我可以使用 sed 来操作 bash 中的变量吗？
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Can I use sed to manipulate a variable in bash?
提问by Leo Chan
In my program, I would like to first get the user input, and insert a \before each /so I write this, but it doesn't work.
在我的程序中，我想首先获取用户输入，并\在每个输入之前插入一个，/所以我写了这个，但它不起作用。
echo "input a website"
read website
sed '/\//i\/' $website
回答by Eugene Yarmash
Try this:
尝试这个：
website=$(sed 's|/|\/|g' <<< $website)
Bash actually supports this sort of replacement natively:
Bash 实际上本身就支持这种替换：
${parameter/pattern/string}— replace the first match of patternwith string.${parameter//pattern/string}— replace all matches of patternwith string.
${parameter/pattern/string}-取代的第一场比赛pattern用string。${parameter//pattern/string}-更换所有比赛pattern用string。
Therefore you can do:
因此你可以这样做：
website=${website////\/}
Explanation:
解释：
website=${website // / / \/}
^ ^ ^ ^
| | | |
| | | string, '\' needs to be backslashed
| | delimiter
| pattern
replace globally
回答by Karoly Horvath
echo $website | sed 's/\//\\//g'
or, for better readability:
或者，为了更好的可读性：
echo $website | sed 's|/|\/|g'
回答by Noam Manos
You can also use Parameter-Expansionto replace sub-strings in variable. For example:
您还可以使用参数扩展来替换变量中的子字符串。例如：
website="https://stackoverflow.com/a/58899829/658497"
echo "${website//\//\/}"
https:\/\/stackoverflow.com\/a\/58899829\/658497
https:\/\/stackoverflow.com\/a\/58899829\/658497
