shell - 如何使用变量读取属性

Question

过去 2/3 天我一直面临这个问题。我有一个简单的脚本，我将在其中读取属性文件并输出一些内容。这是我的脚本

vi testloop.properties
dump_1=abc


vi testloop.sh
. testloop.properties
i=1
dmp="dump_"$i  //as per my understanding dmp=dump_1
echo $dump_1   //abc will echod since dump_1 in property file is abc
echo $dmp      ***i thought here also abc will come but unfortunately it is "dump_1"***
echo $(($dmp)) ***here output is 0*** 


sh testloop.sh
abc
dump_1
0

有人可以帮我指出 echo $dmp 和 echo $(($dmp)) 中发生了什么吗？为什么它不回显 abc

Answer 1

. ./testloop.properties
i=1
dmp="dump_"$i
echo $dump_1   # print abc as expected
echo $dmp      # prints dump_1 since this is what dmp contains
 # ***here output is 0*** 
eval echo \$$dmp # evaluates \$$dmp to $dump1 then echoes abc

至于 echo $(($dmp))，它尝试评估 arithmetic expression结果是0:

Arithmetic expansion allows the evaluation of an arithmetic expression and the substitution of the result. The format for arithmetic expansion is:

$(( expression ))