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awk/sed column replace using column header - help

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Top Forums Shell Programming and Scripting awk/sed column replace using column header - help
# 8  
Old 06-17-2008
Code:
$cat awk.file
NR==1{
for(i=1;i<=NF;i++)
       {
        if($i=="Name")  a=i
        if($i=="Age")   b=i
        if($i=="Sex")   c=i
        if($i=="alias") d=i
        if($i=="nfld")  e=i
        if($i=="xsd")   f=i
       }
     }
{
  print $a,$b,$c,$d,$e,$f
}

Run the script:
Code:
awk -f awk.file log.txt

Regards,
# 9  
Old 06-18-2008
Thanks danmero, but still as I stated I have more than 70 columns, so specifying each and every column name is something difficult for me.

i.e.

Code:
Name Age Sex ............ fldx bloodgrp .....
A 23 M .............. Y A+ ......
D 21 F................ N B+ ........
......
.......

I wanted to replace a column whose header is "Sex" with the column with header "fldx". The field number of both the columns can be dynamic. So I want to replace them based on their column header name.

so that the output will look something like this

Code:
Name Age fldx ................. bloodgrp .......
A 23 Y  .................. A+...............
B 21 N ................... B+..............
...............
.............

so the output will not be having the "Sex" column (will be replaced by "fldx" column)

Thanks
# 10  
Old 06-18-2008
So, something like this maybe?

Code:
awk -v src="alias" -v dst="Lcation" 'NR == 1 {
  for (i=1; i<=NF; ++i) { if ($i == src) scol=i; if ($i == dst) dcol=i; }
  if (scol == "") print "undefined field src=" src;
  if (dcol == "") print "undefined field dst=" dst;
  if (scol == "" || dcol == "") exit 1;
}
{ $dcol=$scol; $scol=""; print }' log.txt

If your awk doesn't support the -v option, see if you can find nawk, mawk, gawk, or XPG4 awk.
# 11  
Old 06-18-2008
Thanks era.
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