linux - sed在bash中使用变量取消注释多行

Question

我遇到以下问题:我正在尝试取消对我确定使用的特定行的注释:

LINESTART=$(grep -nr "matching string" test.conf | cut -d : -f1)

之后，我需要取消注释从 $LINESTART 到 $((LINE+10))

的代码部分

LINEEND=$((LINE+10))

我尝试了以下 sed 语法:

sed -i '${LINESTART},${LINEEND} s/# *//' test.conf

但是我得到以下错误:

sed: -e expression #1, char 4: extra characters after command

示例测试.conf:

 84     #server {
 85     #    listen       8000;
 86     #    listen       somename:8080;
 87     #    server_name  somename  alias  another.alias;
 88 
 89     #    location / {
 90     #        root   html;
 91     #        index  index.php index.html index.htm;
 92     #    }
 93     #}

Answer 1

你能试试这个吗;

sed -i "${LINESTART},${LINEEND} s/# *//" test.conf

测试；

脚本:

#!/bin/bash
LINESTART=$(grep -nr "server {" test | cut -d : -f1 )
LINEEND=$((LINE+10))

sed "${LINESTART},${LINEEND} s/# *//" test

输出:

$ ./test.sh 
    server {
    listen       8000;
    listen       somename:8080;
    server_name  somename  alias  another.alias;

      location / {
      root   html;
      index  index.php index.html index.htm;
      }
      }