In Sed how can I replace starting from the 7th character to the 15th character.

Hi All,

Was wondering how I can do the following....

I have a String as follows

"ACCTRL000005022RRWDKKEEDKDD...."
This string can be in a file called tail.out or in a Variable called $VAR2

Now I have another variable called $VAR1="000004785" (9 bytes long), I need the content of $VAR1 in the string $VAR2

Example:
VAR1=000004785
|
|
VAR2= "ACCTRL000005022RRWDKKEEDKDD...."

So it should look like...
VAR2= "ACCTRL000004785RRWDKKEEDKDD...."

I know this can be done by sed, but I need to know how to replace from the 7th character till the 15th character.

Any help would be appreaciated.

Thanks,

Some ways:

VAR1="000004785" 
echo "$VAR2" | sed "s/\([^0-9]*\)[0-9]*\(.*\)/\1$VAR1\2/"

or:

VAR1="000004785" 
echo "$VAR2" | sed "s/\(......\).........\(.*\)/\1$VAR1\2/"

One more (minor variation of Franklin52's second option):

sed "s/\(.\{6\}\).\{9\}/\1$VAR1/"

Regards,
Alister

bash 4

# VAR1=000004785
# VAR2="ACCTRL000005022RRWDKKEEDKDD...."
# VAR2=${VAR2:0:6}${VAR1}${VAR2:15}
# echo $VAR2
ACCTRL000004785RRWDKKEEDKDD....

Thanks all... it worked as expected.

by awk substr function.

VAR1=000004785
VAR2="ACCTRL000005022RRWDKKEEDKDD...." 

VAR3=$( echo $VAR2|awk -v s=$VAR1 '{sub(substr($0,7,9),s)}1' )