获取名称由变量本身组成的变量的值

Question

脚本：

#!/bin/bash
for SID in db1 db2 db3 db4
do
    "$SID"_fs=$(          
    df -k                           
    | grep "$SID\/data"     
    | tr -s ' '             
    | cut -d' ' -f6         
    | awk  '{print substr([=10=],length([=10=]),1)}' 
    | sort                          
    | tail -1);
    echo "$SID"_fs
done


./test_sm.sh: line 11: db1_fs=7: command not found
db1_fs
./test_sm.sh: line 11: db2_fs=7: command not found
db2_fs
./test_sm.sh: line 11: db3_fs=3: command not found
db3_fs
./test_sm.sh: line 11: db4_fs=2: command not found
db4_fs

变量设置为正确的值，但最终的 "echo" 没有给我变量的值（这是我需要的）。相反，它给了我变量名。

Answer 1

来自http://www.tldp.org/LDP/abs/html/ivr.html：

Indirect referencing in Bash is a multi-step process. First, take the name of a variable: varname. Then, reference it: $varname. Then, reference the reference: $$varname. Then, escape the first $: $$varname. Finally, force a reevaluation of the expression and assign it: eval newvar=$$varname.

所以在你的情况下：

eval echo $"$SID"_fs

Answer 2

使用declare。下面展示了如何设置变量名（使用declare）和如何检索值（使用间接参数扩展）。

for SID in db1 db2 db3 db4
do
    name="${SID}_fs"
    value=$(...)
    declare "$name=$value"

    echo "${!name}"
done

bash 4.3 引入了 namerefs 来简化这个任务。

for SID in db1 db2 db3 db4
do
    declare -n sid_fs=${SID}_fs
    sid_fs=$(...)
    echo "$sid_fs"
done

您也可以考虑使用关联数组，而不是单独的参数。

declare -A fs
for SID in db1 db2 db3 db4; do
    fs[$SID]=$(...)
    echo ${fs[$SID]}
done

获取名称由变量本身组成的变量的值

Get the value of a variable whose name consists of a variable itself

variables

bash

shell

for-loop

sh