awk：将第一行转换为第一列，第二行转换为第二列

Question

我有一个包含这样的数据的文件：

{
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
}

我希望输出为：

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

等等..任何帮助表示赞赏。

我试过的是：

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

sed解决方案：

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

读取并回显：

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

代码格式：

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