Question

我想了解如何将variable.method输出的结果保存到另一个变量中。

我想测试创作一首歌但我仍然坚持如何继续。

目前语音输出本身但不保存在变量中。

他们的逻辑原因是什么？我只学了一个月的powershell。

Add-Type -AssemblyName System.Speech

$speak = New-Object -TypeName System.Speech.Synthesis.SpeechSynthesizer

$man = $speak.Speak("Male Voice") # Outputs the Audio to host, but it does not store the result into Variable
$speak.SelectVoice("Microsoft Zira Desktop")
$woman = $speak.Speak("Female Voice")# Same as above

function song {
$man
$woman
song
}

Answer 1

Speak方法没有返回值

public void Speak（string textToSpeak）

但是可以通过SetOutputToWaveFile将音频输入设置为文件然后，您可以使用SoundPlayer立即播放音频，以获得更好的用户体验。

Add-Type -AssemblyName System.Speech

$soundFilePath = "Test.wav"

# Save the Audio to a file
[System.Speech.Synthesis.SpeechSynthesizer] $voice = $null
Try {
    $voice = New-Object -TypeName System.Speech.Synthesis.SpeechSynthesizer
    $voice.SetOutputToWaveFile($soundFilePath)

    $voice.Speak("Male Voice")
    $voice.SelectVoice("Microsoft Zira Desktop")
    $voice.Speak("Female Voice")
} Finally {
    # Make sure the wave file is closed
    if ($voice) {
        $voice.Dispose()
    }
}


# Load the saved audio and play it back
[System.Media.SoundPlayer] $player = $null
Try {
    $player = New-Object -TypeName System.Media.SoundPlayer($soundFilePath)
    $player.PlaySync()
} Finally {
    if ($player) {
        $player.Dispose()
    }
}

如果您只需要内存中的音频并且不想对任何内容进行后期处理，则可以将数据写入MemoryStream。

Add-Type -AssemblyName System.Speech

[System.IO.MemoryStream] $memoryStream = $null;

Try {
    $memoryStream = New-Object -TypeName System.IO.MemoryStream
    [System.Speech.Synthesis.SpeechSynthesizer] $voice = $null
    Try {
        $voice = New-Object -TypeName System.Speech.Synthesis.SpeechSynthesizer
        $voice.SetOutputToWaveStream($memoryStream)

        $voice.Speak("Male Voice")
        $voice.SelectVoice("Microsoft Zira Desktop")
        $voice.Speak("Female Voice")
    } Finally {
        if ($voice) {
            $voice.Dispose()
        }
    }


    # Load the saved audio and play it back
    [System.Media.SoundPlayer] $player = $null
    Try {
        $memoryStream.Seek(0, [System.IO.SeekOrigin]::Begin) | Out-Null
        $player = New-Object -TypeName System.Media.SoundPlayer($memoryStream)
        $player.PlaySync()
    } Finally {
        if ($player) {
            $player.Dispose()
        }
    }

} Finally {
    if ($memoryStream) {
        $memoryStream.Dispose()
    }
}

Answer 2

您无法将语音输出放入变量中，因为它不是函数返回的值，而是＆＃34;副作用＆＃34;。

但是没有必要存储音频。每次运行脚本时，只存储文本并生成语音输出会更有效。

编辑：我使用.SpeakAsync和sleep来更好地控制时间。调整延迟数，使其适合每条线的流量。

Add-Type -AssemblyName System.Speech

$male = New-Object -TypeName System.Speech.Synthesis.SpeechSynthesizer
$male.SelectVoice("Microsoft David Desktop")

$female = New-Object -TypeName System.Speech.Synthesis.SpeechSynthesizer
$female.SelectVoice("Microsoft Zira Desktop")

function male([string] $text, [int] $duration) {
    $male.SpeakAsync($text) | Out-Null
    sleep -Milliseconds $duration
}

function female([string] $text, [int] $duration) {
    $female.SpeakAsync($text) | Out-Null
    sleep -Milliseconds $duration
}

function song {
    male 'hello' 500
    female 'goodbye' 500
}

song

Powershell - 语音合成保存输出

2 个答案: