I have a bash script like
./script -a filename
And I have the content of a file in a variable $CONTENT
How can I pass the variable into the command. I mean I could write the content of the variable down to disk first, but that seems to be too much overhead.
Is there a better solution?
Something like
./scrip -a << $CONTENT
You need one more <:
./script -a <<<"$CONTENT"
<<< is called a herestring, and takes the following string and passes it as the standard input.
./script -a <<< "$CONTENT"s="a b" (with, say, three spaces between a and b, it doesn't show properly here), then compare cat <<< $s and cat <<< "$s". Also, please refer to the Here String section of the manual. You'll read: The word is expanded and supplied to the command on its standard input. Yes, is expanded. As a general rule: Use More Quotes!
IFS, etc. so better always quote!
xsltproc <<< "$xslt_content" "$xml_path" and file <<< "$content". None of those worked. Does it have to be the last argument? What's wrong with the file call?Kevin's suggestion to use a here string will work, as long as you quote the variable as suggested in the comments to that answer. A more portable solution is to use a here doc:
./script -a << EOF
$CONTENT
EOF
In this case, quotes are not necessary and indeed undesirable.