有没有更好的方式来逃避一倍(或逃跑两次)在BASH用户输入比调用printf的两倍? [英] Is there a better way to double escape (or escape twice) user input in BASH than calling printf twice?
问题描述
该脚本将读取一个古老而来自用户的新值,然后使用SED查找并在文件中替换它们。例如,如果我进入了T * T * Z和B * B *ž它看起来对于T \\ * T \\ * Z在文件中,并为B \\ * B \\ * Z取代。它的工作原理,但我一直在试图使这个更简洁。
This script will read an old and a new value from the user and then use sed to find and replace them in a file. For example if I entered T*T*z and B*B*z it would look for T\*T\*z in the file and replace with B\*B\*z. It works but I've been trying to make this more concise.
我没有任何需要的中间变量$ ESC_OLD_PW和$ ESC_NEW_PW。是否有更多的 明智的办法做到这一点?
I don't have any need for the intermediate variables $ESC_OLD_PW and $ESC_NEW_PW. Is there a more sensible way to do this?
#!/bin/bash
read -sp "Old:" OLD_PW && echo
read -sp "New:" NEW_PW && echo
# Add escape characters to what user entered
printf -v ESC_OLD_PW "%q" "${OLD_PW}"
printf -v ESC_NEW_PW "%q" "${NEW_PW}"
# Escape again for the sed evaluation.
printf -v ESC_ESC_OLD_PW "%q" "${ESC_OLD_PW}"
printf -v ESC_ESC_NEW_PW "%q" "${ESC_NEW_PW}"
sed -i -e s/"${ESC_ESC_OLD_PW}"/"${ESC_ESC_NEW_PW}"/g $1
我试过以下内容:
I've tried the following:
~$ OLD_PW="T*T*z"
~$ printf "%q" $OLD_PW | xargs printf "%q"
printf: %q: invalid conversion specification
~$
和我试过在管道的事情很多变化成的printf ...有什么建议?
And I've tried many variations on piping things into printf... Any suggestions?
推荐答案
中间变量免费电话:
sed -i -e "s/$(printf '%q' $(printf '%q' $OLD_PW))/$(printf '%q' $(printf '%q' $NEW_PW))/g" $1
这篇关于有没有更好的方式来逃避一倍(或逃跑两次)在BASH用户输入比调用printf的两倍?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!