Hi All,

Can any body tell me how to ignore the first and last 3 lines in a source file that needs processing

Thanks
SJNB

Is there any other way of distinguishing these records?

Greg.

if you are taking a file and doing a bulk insert into a table you can specify the starting row to use (same with bcp).

Paul
www.dbmsguy.com

oops, sorry thought i was in a different forum

Paul
www.dbmsguy.com

cat filename | sed 1d |sed 1d |sed 1d

this will delete the first three lines

Paul
www.dbmsguy.com

Hi, snjb!

To ignore the first 3 lines in a source file that needs processing put this statement on first place in awk program:

NR < 4  { next }

For example, this awk script print lines from 4th line to end line:

NR < 4  { next }
{ print }

Bye!

KP.

The problem is with the footer as in awk you don't know
the number of records in a file before the "END" block.
Therefor, you don't know whether you're processing a footer
or you're still in the "body" of the file.

The easiest way is to calculate the number of records/lines
in a file prior to calling awk/nawk and pass the "begin/end"
pivots to awk.

two step approach:

---------- filterHeaderFooter.ksh --------------------------
#!/bin/ksh

let header=1;
let footer=1;
numLines=`cat $* | wc -l | sed -e 's/[ ]*//g'`

let begin=${header}
let end=numLines-footer

echo "begin-> ${begin} end-> ${end}"

nawk -v begin=$begin} -v end=${end} -f filterHeaderFooter.awk $*

-------------------------------------------------------------

---------------filterHeaderFooter.awk ----------------------
BEGIN {

}

FNR > begin && FNR < end { print $0 }

END {

}

Hi sjnb,

This should do it for you:


#!/bin/sh

count=`wc -l $1`
last=`expr $count - 3`
nawk 'NR == 4, NR == '"$last"' {print}' $1 > $1.new

Hope this helps you!


flogrr
flogr@yahoo.com
 
 

If you want to do it all in awk, try this.

# awk program to strip first and last 3 lines from a file.
{
  if (NR == 4)
    hld[0] = $0
  else if (NR == 5)
    hld[1] = $0
  else if (NR == 6)
  {
    hld[2] = $0
    ix = 0
  }
  else if (NR > 6)
  {
    print hld[ix]
    hld[ix] = $0
    ix++
    if (ix > 2) ix=0
  }
}