Question

我应该更正此代码，并使其显示正确的输出。用户输入的数字介于1到10之间，其他任何数字都应提示用户该数字无效。

我尝试将do..while与do..until结合使用，但我要么不断陷入无限循环，要么不显示有效或无效数字。

[int]$UserGuess = Read-Host "pick a number between 1 and 10"

do {
    $UserGuess = Read-Host "Invalid number. Please re-enter (1 - 10)"
} while ($UserGuess -lt 1 -and $UserGuess -gt 10) {
}
do {
    Write-Host "You guessed: $UserGuess"
} until ($UserGuess -gt 1 -and $UserGuess -lt 10) {
}

它显示无限循环。

Answer 1

您应使用-or而不是-and，并检查第一个条目。结果的显示不需要直到循环。并释放空的脚本块。

[int]$UserGuess = Read-Host "pick a number between 1 and 10"

while ($UserGuess -lt 1 -or $UserGuess -gt 10)  
{
    $UserGuess = Read-Host "Invalid number. Please re-enter (1 - 10)"
} 

Write-Host "You guessed: $UserGuess"

Answer 2

就像mhu回答的那样，首先您需要使用-or来代替，因为11不能同时大于10和小于10。同样，do..while loop在这种情况下也不可取，因为它至少每次都必须提供一个输出：

无效的号码。请重新输入（1-10）

即使我输入了正确的数字。如果要使用它，则还必须使用if statement来确保其正常工作，如下所示：

[int]$UserGuess = Read-Host "pick a number between 1 and 10"

do {
    if (($UserGuess -gt 10) -or ($UserGuess -lt 1))
    {
        $UserGuess = Read-Host "Invalid number. Please re-enter (1 - 10)"
    }
} while (($UserGuess -gt 10) -or ($UserGuess -lt 1)) {
}

    Write-Host "You guessed: $UserGuess"

Answer 3

为您的理解提供了另一个示例，还添加了一些注释

$Numbers =1..10
$Userinput =read-host "Please enter number between 1 and 10 "

while(1){ ## loop for eternity

if ( $Numbers -contains $Userinput) ## if your list contains user entered number
{

write-host "You have entered right input, you guessed : $userinput"
$prompt =Read-Host "do you want to still play : Press Y for Yes ,N for No"

Switch($prompt){ ## check for second prompt

'Y' {continue}
'N' {return}
Default { "Wrong option,Press Y or N only " }

              }

}

## if user enters wrong input, he wwill be prompted again
Write-Host "You have entered wrong input,Please enter again"
$Userinput =read-host "Please enter number between 1 and 10 "

}

我的do / while循环代码无法正常工作，我无法弄清楚问题出在哪里

3 个答案: