo $1
one
$ n=1
$ echo ${$n} ## First attempt to echo $1 using brackets fails
bash: ${$n}: bad substitution
$ echo $($n) ## Second attempt to echo $1 using parentheses fails
bash: 1: command not found
$ eval echo \${$n} ## Third attempt to echo $1 using 'eval' succeeds
one
What exactly is happening here and how do the dollar sign and the backslash tie into the problem?
eval takes a string as its argument, and evaluates it as if you'd typed that string on a command line. (If you pass several arguments, they are first joined with spaces between them.)
${$n} is a syntax error in bash. Inside the braces, you can only have a variable name, with some possible prefix and suffixes, but you can't have arbitrary bash syntax and in particular you can't use variable expansion. There is a way of saying “the value of the variable whose name is in this variable”, though:
echo ${!n}
one
$(…) runs the command specified inside the parentheses in a subshell (i.e. in a separate process that inherits all settings such as variable values from the current shell), and gathers its output. So echo $($n) runs $n as a shell command, and displays its output. Since $n evaluates to 1, $($n) attempts to run the command 1, which does not exist.
eval echo \${$n} runs the parameters passed to eval. After expansion, the parameters are echo and ${1}. So eval echo \${$n} runs the command echo ${1}.
Note that most of the time, you must use double quotes around variable substitutions and command substitutions (i.e. anytime there's a $): "$foo", "$(foo)". Always put double quotes around variable and command substitutions, unless you know you need to leave them off. Without the double quotes, the shell performs field splitting (i.e. it splits value of the variable or the output from the command into separate words) and then treats each word as a wildcard pattern. For example:
$ ls
file1 file2 otherfile
$ set -- 'f* *'
$ echo "$1"
f* *
$ echo $1
file1 file2 file1 file2 otherfile
$ n=1
$ eval echo \${$n}
file1 file2 file1 file2 otherfile
$eval echo \"\${$n}\"
f* *
$ echo "${!n}"
f* *
eval is not used very often. In some shells, the most common use is to obtain the value of a variable whose name is not known until runtime. In bash, this is not necessary thanks to the ${!VAR} syntax. eval is still useful when you need to construct a longer command containing operators, reserved words, etc.
Simply think of eval as "evaluating your expression one additional time before execution"
eval echo \${$n} becomes echo $1 after the first round of evaluation. Three changes to notice:
The \$ became $ (The backslash is needed, otherwise it tries to evaluate ${$n}, which means a variable named {$n}, which is not allowed)
$n was evaluated to 1
The eval disappeared
In the second round, it is basically echo $1 which can be directly executed.
So eval <some command> will first evaluate <some command> (by evaluate here I mean substitute variables, replace escaped characters with the correct ones etc.), and then run the resultant expression once again.
eval is used when you want to dynamically create variables, or to read outputs from programs specifically designed to be read like this. See Eval command and security issues for examples. The link also contains some typical ways in which eval is used, and the risks associated with it.
In my experience, a "typical" use of eval is for running commands that generate shell commands to set environment variables.
Perhaps you have a system that uses a collection of environment variables, and you have a script or program that determines which ones should be set and their values. Whenever you run a script or program, it runs in a forked process, so anything it does directly to environment variables is lost when it exits. But that script or program can send the export commands to standard output.
Without eval, you would need to redirect standard output to a temporary file, source the temporary file, and then delete it. With eval, you can just:
eval "$(script-or-program)"
Note the quotes are important. Take this (contrived) example:
# activate.sh
echo 'I got activated!'
# test.py
print("export foo=bar/baz/womp")
print(". activate.sh")
$ eval $(python test.py)
bash: export: `.': not a valid identifier
bash: export: `activate.sh': not a valid identifier
$ eval "$(python test.py)"
I got activated!
