i have the following function in a bash script which does not work?
do_get() {
cmd='<command version="33" cmd="GETINFO" $3</command>'
echo $cmd
}
Now, if i echo $3 right before the cmd variable it echos out 1234 which i am passing as a 3 argument when executing this. BUT it shows just $3 when i do an echo $cmd.
i tried a couple of things like such below thinking its getting striped out
'$3' but it then shows blank
'"$3"' same as above
The variable doesn't expand when inside single quotes. You need to use double quotes instead, but since you have double quotes on the inside, you need to make sure you remember to escape those as well.
do_get() {
cmd="<command version=\"33\" cmd=\"GETINFO\" $3</command>"
echo $cmd
}
Newer versions of bash add a -v flag to the printf command that makes assignments like this a little easier on the eye, in that quoting is reduced.
printf -v cmd '<command version="33" cmd="GETINFO" %d </command' "$3"
The single quote in Bash prevents variable substitution. In order for the third parameter to be substituted, you should enclose your string in double quotes. of course, you have the problem of the double quotes that are part of your string, so they need to be escaped with a backslash:
cmd="<command version=\"33\" cmd=\"GETINFO\" $3 </command>"
Related
I am extracting a part from an existing file and storing it as a string in a variable.The string looks something like this.
var="*a<br>*b<br>*c"
Now as * is a special character in unix it doesnot work in further operations(like sed,grep) until I put an escape character infront of every *
Thats why,I am doing something like this -
echo $var | sed 's/\*/\\*/g'
On running this command in bash we get
echo $var | sed 's/\*/\\*/g'
\*a<br>\*b<br>\*c
which is the desired output,but when I try to store this in a variable, I am getting back my original variable like so
var=`echo $var | sed 's/\*/\\*/g'`
echo $var
*a<br>*b<br>*c
I am assuming this happens because the variable ignores the backslashes interpreting them as escape characters. How can I retain the backslashes and store them as in a variable?
The problem is caused by backticks. Use $( ) instead, and it goes away:
var="*a<br>*b<br>*c"
var=$(printf '%s\n' "$var" | sed 's/\*/\\*/g')
printf '%s\n' "$var"
(Why is this problem caused by backticks? Because the only way to nest them is to escape the inner ones with backslashes, so they necessarily change how backslashes behave; whereas $( ), because it uses different starting and ending sigils, can be nested natively).
That said, if your shell is one (like bash) with ksh-inspired extensions, you don't need sed at all here, as the shell can perform simple string replacements natively via parameter expansion:
var="*a<br>*b<br>*c"
printf '%s\n' "${var//'*'/'\*'}"
For background on why this answer uses printf instead of echo, see Why is printf better than echo? at [unix.se], or the APPLICATION USAGE section of the POSIX specification for echo.
I am trying to use a variable to store the parameters, here is the simple test:
#!/bin/bash
sed_args="-e \"s/aaaa/bbbb/g\""
echo $sed_args`
I expected the output to be
-e "s/aaaa/bbbb/g"
but it gives:
"s/aaaa/bbbb/g"
without the "-e"
I am new to bash, any comment is welcome. Thanks, maybe this is already answered somewhere.
You need an array to construct arguments dynamically:
#!/usr/bin/env bash
sed_args=('-e' 's/aaaa/bbbb/g')
echo "${sed_args[#]}"
When you use the variable without double quotes, it gets word split by the shell even before echo sees the value(s). Then, the bash's builtin echo interprets -e as a parameter for itself (which is normally used to turn on interpretation of backslash escapes).
When you double quote the variable, it won't be split and will be interpreted as a single argument to echo:
echo "$sed_args"
For strings you don't control, it's safer to use printf as it doesn't take any arguments after the format string:
printf %s "$string"
I have an script that echo the input given, into a file as follows:
echo $# > file.txt
When I pass a sting like "\"" I want it to exactly print "\"" to the file however it prints ".
My question is how can I print all characters of a variable containing a string without considering escapes?
When I use echo in bash like echo "\"" it only prints " while when I use echo '"\""' it prints it correctly. I thought maybe that would be the solution to use single quotes around the variable, however I cannot get the value of a variable inside single quotes.
First, note that
echo $# > file.txt
can fail in several ways. Shellcheck identifies one problem (missing quotes on $#). See the accepted, and excellent, answer to Why is printf better than echo? for others.
Second, as others have pointed out, there is no practical way for a Bash program to know exactly how parameters were specified on the command line. For instance, for all of these invocations
prog \"
prog "\""
prog '"'
the code in prog will see a $1 value that consists of one double-quote character. Any quoting characters that are used in the invocation of prog are removed by the quote removal part of the shell expansions done by the parent shell process.
Normally that doesn't matter. If variables or parameters contain values that would need to be quoted when entered as literals (e.g. "\"") they can be used safely, including passing them as parameters to other programs, by quoting uses of the variable or parameter (e.g. "$1", "$#", "$x").
