sed - 将不以特定正则表达式开头的行添加到上一行

Question

sms;deliver;"+99999999999";"";"";"2012.06.23 09:21";"";"xxxxxxxxxxxxx xxxxxxxx,
xxxxxxxxxxxx
xxxxxxxxxxxx
xxxxxxxxxxxx"

我需要任何不以“sms;deliver;”开头的行添加到上一行。即得到这样一行:

sms;deliver;"+99999999999";"";"";"2012.06.23 09:21";"";"xxxxxxxxxxxxx xxxxxxxx, xxxxxxxxxxxx xxxxxxxxxxxx xxxxxxxxxxxx"

^ 这是一行。此外，删除/替换 xxxxx(内容)部分中的任何双引号也会有所帮助。

短信；发送；"+99999999999"；""；""；"2012.06.23 09:21"；""；"xxxxxxxxxxxx xxxxxxxx，xxxxxxxxxxxx xxxxxxxxxxxx "xxxx"xxxxxxx"

因此上面的行将转换为这样(双引号转换为单引号):

短信；发送；"+99999999999"；""；""；"2012.06.23 09:21"；""；"xxxxxxxxxxxx xxxxxxxx，xxxxxxxxxxxx xxxxxxxxxxxx 'xxxx'xxxxxxx"

Answer 1

以下 sed 命令似乎可以满足您的需要(编辑:开头的一个简短的 sed 命令用于过滤引号):

sed '/^sms;deliver;/!'"y/\"/'/" yourfile | sed -n '/^sms;deliver;/!b;:r;${p;b};N;/\nsms;deliver;/!{s/\n//;br};P;s/.*\n//;br'

简短说明:

sed -n '# not print by default
/^sms;deliver;/!b # if line not starting with the pattern, goto end
:r #label r
${p;b} # if last line, print & exit
N # read new line, append to pattern space
/\nsms;deliver;/!{s/\n//;br} # if appended line doesn't start with pattern,
                             # remove newline & goto r
P # print everything up to the newline
s/.*\n//;br # remove what was just printed, goto r'

开头的 sed 仅在与 sms;delivered; 不在同一行时才将 " 更改为 '。