如何将命令行参数传递给函数。范围错误？

Question

如何将命令行参数传递给bash函数？

代码：

#/bin/bash

arg = "$1"

display_success() {
  echo -e "\nSuccessfully created new response chart for the $1 client."
  sleep 1
  echo -e "Set up Firebase Production environment.\n"
}

display_success "$arg"

我不断得到

的结果

Successfully created new response chart for the  client.

注意空格。