shell - 如何从输出中删除换行符？

Question

在 shell (sha512) 中散列密码打破了这一行。
如何在一行中得到结果？

哈希脚本:

password="abc123"
hashPassw="$(/bin/echo -n "${password}" | openssl dgst -binary -sha512 | openssl enc -base64)"
echo "${hashPassw}"

输出是(为什么要断线？):

xwtd2ev7b1HQnUEytxcMnSB1CnhS8AaA9lZY8DEOgQBW5nY8NMmgCw6UAHb1RJXB
afwjAszrMSA5JxxDRpUH3A==

应该是一行:

xwtd2ev7b1HQnUEytxcMnSB1CnhS8AaA9lZY8DEOgQBW5nY8NMmgCw6UAHb1RJXBafwjAszrMSA5JxxDRpUH3A==

Answer 1

来自 OpenSSL Wiki为 enc .

To suppress this you can use in addition to -base64 the -A flag. This will produce a file with no line breaks at all.

所以添加额外的 -A flag 可以解决问题。

password="abc123"
hashPassw="$(/bin/echo -n "${password}" | openssl dgst -binary -sha512 | openssl enc -A -base64)"
echo "${hashPassw}"

哪些输出

xwtd2ev7b1HQnUEytxcMnSB1CnhS8AaA9lZY8DEOgQBW5nY8NMmgCw6UAHb1RJXBafwjAszrMSA5JxxDRpUH3A==