Question

我正在尝试将EditText值与TextView 进行比较，但总是得到“不匹配！”

这是比较两个值的错误方法吗？

例如：我已在 TextView 中存储 win ，并在 EditText 中输入相同的 win ，但获取< strong> 不匹配 ......

    EditText editPassword;
    String strPassword;
    TextView lblPassword;
    String password;
    String strMatch;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.single_list_item);

        editPassword = (EditText) findViewById(R.id.editPassword);

        strPassword = editPassword.getText().toString();

        // getting intent data
        Intent in = getIntent();

        password = in.getStringExtra(TAG_PASSWORD);

        lblPassword = (TextView) findViewById(R.id.password_label);

        lblPassword.setText(password);

        strMatch= lblPassword.getText().toString();        

        btnSubmit = (Button) findViewById(R.id.btnSubmit);
        btnSubmit.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View arg0) {
                // TODO Auto-generated method stub
                if(strPassword.equals(strMatch))
                {
                  Toast.makeText(getApplicationContext(), "Match !",
                                    Toast.LENGTH_LONG).show();
                  editPassword.setText(null);
                }
                else 
                {
                  Toast.makeText(getApplicationContext(), "Does not match !",
                            Toast.LENGTH_LONG).show();
                }
            }
        });        
    }

Answer 1

在onClick

的Buttton方法中添加此行

 strPassword = editPassword.getText().toString();

显示您的edittext值为空字符串表示""

Answer 2

使用

 strPassword = editPassword.getText().toString();

在onClick（）事件中的行

 In your case the edit text value is always to be ""

Answer 3

EditText editPassword;
    String strPassword;
    TextView lblPassword;
    String password;
    String strMatch;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.single_list_item);

        editPassword = (EditText) findViewById(R.id.editPassword);



        // getting intent data
        Intent in = getIntent();

        password = in.getStringExtra(TAG_PASSWORD);

        lblPassword = (TextView) findViewById(R.id.password_label);

        lblPassword.setText(password);



        btnSubmit = (Button) findViewById(R.id.btnSubmit);
        btnSubmit.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View arg0) {
                // TODO Auto-generated method stub
              strPassword = editPassword.getText().toString();
              strMatch= lblPassword.getText().toString(); 

                if(strPassword.equals(strMatch))
                {
                  Toast.makeText(getApplicationContext(), "Match !",
                                    Toast.LENGTH_LONG).show();
                  editPassword.setText(null);
                }
                else 
                {
                  Toast.makeText(getApplicationContext(), "Does not match !",
                            Toast.LENGTH_LONG).show();
                }
            }
        });        
    }

Answer 4

想想你的线的顺序。首先将ONCE分配给strPassword和strMatch，然后更改底层控件（甚至可以通过编辑GUI中的内容），但是onClick（）方法仍会比较最初存储在strPassword和strMatch中的值。 / p>

尝试替换

if(strPassword.equals(strMatch))

与

if(editPassword.getText().equals(lblPassword.getText()))

（或将toString()附加到每个，我不记得是否需要toString()。

Answer 5

尝试以下代码

if(strPassword.trim().equals(strMatch.trim()))

比较TextView和EditText值

5 个答案: