Example
Var = '/etc/sysconfig/..'
export Var
bash script1.sh
in another script1
cat $Var
This is my Problem: The variable does not call the file in this path
Do this:
Var='/etc/sysconfig/..'
bash script1.sh "$Var"
Then in script1.sh:
Var=$1
cat "$Var"
The quotes around "$Var" are required to support paths containing spaces.
Your variable assignment is wrong, it should be:
Var='/etc/sysconfig/..'
No spaces around =.
If you want to send in a environment variable for one script only then you can use:
Var='/etc/sysconfig/..' ./my_script.sh
And inside my_script.sh:
printf "%s\n" "$Var"
# Will print /etc/sysconfig/..
If you want to send arguments to my_script.sh do what #JohnZwinck suggested. What I suggested is only to change environment variable and shouldn't be abused to send/receive regular variables to a command.
I think no need to to more thing
script 1
#!/bin/bash
a="/home/example" ### you can do with export command also export a="/home/example"
sctipt2 ## make effective
. script1;
cd $a
I can't understand the following code in a bash.
set `pwd` ; mfix=$1
It actually get the run directory name.But I don't how does it work.
What is the set command mean?
From the doc for the set:
This builtin is so complicated that it deserves its own section. set
allows you to change the values of shell options and set the
positional parameters, or to display the names and values of shell
variables.
e.g.
set v1 v2 v3 ; echo $1
will print
v1
The comand inside backticks is called as "command substitution". From the docs:
Bash performs the expansion by executing command and replacing the
command substitution with the standard output of the command, with any
trailing newlines deleted.
In your example, it sets the 1st positional argument $1 to the value of the result of execution of command inside the backticks. (called as command substitution). The command is pwd what shows the current working directory.
Anyway, if the path to the directory contains an space, the $1 will get only the first part of the path., e.g.
$ pwd
/some/path with/space
$ set `pwd`
$ echo $1
/some/path
$echo $2
with/space
Finally the all above is strange design, because you can simply:
mfix=$(pwd) #old school: mfix=`pwd`
It is better to use the $(command) instead of the backticks.
This code in bash put the result of the command pwd in the variable mfix.
You can print the result of the mfix variable by running
echo $mfix
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?
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.