Bash (.sh) 文件不生成与 Ubuntu/Linux 中的终端命令相同的任何输出答案

【问题标题】：Bash (.sh) file not generating any output as same as terminal command in Ubuntu/LinuxBash (.sh) 文件不生成与 Ubuntu/Linux 中的终端命令相同的任何输出
【发布时间】：2020-10-27 15:33:07
【问题描述】：

需要帮助！我已经编写了以下 Ubuntu 终端命令并生成输出。现在我想通过编写下面的 (.sh) 脚本检查来做同样的事情：

Ubuntu terminal command:

sudo journalctl | grep -E 'authentication failure' | grep -E 'testusr1' | more

Output:
Jul 05 13:44:56 khUbuntu su[20464]: pam_unix(su-l:auth): authentication failure; logname= uid=1000 euid=0 tty=pts/1 ruser=kh rhost=  user=testusr1
Jul 05 14:56:44 khUbuntu su[23766]: pam_unix(su-l:auth): authentication failure; logname= uid=1000 euid=0 tty=pts/2 ruser=kh rhost=  user=testusr1

然后我写了以下word.sh（shell脚本），但它没有像上面的终端命令那样生成任何输出，我写错了什么？？

#!/bin/bash

# User input:
echo "Enter what you want to Search : "
read inputword

#userinput="authentication failure AND testusr1"

b=$(echo ${inputword%*AND*})   # remove AND from the inputword and get 'authentication failure'
length=${#b}
echo "$length" # get the length of b
removeBack1=${b%%*( )} #remove any white space from tail
length=${#removeBack1} #check length again to verify with the top lenth $b
echo "$length"
echo "This is 1st value:$removeBack1"

a=$(echo ${inputword#*AND})    # removing word AND from inputword and get 'testusr1'
length=${#a}
echo "$length"
removeBack2=${a%%*( )}
length=${#removeBack2}
echo "$length"
echo "This is 2nd value:$removeBack2"

## --- This is not generating any output -----------
sudo journalctl | grep -E '$removeBack1' | grep -E '$removeBack2' | more

【问题讨论】：

标签： linux bash shell ubuntu terminal

【解决方案1】：

我认为您希望在最后一个命令中包含变量，而不是字符串文字：

sudo journalctl | grep -E "$removeBack1" | grep -E "$removeBack2" | more

注意单引号到双引号的变化 - single 将字符串视为原始，而 double 扩展变量。

【讨论】：