根据下一行的第一个字符连接行

Question

我正在寻找基于下一行的客场联系线。到目前为止，我看到的唯一方法是创建一个 shell 脚本，该脚本将逐行读取并按照以下方式执行某些操作：

while read line
    if $line does not start with "," and $curr_line is empty 
        store line in curr_line
    if $line does not start with "," and $curr_line is not empty
        flush $curr_line to file
        store $line in $curr_line
    if $line starts with "," append to $curr_file, flush to file empty curr_line
done < file

所以我试图了解是否可以通过 sed 甚至 grep 重定向来实现。文件的规则很简单。最多只有一行以“,”开头，需要附加到前一行。

前任：

line0
line1
line2
,line3
line4
line5
,line6
line7
,line8
line9
line10
line11

结果文件将是

line0
line1
line2,line3
line4
line5,line6
line7,line8
line9
line10
line11

Answer 1

我会做：

awk -v ORS= '
  NR>1 && !/,/ {print "\n"}
  {print}
  END {if (NR) print "\n"}' < file

也就是说，如果当前行不以 开头，则仅打印分隔前一行的换行符,。

在任何情况下，我都不会使用while readloop。

Answer 2

这是 的经典用例sed，如Sed One-Liners Explained, Part I: File Spacing, Numbering and Text Conversion and Substitution , 40 中所述。如果前一行以等号“=”开头，则在前一行追加一行。（明显修改了,for =）

sed -e :a -e '$!N;s/\n,/,/;ta' -e 'P;D' file
line0
line1
line2,line3
line4
line5,line6
line7,line8
line9
line10
line11

Answer 3

您需要做的就是 slurp 文件并删除逗号前的所有换行符：

$ perl -0777pe 's/\n,/,/g' file
line0
line1
line2,line3
line4
line5,line6
line7,line8
line9
line10
line11