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Martin Hope
aenagy
Asked: 2024-12-23 08:21:05 +0800 CST
  • 6

这是 awk gsub 的延续，没有替换字段中所有句点实例。

我有一些意见：

$ cat ./ipv4settings.txt
ipv4.gateway:                           192.168.2.1
ipv4.routes:                            --
ipv4.route-metric:                      -1
ipv4.route-table:                       0 (unspec)

我能够生成所需的输出（某种程度上）：

$ awk 'BEGIN{FS=":[[:space:]]+";OFS="="}{gsub("[.]|-","_",$1);$1=$1;print $1"="$2 }' ./ipv4settings.txt
ipv4_gateway=192.168.2.1
ipv4_routes=--
ipv4_route_metric=-1
ipv4_route_table=0 (unspec)

接下来我要为每一行输出声明变量。我使用了几种变体declare：

$awk 'BEGIN{FS=":[[:space:]]+";OFS="="}{gsub("[.]|-","_",$1);$1=$1;print $1"="$2 }' ./ipv4settings.txt
ipv4_gateway=192.168.2.1
ipv4_routes=--
ipv4_route_metric=-1
ipv4_route_table=0 (unspec)
$ declare $( awk 'BEGIN{FS=":[[:space:]]+";OFS="="}{gsub("[.]|-","_",$1);$1=$1;print $1"="$2 }' ./ipv4settings.txt )
-bash: declare: `(unspec)': not a valid identifier

awk我尝试引用（declare not a valid identifier bash ）的整行输出（https://www.baeldung.com/linux/awk-print-quote-characters）：

$ awk 'BEGIN{FS=":[[:space:]]+";OFS="="}{gsub("[.]|-","_",$1);$1=$1;print "\042"$1"="$2"\042" }' ./ipv4settings.txt
"ipv4_gateway=192.168.2.1"
"ipv4_routes=--"
"ipv4_route_metric=-1"
"ipv4_route_table=0 (unspec)"
$ declare $( awk 'BEGIN{FS=":[[:space:]]+";OFS="="}{gsub("[.]|-","_",$1);$1=$1;print "\042"$1"="$2"\042" }' ./ipv4settings.txt )
-bash: declare: `"ipv4_gateway=192.168.2.1"': not a valid identifier
-bash: declare: `"ipv4_routes=--"': not a valid identifier
-bash: declare: `"ipv4_route_metric=-1"': not a valid identifier
-bash: declare: `"ipv4_route_table=0': not a valid identifier
-bash: declare: `(unspec)"': not a valid identifier

...或者仅仅是价值部分：

$ awk 'BEGIN{FS=":[[:space:]]+";OFS="="}{gsub("[.]|-","_",$1);$1=$1;print $1"=\042"$2"\042" }' ./ipv4settings.txt
ipv4_gateway="192.168.2.1"
ipv4_routes="--"
ipv4_route_metric="-1"
ipv4_route_table="0 (unspec)"
$ declare $( awk 'BEGIN{FS=":[[:space:]]+";OFS="="}{gsub("[.]|-","_",$1);$1=$1;print $1"=\042"$2"\042" }' ./ipv4settings.txt )
-bash: declare: `(unspec)"': not a valid identifier

如何处理declare带有空格的变量值？

bash
Martin Hope
aenagy
Asked: 2024-12-23 03:45:49 +0800 CST
  • 6

我正在尝试解析一些输出，以便最终可以用来declare创建多个环境变量。我遇到的问题是，awk我的语句没有替换第一个字段 ( $1) 中的所有句点。

输入来自：

etherNIC=$(nmcli --get-values NAME connection show)
nmcli connection show $etherNIC | grep -i  "ipv4.gateway\|ipv4.route" > ./ipv4settings.txt

...看起来像这样：

cat ./ipv4settings.txt
ipv4.gateway:                           192.168.2.1
ipv4.routes:                            --
ipv4.route-metric:                      -1
ipv4.route-table:                       0 (unspec)

这是输出awk：

cat ./ipv4settings.txt | awk '{ split($0, array, ":") ; gsub( "\\.", "_", $1 ) ;  gsub( ":" , "=" , $1 ) ; print $1$2 }'
ipv4_gateway=192.168.2.1
ipv4_routes=--
ipv4_route-metric=-1
ipv4_route-table=0

问题是gsub( "\\.", "_", $1 )不能替换.中的所有出现$1。我想要的输出是：

ipv4_gateway=192.168.2.1
ipv4_routes=--
ipv4_route_metric=-1
ipv4_route_table=0

在这种情况下，第二个.已被正确替换为_。

awk --version
GNU Awk 5.1.0, API: 3.0 (GNU MPFR 4.1.0-p9, GNU MP 6.2.0)
Copyright (C) 1989, 1991-2020 Free Software Foundation.
<snip>

编辑#1

我刚刚意识到的正确输出ipv4.route-table应该0 (unspec)不是0。

awk