There is a problem with variables or parameters that require quoting when entered literally if you need to write them in a way that they can be reused as shell input (e.g. by using eval or source/.). Bash supports the %q format specification to the printf builtin to handle this situation. It's not clear what the OP is trying to do, but one possible solution to the question is:
if (( $# > 0 )) ; then
printf -v quoted_params '%q ' "$#" # Add all parameters to 'quoted_params'
printf '%s\n' "${quoted_params% }" # Remove trailing space when printing
fi >file.txt
That creates an empty 'file.txt' when no positional parameters are provided. The code would need to be changed if that is not what is required.
If you run echo \", the function of the backslash in bash is to escape the character after it. This actually enables you to use the double quotes as an argument. You cannot use a backslash by itself; if you want to have a backslash as an argument you need to use another slash to escape that: echo \\
Now if you want to create a string where these things are not escaped, use single quotes: echo '\'
See for a better explanation this post: Difference between single and double quotes in Bash
I'm writing bash script like this:
VF_ETH=$(command)
if [ -n "$VF_ETH" ] ; then
echo "ixgbevf eth: "$VF_ETH
fi
command is a linux command, my question is:
$VF_ETH is to get value of VF_ETH, why use "" to quote it in line2 in shell test?
if I do not use "" to quote it, will test failed?
if use "" to quote a value is to make it into string, why not use in echo in line3?
Thank you
Assuming you get an actual command stored in VF_ETH variable, which contains spaces. Now if you use if [ -n $VF_ETH ] and when shell expands the variable, there will be multiple parameters to -n whereas it expects only one. Hence you might get something like binary operator expected error.
Also in the echo command, it is not mandatory to have only one parameter. Hence even if you are not using double quotes, it works.
Hence to avoid it, always use double quotes while expanding variables.
Also use https://www.shellcheck.net/ to check your script and it will give you correct information on where your script is wrong/not as per standard.
You should always double quote variables used in command line arguments and within [ ... ] tests, for example:
ls "$var"
echo "$var"
[ -f "$var" ]
test -f "$var"
In all the above examples the commands used will receive different values with and without the double quotes, when the value of $var starts with a bunch of spaces and contains some spaces. For example:
var=" an apple"
echo "$var" # prints " an apple"
echo $var # prints "an apple", without the leading space
You don't need to quote them in simple assignments, for example:
a=$var
a=$var$b
If the assignment contains spaces or other special characters, then you have to quote, or escape those special characters:
a="$var $b"
a=$var\ $b
a=$var" "$b
All the above are equivalent, but the first one is probably the easiest to read, and therefore recommended.
You don't need to quote special variables that never have unsafe values, for example:
test $? = 0
If you're unsure, or not yet confident, then a good role of thumb is to double quote always.
For 1. and 2. If you set $VF_ETH="x -a -z x" and test it with code:
if [ -n $VF_ETH ] ; then
echo yes
else
echo nope
fi
the output will be nope as the the inside of the square brackets would expand to -n x AND -z x (nonempty and empty). Adding quotes around the variable would fix that.
Set $VF_ETH="*". Now if you echo $foo bash would do wildcard expansion and echo would output the contents of your current directory.
Would you please explain why the following shell command wouldn't work:
sh-3.1$ echo $MYPATH
/opt/Application/DATA/CROM/my_application
sh-3.1$ awk '{print substr($MYPATH,3)}'
Thanks
Best Regards
MYPATH is not going to be substituted by the shell since the string uses single quotes. Consider the following:
csh$ echo '{print substr($USER,3)}'
{print substr($USER,3)}
csh$ echo "{print substr($USER,3)}"
{print substr(dshawley,3)}
The usage of single quotes instructs the shell to pass the string argument to the program as-is. Double quotes tell the shell to perform variable expansion on the argument before passing it to the program. This is a basic shell feature that is common amongst shells and some programming languages (e.g., perl).
The next problem that you are going to run into is that awk will want quotes around the first parameter to substr or the parse will fail. You will probably see an "Illegal variable name" warning in this case. This is where I get lost with csh since I have no clue how to properly escape a double-quote within a quoted string. In bash/sh/ksh, you would do the following:
sh$ awk "{print substr(\"$USER\",3)}"
input
^D
hawley
sh$
Just in case you do not already know this, awk will require an input stream before it is going to do anything. I had to type "input" and the EOF character for the little example.
Quoting and escaping
"string" is a weak quote. Enclosed whitespace and wildcards are taken as literals, but variable and command substitutions are still performed.
'string' is a strong quote. The entire enclosed string is taken as a literal.
You can use the -v option to pass variable to awk:
awk -v mypath=$MYPATH 'BEGIN{print substr(mypath, 3)}'