The eval statement tells the shell to take eval’s arguments as commands and run them through the command-line. It is useful in a situation like below:
In your script if you are defining a command into a variable and later on you want to use that command then you should use eval:
a="ls | more"
$a
Output:
bash: command not found: ls | more
The above command didn't work as ls tried to list file with name pipe (|) and more. But these files are not there:
eval $a
Output:
file.txt
mailids
remote_cmd.sh
sample.txt
tmp
Update: Some people say one should -never- use eval. I disagree. I think the risk arises when corrupt input can be passed to eval. However there are many common situations where that is not a risk, and therefore it is worth knowing how to use eval in any case. This stackoverflow answer explains the risks of eval and alternatives to eval. Ultimately it is up to the user to determine if/when eval is safe and efficient to use.
The bash eval statement allows you to execute lines of code calculated or acquired, by your bash script.
Perhaps the most straightforward example would be a bash program that opens another bash script as a text file, reads each line of text, and uses eval to execute them in order. That's essentially the same behavior as the bash source statement, which is what one would use, unless it was necessary to perform some kind of transformation (e.g. filtering or substitution) on the content of the imported script.
I rarely have needed eval, but I have found it useful to read or write variables whose names were contained in strings assigned to other variables. For example, to perform actions on sets of variables, while keeping the code footprint small and avoiding redundancy.
eval is conceptually simple. However, the strict syntax of the bash language, and the bash interpreter's parsing order can be nuanced and make eval appear cryptic and difficult to use or understand. Here are the essentials:
The argument passed to eval is a string expression that is calculated at runtime. eval will execute the final parsed result of its argument as an actual line of code in your script.
Syntax and parsing order are stringent. If the result isn't an executable line of bash code, in scope of your script, the program will crash on the eval statement as it tries to execute garbage.
When testing you can replace the eval statement with echo and look at what is displayed. If it is legitimate code in the current context, running it through eval will work.
The following examples may help clarify how eval works...
Example 1:
eval statement in front of 'normal' code is a NOP
$ eval a=b
$ eval echo $a
b
In the above example, the first eval statements has no purpose and can be eliminated. eval is pointless in the first line because there is no dynamic aspect to the code, i.e. it already parsed into the final lines of bash code, thus it would be identical as a normal statement of code in the bash script. The 2nd eval is pointless too, because, although there is a parsing step converting $a to its literal string equivalent, there is no indirection (e.g. no referencing via string value of an actual bash noun or bash-held script variable), so it would behave identically as a line of code without the eval prefix.
Example 2:
Perform var assignment using var names passed as string values.
$ key="mykey"
$ val="myval"
$ eval $key=$val
$ echo $mykey
myval
If you were to echo $key=$val, the output would be:
mykey=myval
That, being the final result of string parsing, is what will be executed by eval, hence the result of the echo statement at the end...
Example 3:
Adding more indirection to Example 2
$ keyA="keyB"
$ valA="valB"
$ keyB="that"
$ valB="amazing"
$ eval eval \$$keyA=\$$valA
$ echo $that
amazing
The above is a bit more complicated than the previous example, relying more heavily on the parsing-order and peculiarities of bash. The eval line would roughly get parsed internally in the following order (note the following statements are pseudocode, not real code, just to attempt to show how the statement would get broken down into steps internally to arrive at the final result).
eval eval \$$keyA=\$$valA # substitution of $keyA and $valA by interpreter
eval eval \$keyB=\$valB # convert '$' + name-strings to real vars by eval
eval $keyB=$valB # substitution of $keyB and $valB by interpreter
eval that=amazing # execute string literal 'that=amazing' by eval
If the assumed parsing order doesn't explain what eval is doing enough, the third example may describe the parsing in more detail to help clarify what is going on.
