点击按钮打开拨号盘，这个字符串例如String str = "*#06#" 应该在拨号盘中打开。？

Question

我的代码是这样的：

public void onclickdial(View v) {
String str = "*#06#";
startActivity(new Intent(Intent.ACTION_DIAL,Uri.parse("tel:"+str)));
Toast.makeText(getApplicationContext(), str, Toast.LENGTH_SHORT).show(); }

但只有一个问题 * 符号显示在拨号屏幕上：

Answer 1

        String str = "*#06#";
    Intent callIntent = new Intent(Intent.ACTION_DIAL);
    callIntent.setData(Uri.fromParts("tel", str, ","));
    Toast.makeText(getApplicationContext(), str, Toast.LENGTH_SHORT).show();
    startActivity(callIntent);

或：

        String str = "*#06#";
    startActivity(new Intent(Intent.ACTION_DIAL,Uri.parse("tel:"+ Uri.encode(str))));
    Toast.makeText(getApplicationContext(), str, Toast.LENGTH_SHORT).show();

点击按钮打开拨号盘，这个字符串例如String str = "*#06#" 应该在拨号盘中打开。？

click button to open dial pad, this string for example String str = "*#06#" should be open in dial pad.?

java

buttonclick