sed：当找到字符串时，从模式之间的文件中删除多行

Question

我试图删除两个模式之间的行（Beginn：info / End：}），其中匹配一个字符串。

这是我的档案：

# bla bla
# bla bla
# bla bla
# bla bla

nnssjnds nkjdnds "nsrnsnmks" ffsns {
  is on or off at 9:12:43 23/02/2015;
  is nass or trocken at 08:32:12 22/02/2015;
}

info text01text {
  beginn 30/04/2015 10:00:04;
  end    30/04/2015 19:00:04;
  check1 30/04/2015 11:30:04;
  check2 30/04/2015 13:00:04;
  check3 30/04/2015 16:00:04;
  Check4 30/04/2015 18:00:04;
  build top end;
  mix water 0102030456789;
  xim "43ndf392rfhf<DF>3}";
  test space = "ALLFINE";
  eman cpre "ann";
}
info text02text {
  beginn 30/04/2015 10:00:04;
  end    30/04/2015 19:00:04;
  check1 30/04/2015 11:30:04;
  check2 30/04/2015 13:00:04;
  check3 30/04/2015 16:00:04;
  Check4 30/04/2015 18:00:04;
  build top end;
  mix water 0202030456789;
  xim "43ndf392rfhf<DF>3";
  test space2 = "ALLFINE2";
  eman cpre2 "ann2";
}
info text03text {
  beginn 30/04/2015 10:00:04;
  end    30/04/2015 19:00:04;
  check1 30/04/2015 11:30:04;
  check2 30/04/2015 13:00:04;
  check3 30/04/2015 16:00:04;
  Check4 30/04/2015 18:00:04;
  build top end;
  mix water 0302030456789;
  xim "43ndf392rfhf<DF>3";
  test space3 = "ALLFINE3";
  eman cpre3 "ann3";
}

我的sed脚本

:point
/^info/,/^}/ {
   /}/!{
      $!{
         N;
         bpoint
      }
   }
   /0202030456789/d;
}

我的sed脚本使用字符串0202030456789正常工作，并删除从信息text02text {到}的所有行。 尝试使用0102030456789中的字符串text01text，然后从第＆＃34行删除sed到}。 xim＆＃34; 43ndf392rfhf3}＆＃34;;＆＃34;和行

test space = "ALLFINE";
eman cpre "ann";
}

不要删除。

如何删除找到字符串的所有行？

谢谢！