从不一致的文本文件创建csv

Question

我在一个文件中包含结构松散的记录，由 3 或 4 行文本组成（大部分）由空行分隔。并非所有记录都有空行分隔符，但每条记录的最后一行都以“已添加”一词开头。我想生成一个 csv 文件，每条记录在一行上，前面是行号。到目前为止，我只能生成由任意数量的空格和冗余逗号分隔的所有记录的串联。

从逻辑上讲，我试图实现以下目标：

读取行，如果行以“已添加”开头，则在末尾保留换行符，
否则将“换行符”替换为“,”，
或者如果行为空白，则将其删除
endif

样本数据：

Peter Green  
Space Monkey at Area 51  
Joined  
Added by SF 3 weeks ago  
Will Rossiter  
Joined  
Added by SF 3 weeks ago

Dean Matthews  
Guitarist at Blues  
Joined  
Added by SF 3 weeks ago  
Hobbit Mak  
Farnborough, United Kingdom  
Joined  
Added by SF 3 weeks ago  

Keneth W Moorfield  
THE STOREMAN  
Joined  
Added by SF 3 weeks ago  
Mick Georgious  
Software Engineer  
Joined  
Added by SF 3 weeks ago

Answer 1

尝试：

awk '/./{ printf "%s%s", $0, (/Added/?"\n":",") }' data

使用您的样本输入数据：

$ awk '/./{printf "%s%s",$0,(/Added/?"\n":",")}' data
Peter Green,Space Monkey at Area 51,Joined,Added by SF 3 weeks ago
Will Rossiter,Joined,Added by SF 3 weeks ago
Dean Matthews,Guitarist at Blues,Joined,Added by SF 3 weeks ago
Hobbit Mak,Farnborough, United Kingdom,Joined,Added by SF 3 weeks ago
Keneth W Moorfield,THE STOREMAN,Joined,Added by SF 3 weeks ago
Mick Georgious,Software Engineer,Joined,Added by SF 3 weeks ago

这个怎么运作：

• /./{...}

  仅当该行包含字符时，才执行花括号中的命令。换句话说，这会忽略空行。

• printf "%s%s",$0,(/Added/?"\n":",")

  这将打印表示为 的行，$0后跟逗号或换行符，具体取决于该行是否与 regex 匹配Added。