bash - 如何从输出中删除前2行

Question

I have the following output in a text file:

106 pages in list
.bookmarks
20130516 - Daily Meeting Minutes
20130517 - Daily Meeting Minutes
20130520 - Daily Meeting Minutes
20130521 - Daily Meeting Minutes

I'm looking to remove the first 2 lines from my output. This particular shell script that I use to execute, always has those first 2 lines.

This is how I generated and read the file:

#Lists
PGLIST="$STAGE/pglist.lst";
RUNSCRIPT="$STAGE/runPagesToMove.sh";

#Get List of pages
$ATL_BASE/confluence.sh $CMD_PGLIST $CMD_SPACE "$1" > "$PGLIST";

# BUILD executeable script
echo "#!/bin/bash" >> $RUNSCRIPT 2>&1
IFS=''
while read line
  do
     echo "$ATL_BASE/conflunce.sh $CMD_MVPAGE $CMD_SPACE "$1" --title \"$line\" --newSpace \"$2\" --parent \"$3\"" >> $RUNSCRIPT 2>&1
done < $PGLIST

How do I remove those top 2 lines?

Answer 1

You can achieve this with tail :

tail -n +3 "$PGLIST"

  -n, --lines=K output the last K lines, instead of the last 10; or use -n +K to output starting with the Kth 

Answer 2

经典答案将使用sed删除第1行和第2行：

sed 1,2d "$PGLIST"

Answer 3

awk方式：

awk 'NR>2' "$PGLIST"