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 $cmd2
If you're not sure about the operation of a script you could always use the debug mode to see if you can determine the error.
bash -x scriptname
This will run the command and display the output of variable evaluations. Hopefully this will reveal any issues with syntax.
Recently I came across an unusual use of echo to assign variables in a client's ksh scripts.
For example, there are many instances such as the following
a='something'
b='else'
c=`echo "${a} ${b}"`
I have been unable to come up with any reason why someone may have done this.
Could there be some legacy reason for this?
(I've been doing shell for 30+ years, and never before have I seen this)
Or is it just ignorance?
There is no compelling reason whatsoever for this, either in current bash, or its POSIX sh or Bourne predecessors.
c="$a $b"
...is a complete replacement for the code given, and runs far faster (try putting it in a loop; command substitutions, as created by backticks, fork off a new shell as a subprocess and read its stdout -- a high-overhead operation).
What you saw is an example of bad use of echo because c could be declared as:
c="$a $b"
A common use of echo is when you need comands to filter output, for example
$ line="100090 $100,00 Mary"
$ name=`echo "$line" | grep -Eo "[a-zA-Z]+$"`
echo $name
Mary
But it would be more efficient if you don't use echo at all. The same thing above can be done with "read", without creating a new process:
$ line="100090 $100,00 Paul"
$ read -r _ _ name _ <<<"$line"
echo $name
Paul
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'm very new to bash scripting and am running into an issue when using double brackets. I can't seem to get them to work at all in Ubuntu Server 11.10. My script below is in if_test.sh.
#!/bin/bash
if [[ "14"=="14" ]]; then
echo "FOO"
fi
When I run this simple shell script the output I get is: if_test.sh: 5: [[: not found
It seems that I'm running GNU bash version 4.2.10 after running bash --version from the terminal. Any help would be greatly appreciated. Thanks!
The problem lies in your script invocation. You're issuing:
$ sudo sh if_test.sh
On Ubuntu systems, /bin/sh is dash, not bash, and dash does not support the double bracket keyword (or didn't at the time of this posting, I haven't double-checked). You can solve your problem by explicitly invoking bash instead:
$ sudo bash if_test.sh
Alternatively, you can make your script executable and rely on the shebang line:
$ chmod +x if_test.sh
$ sudo ./if_test.sh
Also note that, when used between double square brackets, == is a pattern matching operator, not the equality operator. If you want to test for equality, you can either use -eq:
if [[ "14" -eq "14" ]]; then
echo "FOO"
fi
Or double parentheses:
if (( 14 == 14 )); then
echo "FOO"
fi
My answer doesn't apply to #lots_of_questions's question specifically, but you can also run into this problem if you have the wrong specifier at the top of your script:
#!/bin/sh
if [[ ... ]]
...
You should change that to
#!/bin/bash
if [[ ... ]]
...
Since you are new to scripting, you may be unaware that [[ is a bashism. You may not even know what a bashism is, but both answers given so far are leading you down a path towards a stunted scripting future by promoting their use.
To check if a variable matches a string in any flavor of Bourne shell, you can do test $V = 14 If you want to compare integers, use test $V -eq 14. The only difference is that the latter will generate an error if $V does not look like an integer. There are good reasons to quote the variable (test "$V" = 14), but the quotes are often unnecessary and I believe are the root cause of a common confusion, since "14"=="14" is identical to "14==14" where it is more obvious that '==' is not being used as an operator.
There are several things to note: use a single '=' instead of '==' because not all shells recognize '==', the [ command is identical to test but requires a final argument of ] and many sh coding guidelines recommend using test because it often generates more understandable code, [[ is only recognized by a limited number of shells (this is your primary problem, as your shell does not appear to recognize [[ and is looking for a command of that name. This is surprising if your shebang does indeed specify /bin/bash instead of /bin/sh).
x="a=b"
`echo $x`
echo $a
I expect the second line to generate "a=b", and execute it in the context of the main shell, resulting in a new variable a with value b.
However, what I really get (if I enter the commands manually) is the error message after the second line, bash: a=b: command not found
Why is that so?
Try
eval $x
(And we need 30 characters for this answer to be posted)
What your first echo line does is running in a subshell and returns its value to the callee.. The same result is achieved using $() and is - by the way - easier to use than backticks.
So, what you are doing is first running echo $x (which returns a=b). And, because of the backticks, a=b is returned to the shell that tries to run that line as a command which - obviously - won't work.
Try this in a shell:
$(echo ls)
And you will clearly see what is happening.
It's because of the order in which bash parses the command line. It looks for variable definitions (e.g. a=b) before performing variable and command substitution (e.g. commands in backticks). Because of this, by the time echo $x is replaced by a=b, it's too late for bash to see this as a variable definition and it's parsed as a command instead. The same thing would've happened if you'd just used $x as the command (instead of echo in backticks). As in #mvds's answer, the eval command can be used to force the command to be reparsed from the beginning, meaning that it'll be recognized as a variable definition:
$ x="a=b"
$ `echo $x`
-bash: a=b: command not found
$ $(echo $x) # Exact same thing, but with cleaner syntax
-bash: a=b: command not found
$ $x # This also does the same thing, but without some extra steps
-bash: a=b: command not found
$ eval "$x" # This will actually work
$ echo $a
b
$ a= # Start over
$ eval "$(echo "$x")" # Another way of doing the same thing, with extra steps
$ echo $a
b
Note that when using eval I've put all of the references to $x in double-quotes -- this is to prevent the later phases of bash parsing (e.g. word splitting) from happening twice, since bash will finish its regular parsing process, then recognize the eval command, and then redo the entire parsing process again. It's really easy to get unexpected results from using eval, and this removes at least some of the potential for trouble.
Did you try $x in that funny apostrophes? Without echo, echo seems to be only for displaying string, not execute commands.