In order to get eval to work on commands that contain spaces inside one of the parameters, I have only found this to work so far:
eval 'sed 's/foo/foo'" "'bar/g' filename'
In a hypothetical program where users would enter a command and then the command and arguments to be fed to eval, this isn't a very elegant or robust solution. Are there any other ways to run the eval command so that the interface for my_command can be a little more user friendly? The following is an example of how the program accepts arguments now.
my_command 'sed 's/foo/foo'" "'bar/g' filename'
I would like the interface to work something like this:
my_command sed 's/foo/foo bar/g' filename
edit:
I'll try asking a different question:
How do I get bash to read input from the command line literally? I want the exact input to be preserved, so if there are quotes I want to keep them. I can accomplish what I want to do by using egrep to read from file and then sanitizing the input, like so:
egrep '/.*/' filename |
sed 's/\(.*\)['"'"']\(.*\) \(.*\)['"'"']\(.*\)/\1'"\'"'\2" "\3'"\'"'\4/g'
with "filename" containing this line
sed 's/foo/foo bar/g' file
this gives me the desired output of:
sed 's/foo/foo" "bar/g' file
Problem here is that I can't echo "$#" because bash interprets the quotes. I want the literal input without having to read from file.
Original question
For your preferred use-case, you'd simply write (inside my_command):
"$#"
to execute the command as given.
Your eval line is odd:
eval 'sed 's/foo/foo'" "'bar/g' filename'
Because of the way single quotes don't nest, it is equivalent to:
eval 'sed s/foo/foo" "bar/g filename'
Revised question
Possible solution:
egrep '/.*/' filename | sh
This feeds what is in filename directly to the shell for interpretation. Given file containing:
Some text containing foo; and bar.
More foo bar?
More text; more foo and bar; more foo bar beyond the possibility of unfooing.
The output is:
Some text containing foo bar; and bar.
More foo bar bar?
More text; more foo bar and bar; more foo bar bar beyond the possibility of unfoo baring.
Fixing quotes is hard!
Note that your complex sed script is not complex enough. Given filename containing:
sed 's/foo/foo bar/g' file
sed 's/foo bar/foo bar baz/g' file
the output from:
egrep '/.*/' filename |
sed 's/\(.*\)['"'"']\(.*\) \(.*\)['"'"']\(.*\)/\1'"\'"'\2" "\3'"\'"'\4/g'
is:
sed 's/foo/foo" "bar/g' file
sed 's/foo bar/foo bar" "baz/g' file
which has not solved all the problems for the eval.
I've spent a lot of time, on and off, working on such issues over quite a long period of time (a quarter century is no exaggeration), and it isn't trivial. You can find one discussion in extenso in How to iterate over arguments in bash script. Somewhere, I have another answer which goes through gyrations about this stuff, but I can't immediately find it (where 'immediately' means an hour or so of distracted searching, where the distractions were sets of duplicate questions, etc). It may have been deleted, or I may have looked in the wrong place.
your design is flawed. Create a user interface that doesn't let them input commands directly. give options, or let them enter the parameters only.
At the back end, you do your sanitization check on the parameters before calling sed or other tools desired. You don't have to use eval
Array quoting
The following keeps spaces in arguments by quoting each element of array:
function token_quote {
local quoted=()
for token; do
quoted+=( "$(printf '%q' "$token")" )
done
printf '%s\n' "${quoted[*]}"
}
Example usage:
$ token_quote token 'single token' token
token single\ token token
Above, note the single token's space is quoted as \.
$ set $(token_quote token 'single token' token)
$ eval printf '%s\\n' "$#"
token
single token
token
$
This shows that the tokens are indeed kept separate.
Given some untrusted user input:
% input="Trying to hack you; date"
Construct a command to eval:
% cmd=(echo "User gave:" "$input")
Eval it, with seemingly correct quoting:
% eval "$(echo "${cmd[#]}")"
User gave: Trying to hack you
Thu Sep 27 20:41:31 +07 2018
Note you were hacked. date was executed rather than being printed literally.
Instead with token_quote():
% eval "$(token_quote "${cmd[#]}")"
User gave: Trying to hack you; date
%
eval isn't evil - it's just misunderstood :)
It can actually work as you desire. Use "$#" - this will pass all the arguments exactly as they were given on the command line.
