Bash echo variable in string always has spaces

Question

I am currently working on a Mac OS X software update check script/plugin for Nagios, so that I can monitor my Mac Mini and Mac Pro machines, whether they have any updates available or not.

I created a variable called swUpdateCheck and assigned the check command to it, including some grep and awk, so that I can work with the output at the end. Since I don't always want to call the os update command, I just exported the output to a text file update.txt and am currently using the text file for building the final script.

This is my variable:

swUpdateCheck=`cat update.txt | grep -B1 recommended | grep -v recommended | awk '{gsub(/^[[:cntrl:][:space:]]+|^\-+|\*\s/,"");}NF>0 1' | wc -l`

Content of the Text file:

   * Security Update 2018-004-10.12.6
--
   * Safari11.1.2Sierra-11.1.2
--
   * iTunesX-12.8

Now my issue is, that whenever I call the variable it scans through the file and should give me a number of lines at the end when I echo it. It does give me a number when using a simple echo but as soon as I combine the variable with a string, it adds spaces in front of the number and I don't understand why.

Normal echo of swUpdateCheck:

$ echo $swUpdateCheck
3

Echo swUpdateCheck in a string:

$ echo "There are $swUpdateCheck Updates available."
There are        3 Updates available.

What am I missing or doing wrong?

What do you mean by "call the variable"? Calling is something you do with functions, not variables. — melpomene
– melpomene, Commented Sep 11, 2018 at 13:50
From your output it looks like swUpdateCheck contains a number of spaces followed by 3. What's the problem? — melpomene
– melpomene, Commented Sep 11, 2018 at 13:51
You aren't doing anything wrong. wc -l is padding its output, but leaving the expansion unquoted in echo $swUpdateCheck effectively strips the padding. — chepner
– chepner, Commented Sep 11, 2018 at 13:52
The additional spaces are being added by the wc command. I suggest you collapse your pipeline and do everything in awk. It can keep a counter in a variable that you print in your END block. Alternately, use printf to format your output instead of displaying it with echo. — ghoti
– ghoti, Commented Sep 11, 2018 at 13:52
@melpomene The solution from @Kamil Cuk helped me with my problem. And sorry if I caused any confusion by saying "called", that was my German thinking... @chepner I assumed that wc -l was causing it but was unsure about it. Thanks for the hint — C.Felix
– C.Felix, Commented Sep 11, 2018 at 14:34

KamilCuk · Accepted Answer · 2018-09-11 14:07:51Z

swUpdateCheck has spaces in it, not just 3. As you are not escaping the variable in echo $swUpdateCheck spaces get reinterpreted and 3 get's printed out. If you enclose the variable in ", as in echo ".... $swUpdateCheck ..." spaces will not be ignored. Observe the output with set -x:

$ set -x
$ echo
+ echo

$ swUpdateCheck="       3"  # assigment with spaces in it
+ swUpdateCheck='       3'
$ echo           3  # leading spaces get ignored
+ echo 3
3
$ echo $swUpdateCheck  # just printing '3', this is the same as above line
+ echo 3
3
$ echo "$swUpdateCheck"  # with "
+ echo '       3'
       3
$ swUpdateCheck=${swUpdateCheck// /}  # remove spaces using bash or use sed 's/ //g' or tr -d ' ' or other
+ swUpdateCheck=3
$ echo "$swUpdateCheck"   # now swUpdateCheck is without spaces, it will work
+ echo 3
3
$ echo "There are $swUpdateCheck Updates available."
+ echo 'There are 3 Updates available.'
There are 3 Updates available.

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