regex - (GNU)Sed : how to replace any character from nth character to nth+10?

Question

我需要替换字符串中第 10 到第 20 个字符，如下所示:

123456789012345678901234567890

到目前为止我已经尝试过:

一个)

仅适用于第 10 个字符: echo “123456789012345678901234567890” | sed 's/./X/10'

b)不适用于范围:

echo "123456789012345678901234567890"  | sed 's/./X/10,20'
echo "123456789012345678901234567890"  | sed 's/./X/10\,20'
echo "123456789012345678901234567890"  | sed 's/./X/\{10,20\}'
echo "123456789012345678901234567890"  | sed 's/./X/\{10\,20\}'

不工作，我得到错误

unknown option to `s'

所以 - 问题是 - 我如何让它工作:

回显“123456789012345678901234567890” | sed 's/./X/10,20'

Answer 1

尝试:

$ sed -r "s/^(.{9})(.{11})/\1XXXXXXXXXX/" <<< 123456789012345678901234567890
123456789XXXXXXXXXX1234567890