5条以上数据信息如何生成二维码

Question

问题是，当我尝试将我的字符串放入 qrgEncoder = new QRGEncoder 时，它总是出错，我知道这是错误的，因为我需要将我的所有字符串放在一个字符串或散列中，但我不知道如何将我的哈希.!非常感谢您的回复，先生，谢谢您的预付款

    firstname = findViewById(R.id.firstname);
    phone = findViewById(R.id.phone);
    address = findViewById(R.id.address);
    qr = findViewById(R.id.qr);
    btn = findViewById(R.id.btn);

    btn.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            String first = firstname.getText().toString().trim();
            String phones = phone.getText().toString().trim();
            String textaddress = address.getText().toString().trim();

            if(first.length() > 0 || phones.length() > 0 || textaddress.length() > 0) {

                WindowManager manager = (WindowManager) getSystemService(WINDOW_SERVICE);
                Display display = manager.getDefaultDisplay();
                Point point = new Point();
                display.getSize(point);
                int width = point.x;
                int height = point.y;
                int smallerDimension = width < height ? width : height;
                smallerDimension = smallerDimension * 3 / 4;

                qrgEncoder = new QRGEncoder(
                        first,phones,textaddress, null,
                        QRGContents.Type.TEXT,
                        smallerDimension);
                try {
                    bitmap = qrgEncoder.encodeAsBitmap();
                    qr.setImageBitmap(bitmap);
                } catch (WriterException e) {
                    Log.v(TAG, e.toString());
                }
            }
        }
    });
}

}

Answer 1

如果我查看您的代码以及如何 QRGEncoder implementations，我想您需要先连接字符串输入，然后再将其添加到其构造函数中。

因此，你应该这样称呼它：

StringBuilder qrParam = new StringBuilder();
qrParam.append(first);
qrParam.append(phones);
qrParam.append(textaddress);

qrgEncoder = new QRGEncoder(
                    qrParam.toString(),
                    null,
                    QRGContents.Type.TEXT,
                    smallerDimension);

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