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Top Forums UNIX for Dummies Questions & Answers parsing with multible delimiters Post 17713 by Jimbo on Tuesday 19th of March 2002 04:03:21 PM
Old 03-19-2002
Here is an alternate solution, not using perl:

Code:
cat gillbates.txt | while read string junk
do
echo $string |
  tr "[/]" "[ ]" |
  read words123 word4 rem

if [ "$word4" = Deal30 ] ; then
   word2=`expr $words123 : ".*!\(.*\)!"`
   word3=`expr $words123 : ".*!.*!\(.*\)"`
fi
done

Jimbo
 

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expr(1) 						      General Commands Manual							   expr(1)

Name
       expr - evaluate expressions

Syntax
       expr arg...

Description
       The arguments are taken as an expression.  After evaluation, the result is written on the standard output.  Each token of the expression is
       a separate argument.

       The operators and keywords are listed below.  The list is in order of increasing precedence, with equal precedence operators grouped.

       expr | expr	   Yields the first expr if it is neither null nor 0.  Otherwise yields the second expr.

       expr & expr	   Yields the first expr if neither expr is null or 0.	Otherwise yields 0.

       expr relop expr	   The relop is one of < <= = != >= > and yields 1 if the indicated comparison is true, '0' if false.  The  comparison	is
			   numeric if both expr are integers, otherwise lexicographic.

       expr + expr
	    expr - expr
			   Yields addition or subtraction of the arguments.

       expr * expr
	    expr / expr
	    expr % expr
			   Yields multiplication, division, or remainder of the arguments.

       expr : expr	   The	matching  operator compares the string first argument with the regular expression second argument; regular expres-
			   sion syntax is the same as that of The (...) pattern symbols can be used to select a portion of the  first  argument.
			   Otherwise, the matching operator yields the number of characters matched ('0' on failure).

       ( expr ) 	   parentheses for grouping.

Examples
       The first example adds 1 to the Shell variable a:
       a=`expr $a + 1`
       The second example finds the file name part (least significant part) of the pathname stored in variable a,
       expr $a : '.*/(.*)' '|' $a
       Note the quoted Shell metacharacters.

Diagnostics
       The command returns the following exit codes:

       0    The expression is neither null nor '0'.

       1    The expression is null or '0'.

       2    The expression is invalid.

See Also
       ed(1), sh(1), test(1)

																	   expr(1)
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