If my_command.sh contains:
sed "$#"
Then my_command.sh 's/foo/foo bar/g' filename will do exactly what you expect.
Related
I am having a problem where cmd1 works, but not cmd2 in my Bash script ending in .sh. I have made the Bash script executable.
Additionally, I can execute cmd2 just fine from my Bash terminal. I have tried to make a minimally reproducible example, but my larger goal is to run a complicated executable with command line arguments and pass output to a file that may or may not exist (rather than displaying the output in the terminal).
Replacing > with >> also gives the same error in the script, but not the terminal.
My Bash script:
#!/bin/bash
cmd1="cat test.txt"
cmd2="cat test.txt > a"
echo $cmd1
$cmd1
echo $cmd2
$cmd2
test.txt has the words "dog" and "cat" on two separate lines without quotes.
Short answer: see BashFAQ #50: I'm trying to put a command in a variable, but the complex cases always fail!.
Long answer: the shell expands variable references (like $cmd1) toward the end of the process of parsing a command line, after it's done parsing redirects (like > a is supposed to be) and quotes and escapes and... In fact, the only thing it does with the expanded value is word splitting (e.g. treating cat test.txt > a as "cat" followed by "test.txt", ">", and finally "a", rather than a single string) and wildcard expansion (e.g. if $cmd expanded to cat *.txt, it'd replace the *.txt part with a list of matching files). (And it skips word splitting and wildcard expansion if the variable is in double-quotes.)
Partly as a result of this, the best way to store commands in variables is: don't. That's not what they're for; variables are for data, not commands. What you should do instead, though, depends on why you were storing the command in a variable.
If there's no real reason to store the command in a variable, then just use the command directly. For conditional redirects, just use a standard if statement:
if [ -f a ]; then
cat test.txt > a
else
cat test.txt
fi
If you need to define the command at one point, and use it later; or want to use the same command over and over without having to write it out in full each time, use a function:
cmd2() {
cat test.txt > a
}
cmd2
It sounds like you may need to be able to define the command differently depending on some condition, you can actually do that with a function as well:
if [ -f a ]; then
cmd() {
cat test.txt > a
}
else
cmd() {
cat test.txt
}
fi
cmd
Alternately, you can wrap the command (without redirect) in a function, then use a conditional to control whether it redirects:
cmd() {
cat test.txt
}
if [ -f a ]; then
cmd > a
else
cmd
fi
It's also possible to wrap a conditional redirect into a function itself, then pipe output to it:
maybe_redirect_to() {
if [ -f "$1" ]; then
cat > "$1"
else
cat
fi
}
cat test.txt | maybe_redirect_to a
(This creates an extra cat process that isn't really doing anything useful, but if it makes the script cleaner, I'd consider that worth it. In this particular case, you could minimize the stray cats by using maybe_redirect_to a < test.txt.)
As a last resort, you can store the command string in a variable, and use eval to parse it. eval basically re-runs the shell parsing process from the beginning, meaning that it'll recognize things like redirects in the string. But eval has a well-deserved reputation as a bug magnet, because it's easy for it to treat parts of the string you thought were just data as command syntax, which can cause some really weird (& dangerous) bugs.
If you must use eval, at least double-quote the variable reference, so it runs through the parsing process just once, rather than sort-of-once-and-a-half as it would unquoted. Here's an example of what I mean:
cmd3="echo '5 * 3 = 15'"
eval "$cmd3"
# prints: 5 * 3 = 15
eval $cmd3
# prints: 5 [list of files in the current directory] 3 = 15
# ...unless there are any files with shell metacharacters in their names, in
# which case something more complicated might happen.
BashFAQ #50 discusses some other possible reasons and solutions. Note that the array approach will not work here, since arrays also get expanded after redirects are parsed.
If you pop an 'eval' in front of $cmd2 it should work as expected:
#!/bin/bash
cmd2="cat test.txt > a"
eval $cmd2
If you're not sure about the operation of a script you could always use the debug mode to see if you can determine the error.
bash -x scriptname
This will run the command and display the output of variable evaluations. Hopefully this will reveal any issues with syntax.
My Bash script uses printf to print output of some other commands with formatting applied, in the following manner (note the two leading spaces):
printf " %-16s %s\n" "foo:" "$(bar)"
The two leading spaces are there, because the first line in the file is a comment, and I like to keep things nicely aligned:
# foo
foo: bar
foo: bar
...
This works perfectly fine for commands with one-line output. However, when the output is multiline, the output looses the formatting of the subsequent lines.
For example:
printf " %-16s %s\n" "Contents:" "$(ls -a)"
Results in something like this:
Contents: .
..
foo
bar
Instead, what I am trying to achieve is multiline output, with each line formatted (in columns, if you like), which would look like this (note the lack of "header" in the subsequent lines):
Contents: .
..
foo
bar
My understanding so far is that printf with two arguments (the first one being a string, and the other one the result of the given command) treats the second argument as a single string, which can include newline characters within. Therefore, the output is actually correctly formatted, according to what I asked printf to do, but it is not what I am looking for.
I am aware about some of the pitfalls related to parsing the output of commands like ls, which exhausts my current possibilities of solving this problem. Also, it is possible that printf is not the best facility to do this.
I am considering substituting multiline string into comma-separated list, if no solution to this problem is possible, but this would be a last resort.
What would be the best method to achieve the formatting I need?
Thanks for your help.
printf " %-16s %s\n" "foo:" "$(bar | sed '2,$s/^/ /g')"
You may find it difficult to apply the same print instructions to a range of commands. Also output from ls can be tedious to examine because it selects an output format depending upon environment variables.
I found the following worked well for ls on my Mac:
x=$(ls -a);
x=${x//$'\n'/$'\n' };
printf " %-16s %s" "Contents:" "$x" $'\n'
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 $var
The curious results in option 2 are that we would have passed two parameters as follows:
First parameter: "par1
Second parameter: par2"
How is that for counter intuitive? The additional eval will fix that.
It was adapted from another answer on How can I reference a file for variables using Bash?
I have a variable in a shell script in which I'd like to format the data. The variable stores new data during every iteration of a loop. Each time the new data is stored, I'd like to insert a new line character. Here is how I'm trying to store the data into the variable.
VARIABLE="$VARIABLE '\n' SomeData"
Unfortunately, the output includes the literal '\n' Any help would be appreciative.
Try $'\n':
VAR=a
VAR="$VAR"$'\n'b
echo "$VAR"
gives me
a
b
A common technique is:
nl='
'
VARIABLE="PreviousData"
VARIABLE="$VARIABLE${nl}SomeData"
echo "$VARIABLE"
PreviousData
SomeData
Also common, to prevent inadvertently having your string start with a newline:
VARIABLE="$VARIABLE${VARIABLE:+$nl}SomeData"
(The expression ${VARIABLE:+$nl} will expand to a newline if and only if VARIABLE is set and non-empty.)
VAR="one"
VAR="$VAR.\n.two"
echo -e $VAR
Output:
one.
.two
Other than $'\n' you can use printf also like this:
VARIABLE="Foo Bar"
VARIABLE=$(printf "${VARIABLE}\nSomeData")
echo "$VARIABLE"
OUTPUT:
Foo Bar
SomeData
I had a problem with all the other solutions: when using a # followed by SPACE (quite common when writing in Markdown) both would get split onto a new line.
So, another way of doing it would involve using single quotes so that the "\n" get rendered.
FOO=$'# Markdown Title #\n'
BAR=$'Be *brave* and **bold**.'
FOOBAR="$FOO$BAR"
echo "$FOOBAR"
Output:
# Markdown Title #
Be *brave* and **bold**.
Single quote All special characters between these quotes lose their
special meaning.https://www.tutorialspoint.com/unix/unix-quoting-mechanisms.htm
So the syntax you use does something different that you want to achieve.
This is what you need:
The $'\X' construct makes the -e option in echo unnecessary.
https://linux.die.net/abs-guide/escapingsection.html
echo -e "something\nsomething"
or
echo "something"$'\n'"something"
It's a lot simpler than you think:
VARIABLE="$VARIABLE
SomeData"
Building upon the first two solutions, I'd do like shown below. Concatenating strings with the '+=' operator, somehow looks clearer to me.
Also rememeber to use printf as opposed to echo, you will save yourself so much trouble
sometext="This is the first line"
sometext+=$'\n\n'
sometext+="This is the second line AFTER the inserted new lines"
printf '%s' "${sometext}"
Outputs:
This is the first line
This is the third line AFTER the inserted new line
Your problem is in the echo command, in ash you have to use the option -e to expand special characters. This should work for you:
VAR="First line"
VAR="$VAR\nSecond line"
echo -e $VAR
This outputs
First line
Second line
I'm writing a shell script that should be somewhat secure, i.e., does not pass secure data through parameters of commands and preferably does not use temporary files. How can I pass a variable to the standard input of a command?
Or, if it's not possible, how can I correctly use temporary files for such a task?
Passing a value to standard input in Bash is as simple as:
your-command <<< "$your_variable"
Always make sure you put quotes around variable expressions!
Be cautious, that this will probably work only in bash and will not work in sh.
Simple, but error-prone: using echo
Something as simple as this will do the trick:
echo "$blah" | my_cmd
Do note that this may not work correctly if $blah contains -n, -e, -E etc; or if it contains backslashes (bash's copy of echo preserves literal backslashes in absence of -e by default, but will treat them as escape sequences and replace them with corresponding characters even without -e if optional XSI extensions are enabled).
More sophisticated approach: using printf
printf '%s\n' "$blah" | my_cmd
This does not have the disadvantages listed above: all possible C strings (strings not containing NULs) are printed unchanged.
(cat <<END
$passwd
END
) | command
The cat is not really needed, but it helps to structure the code better and allows you to use more commands in parentheses as input to your command.
Note that the 'echo "$var" | command operations mean that standard input is limited to the line(s) echoed. If you also want the terminal to be connected, then you'll need to be fancier:
{ echo "$var"; cat - ; } | command
( echo "$var"; cat - ) | command
This means that the first line(s) will be the contents of $var but the rest will come from cat reading its standard input. If the command does not do anything too fancy (try to turn on command line editing, or run like vim does) then it will be fine. Otherwise, you need to get really fancy - I think expect or one of its derivatives is likely to be appropriate.
The command line notations are practically identical - but the second semi-colon is necessary with the braces whereas it is not with parentheses.
This robust and portable way has already appeared in comments. It should be a standalone answer.
printf '%s' "$var" | my_cmd
or
printf '%s\n' "$var" | my_cmd
Notes:
It's better than echo, reasons are here: Why is printf better than echo?
printf "$var" is wrong. The first argument is format where various sequences like %s or \n are interpreted. To pass the variable right, it must not be interpreted as format.
Usually variables don't contain trailing newlines. The former command (with %s) passes the variable as it is. However tools that work with text may ignore or complain about an incomplete line (see Why should text files end with a newline?). So you may want the latter command (with %s\n) which appends a newline character to the content of the variable. Non-obvious facts:
Here string in Bash (<<<"$var" my_cmd) does append a newline.
Any method that appends a newline results in non-empty stdin of my_cmd, even if the variable is empty or undefined.
I liked Martin's answer, but it has some problems depending on what is in the variable. This
your-command <<< """$your_variable"""
is better if you variable contains " or !.
As per Martin's answer, there is a Bash feature called Here Strings (which itself is a variant of the more widely supported Here Documents feature):
3.6.7 Here Strings
A variant of here documents, the format is:
<<< word
The word is expanded and supplied to the command on its standard
input.
Note that Here Strings would appear to be Bash-only, so, for improved portability, you'd probably be better off with the original Here Documents feature, as per PoltoS's answer:
( cat <<EOF
$variable
EOF
) | cmd
Or, a simpler variant of the above:
(cmd <<EOF
$variable
EOF
)
You can omit ( and ), unless you want to have this redirected further into other commands.
Try this:
echo "$variable" | command
If you came here from a duplicate, you are probably a beginner who tried to do something like
"$variable" >file
or
"$variable" | wc -l
where you obviously meant something like
echo "$variable" >file
echo "$variable" | wc -l
(Real beginners also forget the quotes; usually use quotes unless you have a specific reason to omit them, at least until you understand quoting.)