Question

我试图找出如何搜索和替换.txt文件中特定行和特定列的密码。这就是它的样子：

Admin1 Pass1 1
Admin2 Pass2 1
User1 Upass1 0
User2 Upass2 0

这是我的代码：

while (true)
do
read -p 'Whose password would you like to change? Enter the corresponding user name.' readUser
userCheck=$(grep $readUser users.txt)

if [ "$userCheck" ]
then
    echo $userCheck > temp2.txt

    read -p 'Enter the old password' oldPass
        passCheck=$(awk '{print$2}' temp2.txt)

    if [ "$passCheck" == "$oldPass" ]
    then  

        read -p 'Enter the new password' newPass                
        sed -i "/^$readUser/ s/$oldPass/$newPass/" users.txt
        break
    else 
        echo 'The username and/or password do not match. Please try again.'
    fi
else 
    echo 'The username and/or password do not match. Please try again.'
fi
done

假设正在用TESTING替换User1的密码，结果就是：

Admin1 Pass1 1 Admin2 Pass2 1 User1 TESTING 0 User2 Upass2 0

我需要的是：

Admin1 Pass1 1
Admin2 Pass2 1
User1 TESTING 0
User2 Upass2 0

Answer 1

您的原始脚本几乎已经完成，只是缺少正确的引用。您可以使用双引号编写：echo "$updatePass" > data以保留换行符。有关报价here

的更多信息

但是，您的脚本还有改进的余地。你可以这样写：

#!/bin/bash

while (true)
do
    read -p 'Whose password would you like to change?' readUser

    # no need for a temporary variable here
    if [ "$(awk -v a="$readUser" '$1==a{print $1}' users.txt)" ] 
    then
        read -p 'Enter the old password' oldPass
        # the awk code checks if the $oldPass matches the recorded password 
        if [ "$oldPass" == "$(awk -v a="$readUser" '$1==a{print $2}' users.txt)" ]
        then 
            read -p 'Enter the new password' newPass
            # the -i flag for sed allows in-place substitution
            # we look for the line begining by $readUser, in case several users have the same password
            sed -i "/^$readUser/ s/$oldPass/$newPass/" users.txt
            break
        else 
            echo 'The username and/or password do not match. Please try again.'
        fi
    else 
        echo 'The username and/or password do not match. Please try again.'
    fi
done

Bash：在.txt文件中搜索并替换密码

1 个答案: