[bash] bash echo number of lines of file given in a bash variable without the file name

I have the following three constructs in a bash script:

NUMOFLINES=$(wc -l $JAVA_TAGS_FILE)
echo $NUMOFLINES" lines"

echo $(wc -l $JAVA_TAGS_FILE)" lines"

echo "$(wc -l $JAVA_TAGS_FILE) lines"

And they both produce identical output when the script is run:

121711 /home/slash/.java_base.tag lines
121711 /home/slash/.java_base.tag lines
121711 /home/slash/.java_base.tag lines

I.e. the name of the file is also echoed (which I don't want to). Why do these scriplets fail and how should I output a clean:

121711 lines

?

I normally use the 'back tick' feature of bash

export NUM_LINES=`wc -l filename`

Note the 'tick' is the 'back tick' e.g. ` not the normal single quote

You can also use awk:

awk 'END {print NR,"lines"}' filename

Or

awk 'END {print NR}' filename

wc can't get the filename if you don't give it one.

wc -l < "$JAVA_TAGS_FILE"

(apply on Mac, and probably other Unixes)

Actually there is a problem with the wc approach: it does not count the last line if it does not terminate with the end of line symbol.

Use this instead

nbLines=$(cat -n file.txt | tail -n 1 | cut -f1 | xargs)

or even better (thanks gniourf_gniourf):

nblines=$(grep -c '' file.txt)

Note: The awk approach by chilicuil also works.

It's a very simple:

NUMOFLINES=$(cat $JAVA_TAGS_FILE | wc -l )

or

NUMOFLINES=$(wc -l $JAVA_TAGS_FILE | awk '{print $1}')