Shell Programming and Scripting

Please look into the example.
My source file is like,

00,57,3,2008-07-24 06:30:06
10,1,8025171,"1M00",17907023,2008-07-23 18:16:58
10,2,8025171,"1M00",17907023,2008-07-23 18:17:01
99,184

What should i do if i want output like...

hello,00,57,3,2008-07-24 06:30:06
hello,10,1,8025171,"1M00",17907023,2008-07-23 18:16:58
hello,10,2,8025171,"1M00",17907023,2008-07-23 18:17:01
hello,99,184

u can use sed..

sed 's/^/hello,/g' filename

The "g" option is unnecesary since there's just one replacement per line.

Regards

thats right, thanks for pointing it out.

Hi, What u wrote is right for the specific case. May be My question was not very clear But in this problem scenario "hello" is coming from a varriable say temp. The content of temp varriable may change every time when I process the script.
In the above code we can not pass the varriabe. I mean we cann't write
sed 's/^/$temp,/g' filename
If we write like this it will give output
$temp,00,57,3,2008-07-24 06:30:06
$temp,10,1,8025171,"1M00",17907023,2008-07-23 18:16:58
$temp,10,2,8025171,"1M00",17907023,2008-07-23 18:17:01
$temp,99,184

So, The resultant question is "How to pass a varriable in the above sed command?".

Hi, What u wrote is right for the specific case. May be My question was not very clear But in this problem scenario "hello" is coming from a varriable say temp. The content of temp varriable may change every time when I process the script.
In the above code we can not pass the varriabe. I mean we cann't write
sed 's/^/$temp,/g' filename
If we write like this it will give output
$temp,00,57,3,2008-07-24 06:30:06
$temp,10,1,8025171,"1M00",17907023,2008-07-23 18:16:58
$temp,10,2,8025171,"1M00",17907023,2008-07-23 18:17:01
$temp,99,184

So, The resultant question is "How to pass a varriable in the above sed command?".

Use double quotes, single quotes prevent the expanding of variables.

Regards