Question

我正在尝试通过单击按钮进行搜索来开始语音识别，因为我使用的是此代码-

<style>
  .speech {border: 1px solid #DDD; width: 300px; padding: 0; margin: 0}
  .speech input {border: 0; width: 240px; display: inline-block; height: 30px;}
  .speech img {float: right; width: 40px }
</style>

<!-- Search Form -->
<form id="labnol" method="get" action="https://www.google.com/search">
  <div class="speech">
    <input type="text" name="q" id="transcript" placeholder="Speak" />
    <img onclick="startDictation()" src="//i.imgur.com/cHidSVu.gif" />
  </div>
</form>

<!-- HTML5 Speech Recognition API -->
<script>
  function startDictation() {

    if (window.hasOwnProperty('webkitSpeechRecognition')) {

      var recognition = new webkitSpeechRecognition();

      recognition.continuous = false;
      recognition.interimResults = false;

      recognition.lang = "en-US";
      recognition.start();

      recognition.onresult = function(e) {
        document.getElementById('transcript').value
                                 = e.results[0][0].transcript;
        recognition.stop();
        document.getElementById('labnol').submit();
      };

      recognition.onerror = function(e) {
        console.log(e);
        recognition.stop();
      }

    }
  }
</script>

我从recognition.onerror-SpeechRecognitionError {isTrusted: true, error: "not-allowed", message: "", type: "error", target: SpeechRecognition, …}

得到了e的值

我正在使用labnol

中的这段代码

我无法弄清问题，为什么我会出错-not allowed谁可以提供帮助。

Answer 1

这是因为来源不受信任。

来源应至少与以下（方案，主机，端口）模式之一匹配：

(https, *, *)
(wss, *, *)
(*, localhost, *)
(*, 127/8, *)
(*, ::1/128, *)
(file, *, —)
(chrome-extension, *, —)

有关更多说明，请访问-https://www.chromium.org/Home/chromium-security/prefer-secure-origins-for-powerful-new-features

使用js启动Google语音识别会产生错误-不允许

1 个答案: