如何让sed替换前一行?我只看到了delete,insert行的例子,但实际上我需要的是,只有在满足了下一行的条件时,我才会对当前行进行替换。我的示例文件如下所示CygwinIs also one of the best environmentsWhy did Sun feel copying Java int
嗨,我编写了以下脚本来更新特定用户的密码 $ldaps_url = "192.168.168.1";
$ldap_conn = ldap_connect($ldaps_url, $port) or die("Sorry! Could not connect to LDAP server ($ip)");