Heredoc: Passing a string with a variable and multiple "$" characters

Question

I want to pass a string in a heredoc that contains a variable, and multiple "$" characters. The command without the heredoc is:

FOO='BAR'
echo ${FOO}: '$NOT_A_VARIABLE'

The FOO variable is correctly substituted and the '$NOT_A_VARIABLE' string avoids substitution because it is single quoted. The command returns as expected:

BAR: $NOT_A_VARIABLE

Now if I do the same inside the heredoc:

sh << EOF
   echo ${FOO}: '$NOT_A_VARIABLE'
EOF

The output is not what I desired:

BAR:

It seems, that the '$NOT_A_VARIABLE' is substituted even though it is surrounded by single quotes.

I know that this can be avoid by escaping the $ with a backslash: \$ but this solution is annoying since there might be multiple $ characters randomly distributed in the string.

I wonder if there is a smart use of quotes (single/double) or other solutions that might fix it.

bash doesn't parse the heredoc, so it can either expand all variables, or none. — choroba
– choroba, Commented Jan 19, 2022 at 17:40

chepner · Accepted Answer · 2022-01-19 22:50:41Z

4

The contents of your here document are treated as a double-quoted string, just as if you had written

echo "$FOO: '$NOT_A_VARIABLE'" | sh

If you want to suppress all parameter expansion in the here document, you can quote the delimiter:

sh << 'EOF'
    echo ${FOO}: '$NOT_A_VARIABLE'
EOF

which will pass

echo ${FOO}: '$NOT_A_VARIABLE'

as the input to sh.

If you only want to suppress a specific expansion, escape the $ just as you would in a double-quoted string.

# echo "$FOO: \$NOT_A_VARIABLE" | sh
sh << EOF
   echo ${FOO}: \$NOT_A_VARIABLE
EOF

edited Jan 19, 2022 at 22:50

answered Jan 19, 2022 at 18:22

chepner

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4 Comments

glenn jackman Over a year ago

Actually, the quoted heredoc outputs BAR: $NOT_A_VARIABLE if FOO is exported and : $NOT_A_VARIABLE otherwise. The quoted heredoc suppresses variable expansion for the current shell, but the child sh process will evaluate its input as code, repecting the unquoted variable and the single quoted one.

chepner Over a year ago

Reworded the answer to emphasize what the shell's input will be without worrying about the output of the shell.

glenn jackman Over a year ago

Nice. I suppose the OP needs to decide at what level that variable ought to be expanded: in the current shell or the child shell.

Marcos Over a year ago

Thanks for the explanations. I was hoping for a solution more elegant and compact than escaping every $ with \ . I will al least escape all $ instances automatically with sed 's/\$/\\$/g'.

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