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Bash声明不喜欢带空格的值

declare awk shell bash

aenagy · 技术社区 · 5 月前

这是一个延续 awk gsub not replacing all instances of period in field .

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(在一定程度上)生成所需的输出:

$ 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)

接下来我要做的是为每一行输出声明变量。我使用了几个变体 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

我试着引用了 awk ( declare not a valid identifier bash )( 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

…或者只是价值部分:

$ 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

我怎么得到 声明 使用带空格的变量值?

2 回复 | 直到 5 月前

markp-fuso 5 月前

Another answer 您之前的问题解决了这个(引用)问题:

$ cat ipv4.awk
BEGIN                     { sq = "\x27" }                  # define single quote
/^ipv4[.](gateway|route)/ {                                # lines that start with ipv4.gateway or ipv4.route
                            gsub(/[.-]/,"_",$1)            # 1st field: replace period/hyphens with underscore
                            sub(/:[[:space:]]+/,"=" sq)    # replace colon + white space with equal sign and single quote
                            sub(/$/,sq)                    # append single quote on end of line
                            print                          # print current line

$ awk -f ipv4.awk ipv4settings.txt
ipv4_gateway='192.168.2.1'
ipv4_routes='--'
ipv4_route_metric='-1'
ipv4_route_table='0 (unspec)'

要将这些变量加载到当前环境中,您可以重定向 awk 结果到 source 例如:

$ source <(awk -f ipv4.awk ipv4settings.txt)
$ typeset -p ipv4_gateway ipv4_routes ipv4_route_metric ipv4_route_table
declare -- ipv4_gateway="192.168.2.1"
declare -- ipv4_routes="--"
declare -- ipv4_route_metric="-1"
declare -- ipv4_route_table="0 (unspec)"

ticktalk 5 月前

下面所示的内容是否合适?

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

$ cat fixme.awk
{printf("%s=\"%s\"\n",gensub(/[\.-]/,"_","g",$1),gensub(/ {2,}/,"","g",$2))}

$ awk -F: -f fixme.awk ipv4settings.txt
ipv4_gateway="192.168.2.1"
ipv4_routes="--"
ipv4_route_metric="-1"
ipv4_route_table="0 (unspec)"