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Bash:如何从多维数组中的命名键获取值

multidimensional-array loops bash arrays

Genki · 技术社区 · 1 年前

当试图从动态引用的数组中获取命名键时,我遇到了一个“错误替换”错误。

GNU Bash版本4.1.2(2)

作品:

declare -rA DBONE=([system_name]='ABC' [dbname]='abc_database' [dbusername]='user1')
declare -rA DBTWO=([system_name]='DEF' [dbname]='def_database' [dbusername]='user2')
declare -rA DBTHREE=([system_name]='XYZ' [dbname]='xyz_database' [dbusername]='user3')

# works: get value when using the real name of the array, e.g. "DBTWO"
echo ${DBTWO["dbname"]}

输出:“def_database”(正确)

错误:替换错误

declare -rA DBONE=([system_name]='ABC' [dbname]='abc_database' [dbusername]='user1')
declare -rA DBTWO=([system_name]='DEF' [dbname]='def_database' [dbusername]='user2')
declare -rA DBTHREE=([system_name]='XYZ' [dbname]='xyz_database' [dbusername]='user3')

# create an array of the arrays
declare -a databases=(DBONE DBTWO DBTHREE)

# get the first array (e.g. DBONE)
database="${databases[0]}"
echo "The dbname is ${$database[dbname]}"
echo "The dbname is ${$database}[dbname]"

输出

./myscript.sh: ${$database[dbname]}: bad substitution
./myscript.sh: ${$database}[dbname]: bad substitution

最终,我的最终目标是在数组上循环,如下所示:

# create an array of the arrays
declare -a databases=(DBONE DBTWO DBTHREE)
for ((i=0; i<"${#databases[*]}"; i++)); do
    database="${databases[$i]}"
    dbname="${database[dbname]}"
    echo "The database is $database, and the dbname is $dbname"
done

注意:我的bash版本(GNU bash版本4.1.2(2))不支持 declare "-n" 。它允许以下选项: declare: usage: declare [-aAfFilrtux] [-p] [name[=value] ...]

1 回复 | 直到 1 年前

Philippe 1 年前

您可以使用间接引用:

declare -rA DBONE=([system_name]='ABC' [dbname]='abc_database' [dbusername]='user1')
declare -rA DBTWO=([system_name]='DEF' [dbname]='def_database' [dbusername]='user2')
declare -rA DBTHREE=([system_name]='XYZ' [dbname]='xyz_database' [dbusername]='user3')

declare -a databases=(DBONE DBTWO DBTHREE)
for ((i=0; i<"${#databases[*]}"; i++)); do
    database="${databases[$i]}"
    ref="$database[dbname]"
    dbname="${!ref}"
    echo "The database is $database, and the dbname is $dbname"
done

Paul Hodges 1 年前

bash 不做多维数组,但关联数组允许任意复杂的字符串作为键,所以如果你只重构一点设计,实际上也是一样的。

只需为每个字段使用一个标准名称,并附加一个关键字来区分它们。

#! /usr/bin/env bash

declare -rA db=( [system_name:DB1]='ABC' [dbname:DB1]='abc_database' [dbusername:DB1]='user1'
                 [system_name:DB2]='DEF' [dbname:DB2]='def_database' [dbusername:DB2]='user2'
                 [system_name:DB3]='XYZ' [dbname:DB3]='xyz_database' [dbusername:DB3]='user3' )

declare -ra databases=( DB1 DB2 DB3 )

setdata(){
  local dbkey="$1";
  declare -g dbname=${db["dbname:$dbkey"]}
  declare -g username=${db["dbusername:$dbkey"]}
  declare -g system_name=${db["system_name:$dbkey"]}
}

echo "With key variables -"
field=dbname; database="${databases[0]}"; key="$field:$database"
printf '\n%s\n' "The $field for $database is ${db[$key]}"

printf '\nRaw access -\n'
printf '\n%s\n' "The system_name for ${databases[1]} is ${db[system_name:${databases[1]}]}"

printf '\nLooping with function:\n\n'

for database in "${databases[@]}"
do setdata $database
   echo "$database: dbname=$dbname username=$username system_name=$system_name"
done
echo

使用中:

$: ./tst
With key variables -

The dbname for DB1 is abc_database

Raw access -

The system_name for DB2 is DEF

Looping with function:

DB1: dbname=abc_database username=user1 system_name=ABC
DB2: dbname=def_database username=user2 system_name=DEF
DB3: dbname=xyz_database username=user3 system_name=XYZ

glenn jackman 1 年前

如果您可以使用较新的bash版本,那么使用namerefs将变得非常容易:

for name in DB{ONE,TWO,THREE}; do
  declare -n ref="$name"
  printf 'array: %s; system: %s; db: %s\n' "$name" "${ref[system_name]}" "${ref[dbname]}"
done

array: DBONE; system: ABC; db: abc_database
array: DBTWO; system: DEF; db: def_database
array: DBTHREE; system: XYZ; db: xyz_database

我不记得namerefs是在4.4版还是5.0版中引入的