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Full Discussion: ksh pattern matching
Top Forums Shell Programming and Scripting ksh pattern matching Post 302165964 by ripat on Sunday 10th of February 2008 06:13:47 AM
Old 02-10-2008
ksh pattern matching

I try to use the pattern substitution operators as documented in the O'Reilley "Learning the Korn Shell" but it doesn't seem to work as advertised.

This works all right:
Code:
var='Regular expressions rules!'
$ echo ${var//e/#}
R#gular #xpr#ssions rul#s!

The docs says that using !(expr) matches anything that does't match expr but if I try to replace all but the "e" character, it does not seem to work:
Code:
var='Regular expressions rules!'
$ echo ${var//!(e)/#}
#

Any idea?

Last edited by ripat; 02-10-2008 at 08:05 AM..
 

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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)
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