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# 1  
Old 04-09-2010
problem with awk usage
I have to change the 4th feild of a line; Most of the time, the line has only 4 feilds, but sometimes, it has more than 4 feilds.

I am using the below command:
Code:
NEW_LINE=`echo $Line | awk '{sub ($4, "2010\\\/04\\\/07"); printf "# %-7s %9s %18s\n", $2,$3,$4}'`

This works fine but if the line has more than 4 feilds, it eats the other feilds following 4th feild.

Ex: I have a line as shown below:
Code:
# seh         Lane        2010/02/17      retrieve 0 rows from US FIW on 2009/06/11

I am getting the output as:
Code:
# seh         Lane        2010/04/07

Is there a way to have the entire line. Please note that I have to format the other feilds before printing.

Thanks,
Sugan
# 2  
Old 04-09-2010
Assign the value to the 4th field as:
Code:
awk '{$4="2010/04/07"; printf "# %-7s %9s %18s\n", $2,$3,$4}'

# 3  
Old 04-09-2010
Sorry. I cant understand what you are saying.

I have a line with many feilds; I need to modify the 4th feild, format the 2, 3, and 4th feild and want to print the rest of the line (other feilds after 4th feild) as such.

Is there a way to do this.
# 4  
Old 04-09-2010
Quote:
Originally Posted by sugan
Sorry. I cant understand what you are saying.

I have a line with many feilds; I need to modify the 4th feild, format the 2, 3, and 4th feild and want to print the rest of the line (other feilds after 4th feild) as such.

Is there a way to do this.
One way:
Code:
awk '{$4="2010/04/07"; printf "# %-7s %9s %18s", $2,$3,$4}{$1=$2=$3=$4=""; sub("   ",""); print}'

# 5  
Old 04-09-2010
Thanks very much. It works. But I got into another problem.
Here is my code:
Code:
NEW_LINE=`echo $Line | awk '{sub ($4, "2010\\\/04\\\/07"); printf "# %-7s %9s %18s\n", $2,$3,$4}' {$1=$2=$3=$4=""; sub("   ",""); print}'`
      sed $n" s/.*/$NEW_LINE/" kfile > tmp

Most of the time, when the number of feilds exceed, it has a date part in it in the format, 'yyyy/mm/dd'. The sed command that comes next, fails due to this date format.

Any idea how to handle this.
# 6  
Old 04-09-2010
Given this as input:
Code:
# seh         Lane        2010/02/17      retrieve 0 rows from US FIW on 2009/06/11

What should be the output (end result) in the file tmp?
# 7  
Old 04-09-2010
The output in the tmp file should be the same with the 4th feild changed. I am using sed to replace the existing line with the new line. When the new line has dates in the format 'yyyy/mm/dd' sed throws error.

Here is my code:
Code:
 elif [ $1 != 'Cindy' -a  `echo $Line | grep -ce 'Larry' -e 'Joe' -e 'Bryan' -e 'Lane'` -gt 0 ]
    then
      NEW_LINE=`echo $Line | awk '{sub ($4, "2010\\\/04\\\/07"); printf "# %-7s %9s %18s\n", $2,$3,$4} {$1=$2=$3=$4=""; sub("   ",""); print}'`
      sed $n" s/.*/$NEW_LINE/" kfile > tmp
      mv tmp kfile
    fi

Here is the error msg I get:
Code:
sed: 0602-404 Function 167 s/.*/# seh          Lane       2010\/04\/07 retrieve 0 rows from US FIW on 2009/06/11/ cannot be parsed.

In the above code, when the new line has dates in the format 'yyyy/mm/dd' after 4th feild, I have a problem with sed.
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