用逗号分隔符替换换行符

Question

我有一个文本文件，该记录采用以下格式。请注意，名称，ID和等级部分中没有空文件。

"NAME","STUDENT1"
"ID","123"
"RANK","10"
"NAME","STUDENT2"
"ID","124"
"RANK","11"

我必须将上述文件转换为以下格式

"STUDENT1","123","10"
"STUDENT2","124","11"

据我所知，这可以通过读取记录并将其写入另一个输出文件来使用shell脚本来实现。但这可以使用awk或sed吗？

Answer 1

$ awk -F, '{ORS=(NR%3?FS:RS); print $2}' file
"STUDENT1","123","10"
"STUDENT2","124","11"

Answer 2

使用awk：

awk -F, '$1=="\"RANK\""{print $2;next}{printf "%s,",$2}' file

Answer 3

使用awk，每3行打印换行符：

awk -F, '{printf "%s",$2;if (NR%3){printf ","}else{print""};}'

Answer 4

关注awk也可以帮助您。

awk -F, '{ORS=$0~/^"RANK/?"\n":FS;print $NF}'  Input_file

Answer 5

使用sed

sed -E 'N;N;;y/\n/ /;s/([^,]*)(,[^ ]*)/\2/g;s/,//' infile