CODEWITHSUNDEEP
bashmacos
user12926848
user12926848

Reputation: 23

avoid spaces when concatenating variables in bash under macos

I am having a strange issue with bash under macos. When I concatenate two variables, it adds an extra space between them which I cannot get rid of.

~/testdrive/dir1 $ curd=$(pwd)
~/testdrive/dir1 $ echo $curd
/xx/xx/testdrive/dir1
~/testdrive/dir1 $ fcount=$(ls -l | wc -l)
~/testdrive/dir1 $ echo $fcount
5          # notice no space in front of the 5 
~/testdrive/dir1 $ echo $curd$fcount
/xx/xx/testdrive/dir1 5 # space between directory name and 5

I am using GNU bash, Version 5.0.16(1)-release (x86_64-apple-darwin19.3.0). I tried newd="$curd$fcount" and newd=${curd}${fcount} with the same result. In some directories it adds 5 or more spaces between the variables.

However,

~/testdrive/dir1 $ var1=abc
~/testdrive/dir1 $ var2=def
~/testdrive/dir1 $ echo $var1$var2
abcdef   # expected behavior

Then, again,

~/testdrive/dir1 $ echo $var1$fcount
abc 5  # space between

I've seen many tips how to remove whitespace from strings, however I do not understand why it is there in the first place. I am assuming it has to do with fcount=$(ls -l | wc -l) but how? Any ideas?

Upvotes: 2

Views: 67

Answers (2)

Mark Setchell
Mark Setchell

Reputation: 207708

Bash variables are untyped. Try this:

fcount=$(ls | wc -l)

echo $fcount          # check it
469                   # it looks ok, but...

echo "$fcount"        # ... when quoted properly
     469              # it has leading spaces! UGH!

Try again, but this time tell bash it's an integer:

declare -i fcount     # Tell bash it's an integer
fcount=$(ls | wc -l)  # set it
echo "$fcount"        # check value with correct quoting
469                   # and it's correct

Or, if you don't like that method, you can tell bash to replace all spaces with nothing/emptiness.

string="   abc  |"

echo "$string"           # display string as is
   abc  |

echo "${string// /}"     # delete all spaces in string
abc|

Upvotes: 2

Matias Barrios
Matias Barrios

Reputation: 5056

It works for me :

 $ name='john'
 $ age='5'
 $ echo "${name}${age}"
john5

I am using :

GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin18)
Copyright (C) 2007 Free Software Foundation, Inc.

Please try to replicate this in your version. Otherwise this might work too :

$ newvalue=$( printf "%s%s\n" $name $age )
$ echo "$newvalue" 
john5

Upvotes: 0

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