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

Question

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 

?

Answer 1

An Example Using Your Own Data

You can avoid having your filename embedded in the NUMOFLINES variable by using redirection from JAVA_TAGS_FILE, rather than passing the filename as an argument to wc. For example:

NUMOFLINES=$(wc -l < "$JAVA_TAGS_FILE") 

Explanation: Use Pipes or Redirection to Avoid Filenames in Output

The wc utility will not print the name of the file in its output if input is taken from a pipe or redirection operator. Consider these various examples:

# wc shows filename when the file is an argument $ wc -l /etc/passwd 41 /etc/passwd  # filename is ignored when piped in on standard input $ cat /etc/passwd | wc -l 41  # unusual redirection, but wc still ignores the filename $ < /etc/passwd wc -l 41  # typical redirection, taking standard input from a file $ wc -l < /etc/passwd 41 

As you can see, the only time wc will print the filename is when its passed as an argument, rather than as data on standard input. In some cases, you may want the filename to be printed, so it's useful to understand when it will be displayed.