批量替换配置文件指定行，这个需求怎么实现？谢谢

realcriss

配置文件片段，蓝色部分是需要替换的行

    server {
        listen       80;
        server_name  zzss.example.com;
        access_log   syslog:local2:info:log_server:6010:zzss.example.com  proxy;
    }
    server {
        listen       80;
        server_name  bcact.example.com;
        access_log   syslog:local2:info:log_server:6010:act.example.com  proxy;
    }

需求：
将上面Server段里面，access_log行，替换成下面行，

access_log   syslog:server=1.1.1.1:6010,facility=local2,tag="对应各自server段落的server_name",severity=info itc;

例如

access_log   syslog:server=1.1.1.1:6010,facility=local2,tag=zzss.example.com,severity=info itc;

zxy877298415

1. awk '/server_name/{p=gensub(";","","1",$2)}/access_log/{$0="\t "$1"   syslog:server=1.1.1.1:6010,facility=local2,tag=\""p"\",severity=info itc;";p="";}1' file

复制代码

回复 1# realcriss


   

陈卓文

awk '/access_log/{split($2,a,":");printf("\t%s %s:server=1.1.1.1:%s,facility=%s,tag=%s,severity=%s itc\n" ,$1 ,a[1] ,a[5], a[2], a[6], a[3]);next}{print}'