How do I do an exact word match on a variable in ksh?

Question

I used the following syntax as part of a ksh script to verify if the word Validation exists in LINE_FROM_FILE.

[[ "${LINE_FROM_FILE##*Validation}" != "${LINE_FROM_FILE}" ]] && print "match Validation"

The problem of this syntax is that it is also matching words like Valid or ValidationVALID etc. and my goal is to exactly match the word Validation in the variable $LINE_FROM_FILE.

I ask if it is also possible to use Perl syntax in my script to exactly match the word Validation, for example:

[[ ` some perl command ` = Validation  ]] && print "match Validation"

Answer 1

if your using bash, you're in luck:

bash regexs

Example:

input=$1


if [[ "$input" =~ "[0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9][0-9][0-9]" ]]
#                 ^ NOTE: Quoting not necessary, as of version 3.2 of Bash.
# NNN-NN-NNNN (where each N is a digit).
then
  echo "Social Security number."
  # Process SSN.
else
  echo "Not a Social Security number!"
  # Or, ask for corrected input.
fi

Answer 2

Test whether the line contains the word between word-separating characters. A helpful trick is to add a word-separating character at the beginning and at the end of the string, in case the word is at the beginning or end of the string.

[[ " $LINE_FROM_FILE " == *[![:alnum:]]Validation[![:alnum:]]* ]]

This assumes that words consist of letters and digits only. Adjust the pattern if you have a different definition.

Note that the test you wrote, [[ "${LINE_FROM_FILE##*Validation}" != "${LINE_FROM_FILE}" ]], is a complicated way of writing [[ $LINE_FROM_FILE = *Validation* ]] (i.e., checking for Validation as a substring).