将Android上的语音输入与单词进行比较

时间:2015-02-01 11:46:44

标签: android speech-recognition

我这是为了检查人们说的是否包含“recherche”,但它没有显示吐司:

public void bn6(View view ){
    Intent intent = new Intent(RecognizerIntent.ACTION_RECOGNIZE_SPEECH);
    intent.putExtra(RecognizerIntent.EXTRA_LANGUAGE_MODEL,
            RecognizerIntent.LANGUAGE_MODEL_FREE_FORM);
    intent.putExtra(RecognizerIntent.EXTRA_PROMPT, "Speech to text");
    startActivityForResult(intent, 1);
}

public void onActivityResult(int requestCode, int resultCode, Intent data) {
    if (requestCode == 1 && resultCode == RESULT_OK) {
        ArrayList<String> matches = data.getStringArrayListExtra(
                RecognizerIntent.EXTRA_RESULTS);
        TextView speechText = (TextView)findViewById(R.id.speechText);
        speechText.setText(matches.get(0).toString());
        String spec =  speechText.getText().toString();
        if (spec.toLowerCase().contains("Recherche")) {
            Toast.makeText(getApplicationContext(), "bravo", Toast.LENGTH_LONG).show();
        }
    }
    super.onActivityResult(requestCode, resultCode, data);
}

因此它会更改textview,但它不会覆盖任何内容

2 个答案:

答案 0 :(得分:0)

  。

spec.toLowerCase()包含(&#34; RECHERCHE&#34)

如果您比较较低的&#39; r&#39;您的文字将始终为false。到资本&#39; R&#39;

使用&#34; recherche&#34;小写

答案 1 :(得分:0)

在这部分:

spec.toLowerCase().contains("Recherche")

您将规范转换为小写,但 Recherche R 开头,这是一个大写!请改用:

spec.toLowerCase().contains("recherche")