Shell Programming and Scripting

poor-man apprroach of in-place editing:

Code:

for f in file* ; do { rm $f; awk '/headers/{p=1;next}p' > $f'} < $f ; done

Try also

Code:

awk '/headers/,EOF {print > FILENAME".res"}' file*

If your first data line always starts with a date e.g.

Code:

26/2/2015 ...

then you can search this line and print up to the end.

Code:

ls -tr file* |
while read f
do
  sed -i -n '\m^ *[0-9]\{1,2\}/[0-9]\{1,2\}/[0-9]\{4\}m,$ p' "$f"
done

Or without GNU sed, the poor man approach

Code:

ls -tr file* |
while read f
do
  {
     rm -f "$f"
     sed -n '\m^ *[0-9]\{1,2\}/[0-9]\{1,2\}/[0-9]\{4\}m,$ p' > "$f"
  } < "$f"
done

the below seem to work...could you please explain what you are doing ?

ls -tr file* |
while read f
do
sed -i -n '\m^ *[0-9]\{1,2\}/[0-9]\{1,2\}/[0-9]\{4\}m,$ p' "$f"
done

If the string headers is found then set variable p to 1 and proceed with the next cycle (line).
If p is not zero (or unset) then print the line (default action).
By switching the order one can save the next:

Code:

'p==1{print} /headers/{p=1}'

Or

Code:

'p==1; /headers/{p=1}'

Or

Code:

'p; /headers/{p=1}'

---------- Post updated at 05:00 PM ---------- Previous update was at 04:36 PM ----------

Code:

ls -tr file* |
while read f
  ...
done

processes the oldest files first, like in your initial post.
The file loop is an alternative to a for loop.
sed:
-i in place replacement
-n noprint mode, so it needs explicit printing.
/search/,$ p from the line where search is found thru the $=last line print.
search is a RE (regular expression), here ^ *[0-9]\{1,2\}/[0-9]\{1,2\}/[0-9]\{4\}. Because the expression contains a / that collides with the / delimiter, I have replaced the delimiters to another character m that is not in the RE\msearchm.
The RE means: beginning of line, optional spaces, a digit 1 or 2 times, a /, a digit 1 or 2 times, a /, a digit 4 times.