Powershell检查int和double的类型

Question

我需要用户填写一个整数,但是下面显示的代码也允许使用双精度数.如何将其更改为仅允许整数？

do{

    $opgegevenGetal = read-host "Enter an integer"

    if(![bool]($opgegevenGetal -as [int])){
        write-host "Only integers please"
    }

}

until ([bool]($opgegevenGetal -as [int]))

Answer 1

尝试：

do{
    $opgegevenGetal = read-host "Enter an integer"
    $a = ""
    if(  ![int32]::TryParse( $opgegevenGetal , [ref]$a ))
      {
        write-host "Only integers please"
      }
  } until ($a -gt 0)

接受0作为输入的代码：

do{
    $opgegevenGetal = read-host "Enter an integer"
    $a = ""
    if(  ![int32]::TryParse( $opgegevenGetal , [ref]$a ))
      {
        write-host "Only integers please"
      }
  } until ($a -gt 0 -or $opgegevenGetal -eq '0')

或者：

do{
    $opgegevenGetal = read-host "Enter an integer"
    $a = ""
    $b = [int32]::TryParse( $opgegevenGetal , [ref]$a )
    if(  !$b)
      {
        write-host "Only integers please"
      }
  } until ($b)

Answer 2

克里斯蒂安答案的更紧凑版本:

$value = 0
$read = Read-Host 'Enter an integer'
while( ![int]::TryParse( $read, [ref]$value ) ) {
  $read = Read-Host 'Only integers please'
}