Bash脚本 - unix - 找不到命令

Question

我是bash脚本的新手，我想知道为什么我收到上述消息。我尝试一个来自for循环的算术值，然后我想打印数组。任何人都可以帮助我吗？

提前谢谢!!

#!/bin/bash
declare -a SCORES
for j in `seq 0 5`;
do
SCORES$j="$(sh myscript.sh $DSLAM $j | grep "" -c )"
done
for k in "${SCORES[@]}"
do
    echo "message $'\t' $SCORES$k"
done
echo ${#SCORES}

=======

输出

abcd.sh: line 16: SCORES0=3: command not found
abcd.sh: line 16: SCORES1=135: command not found
abcd.sh: line 16: SCORES2=826: command not found
abcd.sh: line 16: SCORES3=107: command not found
abcd.sh: line 16: SCORES4=3: command not found
abcd.sh: line 16: SCORES5=3: command not found
0

Answer 1

您不能使用在运行时生成的名称来分配变量;至少不是你尝试的方式。

您有以下选项：

declare "SCORES$j=$(sh myscript.sh $DSLAM $j | grep '' -c )" # creates new variables like SCORES1, SCORES2 etc.

eval "SCORES$j=$(sh myscript.sh $DSLAM $j | grep '' -c )" #Definitely not preferred.

SCORES[$j]="$(sh myscript.sh $DSLAM $j | grep '' -c )" #uses array you have created. 

最有可能的是，选项3就是你想要的。