Example 4:
Discover whether vars, whose names are contained in strings, themselves contain string values.
a="User-provided"
b="Another user-provided optional value"
c=""
myvarname_a="a"
myvarname_b="b"
myvarname_c="c"
for varname in "myvarname_a" "myvarname_b" "myvarname_c"; do
eval varval=\$$varname
if [ -z "$varval" ]; then
read -p "$varname? " $varname
fi
done
In the first iteration:
varname="myvarname_a"
Bash parses the argument to eval, and eval sees literally this at runtime:
eval varval=\$$myvarname_a
The following pseudocode attempts to illustrate how bash interprets the above line of real code, to arrive at the final value executed by eval. (the following lines descriptive, not exact bash code):
1. eval varval="\$" + "$varname" # This substitution resolved in eval statement
2. .................. "$myvarname_a" # $myvarname_a previously resolved by for-loop
3. .................. "a" # ... to this value
4. eval "varval=$a" # This requires one more parsing step
5. eval varval="User-provided" # Final result of parsing (eval executes this)
Once all the parsing is done, the result is what is executed, and its effect is obvious, demonstrating there is nothing particularly mysterious about eval itself, and the complexity is in the parsing of its argument.
varval="User-provided"
The remaining code in the example above simply tests to see if the value assigned to $varval is null, and, if so, prompts the user to provide a value.
I originally intentionally never learned how to use eval, because most people will recommend to stay away from it like the plague. However I recently discovered a use case that made me facepalm for not recognizing it sooner.
If you have cron jobs that you want to run interactively to test, you might view the contents of the file with cat, and copy and paste the cron job to run it. Unfortunately, this involves touching the mouse, which is a sin in my book.
Lets say you have a cron job at /etc/cron.d/repeatme with the contents:
*/10 * * * * root program arg1 arg2
You cant execute this as a script with all the junk in front of it, but we can use cut to get rid of all the junk, wrap it in a subshell, and execute the string with eval
eval $( cut -d ' ' -f 6- /etc/cron.d/repeatme)
The cut command only prints out the 6th field of the file, delimited by spaces. Eval then executes that command.
I used a cron job here as an example, but the concept is to format text from stdout, and then evaluate that text.
The use of eval in this case is not insecure, because we know exactly what we will be evaluating before hand.
I've recently had to use eval to force multiple brace expansions to be evaluated in the order I needed. Bash does multiple brace expansions from left to right, so
xargs -I_ cat _/{11..15}/{8..5}.jpg
expands to
xargs -I_ cat _/11/8.jpg _/11/7.jpg _/11/6.jpg _/11/5.jpg _/12/8.jpg _/12/7.jpg _/12/6.jpg _/12/5.jpg _/13/8.jpg _/13/7.jpg _/13/6.jpg _/13/5.jpg _/14/8.jpg _/14/7.jpg _/14/6.jpg _/14/5.jpg _/15/8.jpg _/15/7.jpg _/15/6.jpg _/15/5.jpg
but I needed the second brace expansion done first, yielding
xargs -I_ cat _/11/8.jpg _/12/8.jpg _/13/8.jpg _/14/8.jpg _/15/8.jpg _/11/7.jpg _/12/7.jpg _/13/7.jpg _/14/7.jpg _/15/7.jpg _/11/6.jpg _/12/6.jpg _/13/6.jpg _/14/6.jpg _/15/6.jpg _/11/5.jpg _/12/5.jpg _/13/5.jpg _/14/5.jpg _/15/5.jpg
The best I could come up with to do that was
xargs -I_ cat $(eval echo _/'{11..15}'/{8..5}.jpg)
This works because the single quotes protect the first set of braces from expansion during the parsing of the eval command line, leaving them to be expanded by the subshell invoked by eval.
There may be some cunning scheme involving nested brace expansions that allows this to happen in one step, but if there is I'm too old and stupid to see it.
You asked about typical uses.
One common complaint about shell scripting is that you (allegedly) can't pass by reference to get values back out of functions.
But actually, via "eval", you can pass by reference. The callee can pass back a list of variable assignments to be evaluated by the caller. It is pass by reference because the caller can allowed to specify the name(s) of the result variable(s) - see example below. Error results can be passed back standard names like errno and errstr.
