Bash 变量分配错误-“；“未找到命令”；

当我运行脚本时，我得到以下错误。请问我做错了什么？感谢任何帮助-Bash新手

count_raw=0
avg_raw=0

$count_raw=1
$avg_raw=1

echo "count_raw=$count_raw"
echo "avg_raw=$avg_raw"

错误：

line 12: 0=1: command not found
line 13: 0=1: command not found

count_raw=0
avg_raw=0

$count_raw=1
$avg_raw=1

echo "count_raw=$count_raw"
echo "avg_raw=$avg_raw"

我的剧本：

count_raw=0
avg_raw=0

$count_raw=1
$avg_raw=1

echo "count_raw=$count_raw"
echo "avg_raw=$avg_raw"

---

=

是一个赋值运算符，当被发现为空闲时，

$

保存变量的值（不仅在美国，在bash中也是如此）

count_raw=0
avg_raw=0

$count_raw=1
$avg_raw=1

echo "count_raw=$count_raw"
echo "avg_raw=$avg_raw"

所以当你说：

$var=1

时，你实际上是在尝试在bash中键入一个随机字符串（在你的例子中是

0=1

），而bash不喜欢这样。请看下面的一行代码，它显示了一个示例，您可以在其中键入

$var=1

，bash将能够处理它：

count_raw=0
avg_raw=0

$count_raw=1
$avg_raw=1

echo "count_raw=$count_raw"
echo "avg_raw=$avg_raw"

var=1; if [[ $var=1 ]]; then printf "Congrats! You have learned the difference between variable assignment and variable comparison in the ${var}st attempt.\n"; fi;

---

在

count_raw

和

avg_raw

之前移除美元，修复了它。非常好，很有趣，还回答了T的问题。谢谢Donbhupi