10-18-2006
Searching Files
Hi,
I have a file
/db01/dat/march 2006/7001DW06.03B
Please note, between "march 2006" there is a space/tab.
While running the following script, it identifies
/db01/dat/march ----> as first file
2006/7001DW06.03B ---> as second file.
SRC_PATH = /db01/dat
SEARCH_FILENAME = 7001DW06.03B
for i in `find $SRC_PATH -name $SEARCH_FILENAME`
do
IDX=`expr index $i .`
if [ $IDX -ne 0 ]
then
STRTPOS=`expr $IDX - 8`
FILENAME=`expr substr $i $STRTPOS 12`
FILETYPE=`expr substr $FILENAME 5 2`
if [ "$FILETYPE" = "DW" ] || [ "$FILETYPE" = "dw" ]
then
echo Extracting $FILENAME ...
# unzip -jo $i -d $DEST_PATH/$1
fi
fi
done
echo Done
I need to get this as a single file instead of two files.
Regards,
Ronald.
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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)