java - Android 速率按钮语法错误

Question

我收到错误: token “marketrate”上的语法错误，此 token 后预期的 VariableDeclaratorId... Uri.parse("https://market.android.com/details?id=com.synamegames.giveaway");

和

token 语法错误、构造错误...startActivity(market)；

        private void makeDialog() {     

        AlertDialog.Builder about = new AlertDialog.Builder(this);      

        about.setMessage("About The Giveaway");

        about.setPositiveButton("Rate", new DialogInterface.OnClickListener() {
            Intent market = new Intent(
                    "android.intent.action.VIEW", 

                startActivity(market);
            public void onClick(DialogInterface arg0, int arg1) {
//action
            }
        });

        about.setNegativeButton("Close", new DialogInterface.OnClickListener() {
            public void onClick(DialogInterface arg0, int arg1) {}
        });

        about.show();
    }

}

Answer 1

我认为你有两个问题。

首先，这里的括号不匹配:

Intent market = new Intent(
    "android.intent.action.VIEW", 
    Uri.parse("https://market.android.com/details?id=com.synamegames.giveaway");
startActivity(market);

我认为在 Uri.parse 调用之后需要一个提取括号，以关闭 Intent 构造函数调用:

Intent market = new Intent(
    "android.intent.action.VIEW", 
    Uri.parse("https://market.android.com/details?id=com.synamegames.giveaway"));
startActivity(market);

或者提取 Uri.parse 调用以开始:

Uri uri = Uri.parse("https://market.android.com/details?id=com.synamegames.giveaway");
Intent market = new Intent("android.intent.action.VIEW", uri);
startActivity(market);

其次，您的 OnClickListener() 内部子类中实际上有一段代码，而不是初始化程序 block 或方法中。您是否打算将其放入 onClick 处理程序中？

about.setPositiveButton("Rate", new DialogInterface.OnClickListener() {
    public void onClick(DialogInterface arg0, int arg1) {
        Uri uri = Uri.parse("...");
        Intent market = new Intent("android.intent.action.VIEW", uri);
        startActivity(market);
    }
});