Bash 4.1.2如何从动态引用的关联数组中获取指定键值

问题背景 #

GNU Bash 4.1.2版本不支持declare -n（命名引用）特性，当尝试通过存储数组名称的变量动态获取关联数组的键值时，会触发bad substitution错误。

正常取值示例 #

直接使用关联数组名称可以正常获取对应键的值：

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')

# 直接使用数组名称取值
echo ${DBTWO["dbname"]}

输出：def_database（结果正确）

错误示例及原因 #

尝试通过变量间接引用关联数组时，错误的写法会导致语法报错：

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')

# 存储数组名称的数组
declare -a databases=(DBONE DBTWO DBTHREE)

# 获取第一个数组名称
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

原因：Bash 4.1.2不支持${$variable}这种嵌套变量替换语法，且没有命名引用特性来简化动态数组引用。

遍历动态数组的解决方案 #

针对遍历存储数组名称的数组这一需求，可使用间接扩展特性实现，无需依赖declare -n：

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 db in "${databases[@]}"; do
    # 构造关联数组的引用字符串
    ref="$db[dbname]"
    # 使用间接扩展获取对应键的值
    dbname="${!ref}"
    echo "The database is $db, and the dbname is $dbname"
done

输出：

The database is DBONE, and the dbname is abc_database
The database is DBTWO, and the dbname is def_database
The database is DBTHREE, and the dbname is xyz_database

也可以使用eval实现（注意：使用eval时需确保变量内容安全，避免注入风险）：

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 db in "${databases[@]}"; do
    eval dbname=\"\${$db[dbname]}\"
    echo "The database is $db, and the dbname is $dbname"
done

内容的提问来源于stack exchange，提问作者Genki