linux - 使用 sed 替换 bash 文件中的变量 - 输出滞后

Question

I'm trying to write a bash script that search and replace a specific user input saved in config.sh using sed. This does work; however it only works partially as shown below.

config.sh

#!/bin/bash
#
#UserName to be deleted
delUserName=""
#Source
delUserSrc=/Users/"$delUserName"
#Destination
delUserDest=/Users/John/BackUp/"$delUserName"/"$delUserName".zip

main.sh

#!/bin/bash
#
source scripts/config.sh

echo -e "\nEnter user you wish to delete: \c" 
read -r UserName
sed -i '' -e "s/delUserName=.*/delUserName=$UserName/g" scripts/config.sh
echo -e "delUserName: $delUserName"
echo -e "delUserSrc: $delUserSrc"
echo -e "delUserDest: $delUserDest"

output1:

Enter user you wish to delete: Test
delUserName: 
delUserSrc:/Users/
delUserDest:/Users/John/BackUp/ / .zip

output2:

Enter user you wish to delete: Test1
delUserName:Test
delUserSrc:/Users/Test
delUserDest:/Users/John/BackUp/Test/Test.zip

output3:

Enter user you wish to delete: Test1
delUserName:Test1
delUserSrc:/Users/Test1
delUserDest:/Users/John/BackUp/Test1/Test1.zip

expected output1:

Enter user you wish to delete: Test
delUserName:Test
delUserSrc:/Users/Test
delUserDest:/Users/John/BackUp/Test/Test.zip

expected output2:

Enter user you wish to delete: Test1
delUserName:Test1
delUserSrc:/Users/Test1
delUserDest:/Users/John/BackUp/Test1/Test1.zip

脚本滞后。 sed 立即更改了 $delUserName 的值，但是 $delUserName、$delUserSrc 和 $delUserDest 的正确值仅在第二次运行时出现。当所有变量都在 main.sh 中时，脚本运行良好，除非我必须这样做。将用户输入保存到 $UserName 中。知道为什么第一次运行时不显示这些值吗？ 谢谢

Answer 1

这是我认为正在发生的事情。

sed 命令替换文件中的文本。它不会修改内存中变量的值。这些值是在您 source config.sh 时分配的。

所以在你的 sed 行之后，你需要加入这一行:

source scripts/config.sh

它与您的脚本中的上述行相同。这也是必需的，以便您新替换的值将加载到变量中，以便您可以显示它们。一旦新值加载到内存中，echo 语句将能够将变量扩展为该新值。