将不以特定正则表达式开头的行添加到前一行

Question

短消息；传递；“+99999999999”；“2012.06.23:21”；“”；“

xxxxxxxxxxxx

xxxxxxxxxxxx

xxxxxxxxxxxx“

我需要任何不以“sms；交付；”开头的行来添加到前面的行中。也就是说，得到这样的一条线：

短信；递送；“+99999999999”；“”；“2012.06.23:21”；“”；“

那是一行。另外，删除/替换xxxxx(内容)部分中的任何双引号也是有帮助的。

短信；递送；“+99999999999”；“”；“2012.06.23 09:21";"";"xxxxxxxxxxxxx，

因此，上面的行将被转换为这个(双引号转换为单引号)：

短信；递送；“+99999999999”；“”；“2012.06.23 09:21";"";"xxxxxxxxxxxxx，

Answer 1

下面的sed命令似乎可以满足您的需要(编辑:一个简短的sed命令，用于筛选引号)：

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

简短的解释：

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'

最初的sed只在不与sms;delivered;联机时才将"更改为'

Answer 2

这可能对你有用：

sed ':a;$!N;/\nsms;deliver;/!s/\n//;ta;:b;s/\(;".*\)"\([^";]*\)"\([^";]*"\)$/\1'\''\2'\''\3/;tb;P;D' file

编辑：

"问题的测试数据：

echo 'sms;deliver;"+99999999999";"";"";"2012.06.23 09:21";"";"xxxxxxxxxxxxx xxxxxxxx, xxxxxxxxxxxx xxxxxxxxxxxx "xxxx"xxxxxxxx"' >/tmp/a
sed ':a;$!N;/\nsms;deliver;/!s/\n//;ta;:b;s/\(;".*\)"\([^";]*\)"\([^";]*"\)$/\1'\''\2'\''\3/;tb;P;D' /tmp/a
sms;deliver;"+99999999999";"";"";"2012.06.23 09:21";"";"xxxxxxxxxxxxx xxxxxxxx, xxxxxxxxxxxx xxxxxxxxxxxx 'xxxx'xxxxxxxx"
sed 's/xx/"&"/g' /tmp/a >/tmp/b
sed ':a;$!N;/\nsms;deliver;/!s/\n//;ta;:b;s/\(;".*\)"\([^";]*\)"\([^";]*"\)$/\1'\''\2'\''\3/;tb;P;D' /tmp/b
sms;deliver;"+99999999999";"";"";"2012.06.23 09:21";"";"'xx''xx''xx''xx''xx''xx'x 'xx''xx''xx''xx', 'xx''xx''xx''xx''xx''xx' 'xx''xx''xx''xx''xx''xx' ''xx''xx'''xx''xx''xx''xx'"