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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.

masec
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    You can [sort of do this](https://stackoverflow.com/questions/23916410/how-to-get-exact-command-line-string-from-shell), but it's neither practical nor useful. This question generally only comes up because people misunderstand quotes and arguments, so if you explain the problem that made you want to do this, you can probably get a much simpler and more helpful answer. – that other guy Oct 18 '18 at 22:24

2 Answers2

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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.

pjh
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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

rje
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  • That is my question. I know backslash is for escaping. But what I want to do is to print any given input literally without considering the escape. Echo does that when I put the input inside single quotes. However I cannot do it in my script because the input is inside a variable. – masec Oct 18 '18 at 21:27
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    Bash passes the arguments to your script **after** interpolation and substitution, in other words after all the escapes have been done and after the values of the variables have been expanded etc. etc. From inside your script it is impossible to know exactly what the user typed on the command line. – rje Oct 18 '18 at 21:29
  • Any reference to that? If that is the case then I guess as you said it is impossible. But I doubt. – masec Oct 18 '18 at 21:32