bash variable interpolation separate variables by a hyphen or underscore

Question

This is a simple script just to see if the file has been downloaded. On this script the find command always evaluated to zero - even if it didn\'t find anything. So I commen

Answer 1

Several possibilities:

• The most natural one: enclose your variable name in curly brackets (Ignacio Vazquez-Abrams's solution):

  echo "${filename}_$yesterday.CSV"
  

• Since your separator is a rather special character, you may use a backslash (Sriharsha's Kallury's solution):

  echo "$filename\_$yesterday.CSV"
  

• (Ab)use quotes:

  echo "$filename""_$yesterday.CSV"
  

  or

  echo "$filename"_"$yesterday.CSV"
  

• Use an auxiliary variable for the separator:

  sep=_
  echo "$filename$sep$yesterday.CSV"
  

• Use an auxiliary variable for the final string, and build it step by step:

  final=$filename
  final+=_$yesterday.CSV
  echo "$final"
  

  or in a longer fashion:

  final=$filename
  final+=_
  final+=$yesterday
  final+=.CSV
  echo "$final"
  

• Use an auxiliary variable for the final string, and build it with printf:

  printf -v final "%s_%s.CSV" "$filename" "$yesterday"
  echo "$final"
  

(feel free to add other methods to this post).

Answer 2

By telling bash where the variable name ends.

"${filename}_$yesterday.CSV"

Answer 3

You can use backslash to do that.

# filename=test
# yesterday=somedate
# echo $filename_$yesterday.csv
somedate.csv
# echo $filename\_$yesterday.csv
test_somedate.csv
#