Android AlertDialog 显示时出错

Question

我正在做一个 android 项目，在这一点上，我想要这个 'fab' onclick 时，它会弹出一个 window文本输入。和两个按钮。但它带来了一个我试图修复的错误，最后我想我可以在这里找到帮助。

这一行有错误：

AlertDialog.Builder builder = new AlertDialog.Builder(this); with this one too: final EditText input = new EditText(this);

 private String m_Text = "";


@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    Toolbar toolbar = (Toolbar) findViewById(R.id.toolbar);
    setSupportActionBar(toolbar);

    FloatingActionButton fab = (FloatingActionButton) findViewById(R.id.fab);
    fab.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {

            AlertDialog.Builder builder = new AlertDialog.Builder(this);
            builder.setTitle("QuickSearch:");
            //input initial
            final EditText input = new EditText(this);
            //
            input.setInputType(InputType.TYPE_CLASS_TEXT );
            builder.setView(input);
            // Set of  buttons to be displayed.
            builder.setPositiveButton("Go", new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialog, int which) {
                    m_Text = input.getText().toString();
                }
            });
            builder.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialog, int which) {
                    dialog.cancel();
                }
            });

            builder.show();


            //Search action here.
            //Snackbar.make(view, "Replace with your own action", Snackbar.LENGTH_LONG)
                  //  .setAction("Action", null).show();
        }
    });

Answer 1

您在每次单击按钮时都在创建新的 Alertdialog 实例..而不删除视图

试试这个：

    AlertDialog.Builder builder = new AlertDialog.Builder(this);
   final AlertDialog alert = builder.create();


  fab.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {


            builder.setTitle("QuickSearch:");
            //input initial
            final EditText input = new EditText(MainActivity.this);
            //
            input.setInputType(InputType.TYPE_CLASS_TEXT );
            builder.setView(input);
            // Set of  buttons to be displayed.
            builder.setPositiveButton("Go", new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialog, int which) {
                    m_Text = input.getText().toString();
                }
            });
            builder.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
                @Override
                public void onClick(DialogInterface dialog, int which) {
                    alert.dismiss();
                }
            });

            alert.show();


            //Search action here.
            //Snackbar.make(view, "Replace with your own action", Snackbar.LENGTH_LONG)
                  //  .setAction("Action", null).show();
        }
    });

Answer 2

在您的代码中，this 指的是 onClick 而不是 context of that Activity。

代码应该是这样的

AlertDialog.Builder builder = new AlertDialog.Builder(YourActivityName.this);
final EditText input = new EditText(YourActivityName.this);

What is the meaning of “this” in Java?

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