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Top Forums Shell Programming and Scripting Use of expr to calc differenc ein to epoch values Post 303016701 by RudiC on Wednesday 2nd of May 2018 12:17:38 PM
Old 05-02-2018
Does the error occur immediately, i.e. right at the begin of the program working on the first input line, or more towards the end of it?
In the code in post#8, shift the DATE_DIFF calculation after the four commented out echo lines, and remove the four comment signs. You may want to add and output a loop counter.
 

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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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