如何用*替换字符串中的中间字符? [英] How to replace middle chars with * in a string?

问题描述

我有以下内容

$builder = [System.Data.SqlClient.SqlConnectionStringBuilder]::New('Connection Timeout=120;User Id=UID1;Data Source=datasource.com;Password=password12!553;')

$builder

这将按原样打印出连接字符串.

This will print out the connection string as is.

我想打印出带有密码的连接字符串，只显示第一个和最后一个字符，中间的其余字符为 *

i want to print out the connection string with password showing only the 1st and last chars, and rest of chars in middle as *

Connection Timeout=120;User Id=UID1;Data Source=datasource.com;Password=p************3; 

我怎样才能动态完成(即知道密码值的长度，然后用 * 替换中间字符)?

how can i accomplish that dynamically (i.e. knowing the length of the password value and then replacing the middle chars with *)?

pseudocode: $builder.Password -replace($_ middle_chars, "*") 

推荐答案

function Hide-ConnectionStringPassword {
    param(
       [parameter(Mandatory,ValueFromPipeline)]
       [System.Data.SqlClient.SqlConnectionStringBuilder]$ConnectionString
    )
    [string]$FistChar = $ConnectionString.Password[0]
    [string]$LastChar = $ConnectionString.Password[($ConnectionString.Password.Length - 1)]
    [string]$Stars = '*' * ($ConnectionString.Password.Length - 2)
    $ConnectionString.Password = $FistChar + $Stars + $LastChar 
    return $ConnectionString.ConnectionString
}

Hide-ConnectionStringPassword 'Connection Timeout=120;User Id=UID1;Data Source=datasource.com;Password=password12!553;'

输出:

Data Source=datasource.com;User ID=UID1;Password=p************3;Connect Timeout=120

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