In bash, feed output of one command line-by-line into another command

Question

I have the following bash command which fetches and block XML from a single column in mysql and feeds it into xmllint to prettify it.

mysql --format csv --skip-column-names -e "select xmldata from mytable limit 1;" | xmllint --format -

So far so good, but it only works for one row at a time. I would like to adapt this so each line in the SQL query output is fed in to a separate call to xmlint.

I know this can be done with a loop in a bash script -- but I'd like to do it in a one-liner if possible. Supposedly this can be done with xargs but I can't figure it out.

Answer 1

mysql --format csv --skip-column-names \
    -e "select xmldata from mytable limit 1;" \
    | while IFS= read -r line; do xmllint --format - <<<"$line"; done

Split for easier reading, obviously you don't need to. The <<<$line is a herestring, which feeds the contents of $line to xmllint as stdin.

Answer 2

This is possible with xargs, but xargs would pass each line as an argument to xmllint, not on its standard input. You would need to have xargs invoke a shell to feed the text to xmllint.

mysql … | xargs -L 1 sh -c 'echo "$0" | xmllint --format -'

A shell loop is the easiest method. The canonical way to loop over lines is while IFS= read -r line; do ….

mysql … |
while IFS= read -r line; do printf '%s\n' "$line" | xmllint --format -; done