Here is an example of passing by reference in bash:
#!/bin/bash
isint()
{
re='^[-]?[0-9]+$'
[[ $1 =~ $re ]]
}
#args 1: name of result variable, 2: first addend, 3: second addend
iadd()
{
if isint ${2} && isint ${3} ; then
echo "$1=$((${2}+${3}));errno=0"
return 0
else
echo "errstr=\"Error: non-integer argument to iadd $*\" ; errno=329"
return 1
fi
}
var=1
echo "[1] var=$var"
eval $(iadd var A B)
if [[ $errno -ne 0 ]]; then
echo "errstr=$errstr"
echo "errno=$errno"
fi
echo "[2] var=$var (unchanged after error)"
eval $(iadd var $var 1)
if [[ $errno -ne 0 ]]; then
echo "errstr=$errstr"
echo "errno=$errno"
fi
echo "[3] var=$var (successfully changed)"
The output looks like this:
[1] var=1
errstr=Error: non-integer argument to iadd var A B
errno=329
[2] var=1 (unchanged after error)
[3] var=2 (successfully changed)
There is almost unlimited band width in that text output! And there are more possibilities if the multiple output lines are used: e.g., the first line could be used for variable assignments, the second for continuous 'stream of thought', but that's beyond the scope of this post.
In the question:
who | grep $(tty | sed s:/dev/::)
outputs errors claiming that files a and tty do not exist. I understood this to mean that tty is not being interpreted before execution of grep, but instead that bash passed tty as a parameter to grep, which interpreted it as a file name.
There is also a situation of nested redirection, which should be handled by matched parentheses which should specify a child process, but bash is primitively a word separator, creating parameters to be sent to a program, therefore parentheses are not matched first, but interpreted as seen.
I got specific with grep, and specified the file as a parameter instead of using a pipe. I also simplified the base command, passing output from a command as a file, so that i/o piping would not be nested:
grep $(tty | sed s:/dev/::) <(who)
works well.
who | grep $(echo pts/3)
is not really desired, but eliminates the nested pipe and also works well.
In conclusion, bash does not seem to like nested pipping. It is important to understand that bash is not a new-wave program written in a recursive manner. Instead, bash is an old 1,2,3 program, which has been appended with features. For purposes of assuring backward compatibility, the initial manner of interpretation has never been modified. If bash was rewritten to first match parentheses, how many bugs would be introduced into how many bash programs? Many programmers love to be cryptic.
As clearlight has said, "(p)erhaps the most straightforward example would be a bash program that opens another bash script as a text file, reads each line of text, and uses eval to execute them in order". I'm no expert, but the textbook I'm currently reading (Shell-Programmierung by Jürgen Wolf) points to one particular use of this that I think would be a valuable addition to the set of potential use cases collected here.
For debugging purposes, you may want to go through your script line by line (pressing Enter for each step). You could use eval to execute every line by trapping the DEBUG signal (which I think is sent after every line):
trap 'printf "$LINENO :-> " ; read line ; eval $line' DEBUG
I like the "evaluating your expression one additional time before execution" answer, and would like to clarify with another example.
var="\"par1 par2\""
echo $var # prints nicely "par1 par2"
function cntpars() {
echo " > Count: $#"
echo " > Pars : $*"
echo " > par1 : $1"
echo " > par2 : $2"
if [[ $# = 1 && $1 = "par1 par2" ]]; then
echo " > PASS"
else
echo " > FAIL"
return 1
fi
}
# Option 1: Will Pass
echo "eval \"cntpars \$var\""
eval "cntpars $var"
# Option 2: Will Fail, with curious results
echo "cntpars \$var"
cntpars $var
The curious results in option 2 are that we would have passed two parameters as follows:
First parameter: "par1
Second parameter: par2"
How is that for counter intuitive? The additional eval will fix that.
It was adapted from another answer on How can I reference a file for variables using Bash?