awk：将第一行转换为第一列，第二行转换为第二列

Question

I have a file with data something like this :

{
MAG 121/002
Wed Mar 14 00:00:00 2018
MAG 121/003
Wed Mar 14 00:00:00 2018
MAG 121/004
Wed Mar 14 00:00:00 2018
}

I want the output as :

{
MAG 121/002 | Wed Mar 14 00:00:00 2018
MAG 121/003 | Wed Mar 14 00:00:00 2018
}

and so on.. Any help is appreciated.

What I tried was:

cat <filename> | awk '{printf "%s" (NR%2==0? RS:FS), $1}'

Answer 1

您能否尝试遵循并让我知道是否有帮助。

awk '/{/||/}/{print;next} /MAG/{val=$0;getline;print val OFS $0}' OFS=" | "  Input_file

Answer 2

Solution with sed:

echo "MAG 121/002
Wed Mar 14 00:00:00 2018
MAG 121/003
Wed Mar 14 00:00:00 2018
MAG 121/004
Wed Mar 14 00:00:00 2018" | tr "\n" "|" | sed 's/|/ | /g' | sed -r 's/([^|]+\|[^|]+)\| /\1\n/g'
MAG 121/002 | Wed Mar 14 00:00:00 2018 
MAG 121/003 | Wed Mar 14 00:00:00 2018 
MAG 121/004 | Wed Mar 14 00:00:00 2018 

Answer 3

Read and echo:

echo "MAG 121/002
Wed Mar 14 00:00:00 2018
MAG 121/003
Wed Mar 14 00:00:00 2018
MAG 121/004
Wed Mar 14 00:00:00 2018" | while read line ; do case $line in MAG*) echo -n $line "| " ;; *) echo $line ;; esac ; done 
MAG 121/002 | Wed Mar 14 00:00:00 2018
MAG 121/003 | Wed Mar 14 00:00:00 2018
MAG 121/004 | Wed Mar 14 00:00:00 2018

code formatted:

while read line
do
  case $line in 
    MAG*) echo -n $line "| " ;; 
       *) echo $line ;; 
  esac
done