02-24-2010
help required for 'expr substr' function
hi
iam trying to extract a certain portion of the string whose value is stored below,but am getting syntax eror.The command is shown below
for file in GMG_BASEL2*.txt
do
m1= cat reporting_date.txt
year= expr substr $m1 1 2
echo $year
done
m1 has date 10/31/2009
but this vale is not passed to next variable where am using substr function.
can some one please help on this?
Thanks in advance
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EXPR(1) General Commands Manual EXPR(1)
NAME
expr - evaluate arguments as an expression
SYNOPSIS
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
where relop is one of < <= = != >= >, 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
addition or subtraction of the arguments.
expr * expr
expr / expr
expr % expr
multiplication, division, or remainder of the arguments.
expr : expr
The matching operator compares the string first argument with the regular expression second argument; regular expression syntax is
the same as that of ed(1). 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:
To add 1 to the Shell variable a:
a=`expr $a + 1`
To find the filename part (least significant part) of the pathname stored in variable a, which may or may not contain `/':
expr $a : '.*/(.*)' '|' $a
Note the quoted Shell metacharacters.
SEE ALSO
sh(1), test(1)
DIAGNOSTICS
Expr returns the following exit codes:
0 if the expression is neither null nor `0',
1 if the expression is null or `0',
2 for invalid expressions.
7th Edition April 29, 1985 EXPR(1)