Dear Members,

Following is the code which i am using:

integer i=7
while ((i <= 10 ));
do
param[i]=`echo $TEST_OUT | cut -d"^" -f$i`

a=`echo ${param[7]}`
echo `echo $a | sed 's/+/ /g'`

(( i = i + 1));
done

From the above code TEST_OUT is a variable which has the following value:

TEST_OUT='12345++++++TEST COMPANY+++++++++02182009+++++++01232009+++++++FREIGHT PAYMENT+++5.64'

as you can see all the values in TEST_OUT are separated by '+' sign.

MY aim is to replace the '+' sign with a white space, so i am using the following command:

echo `echo $a | sed 's/+/ /g'`

If i do this it is replacing '+' with only one white space. The result looks like

'12345 TEST COMPANY 02182009 01232009 FREIGHT PAYMENT 5.64'

which i don't want. Suppose we have 7 '+' sign then i would like to see 7 white spaces which is not happening. Its only putting one space.

How can i avoid this.

I am using Korn Shell.

Thanks
Sandeep

$ TEST_OUT='12345++++++TEST COMPANY+++++++++02182009+++++++01232009+++++++FREIGHT PAYMENT+++5.64'
$ echo "$TEST_OUT" | sed 's/+/ /g'
12345      TEST COMPANY         02182009       01232009       FREIGHT PAYMENT   5.64
$ echo `echo "$TEST_OUT" | sed 's/+/ /g'`
12345 TEST COMPANY 02182009 01232009 FREIGHT PAYMENT 5.64