在bash脚本中导出环境变量 - 它不起作用

Question

我写了一个小的bash脚本来导出环境变量:

#!/bin/bash

echo "Pass a path:"
read path
echo $path

defaultPath = /home/katie/Desktop

if [ -n "$path" ]; then
    echo "Path is empty! Exporting default path ..."
    export my_var=$defaultPath
else
    export my_var=$path
fi

但是我得到了错误:

找不到defaultPath:命令

怎么解决？

WORKNG版本:

#!/bin/bash

echo "Pass a path:"
read path
echo $path

defaultPath=/home/user/Desktop

if [ -n "$path" ]; then
    export my_var=$path
else
    echo "Path is empty! Exporting default path ..."
    export my_var=$defaultPath
fi

Answer 1

在变量赋值中,=周围不允许有空格:

defaultPath=/home/katie/Desktop

对于空格,该行被解释为一个简单的命令,它试图defaultPath用两个参数执行命令,=和/home/katie/Desktop.