TextView的比较和价值观的EditText [英] Compare TextView and EditText Values

问题描述

我想的与TextView的的比较值的EditText，但总是得到不匹配！

这是错误的方式来比较两个值？

有关的例子：我已存储的赢中的的TextView 的并输入相同的赢中的的EditText 的，但得到< STRONG>的不匹配的 ......

 的EditText editPassword;
    字符串strPassword;
    TextView的lblPassword;
    字符串密码;
    串strMatch;    @覆盖
    公共无效的onCreate（捆绑savedInstanceState）{
        super.onCreate（savedInstanceState）;
        的setContentView（R.layout.single_list_item）;        editPassword =（EditText上）findViewById（R.id.editPassword）;        。strPassword = editPassword.getText（）的toString（）;        //获取数据的意图
        在意向= getIntent（）;        密码= in.getStringExtra（TAG_PASSWORD）;        lblPassword =（的TextView）findViewById（R.id.password_label）;        lblPassword.setText（密码）;        。strMatch = lblPassword.getText（）的toString（）;        btnSubmit按钮=（按钮）findViewById（R.id.btnSubmit）;
        btnSubmit.setOnClickListener（新OnClickListener（）{            @覆盖
            公共无效的onClick（查看为arg0）{
                // TODO自动生成方法存根
                如果（strPassword.equals（strMatch））
                {
                  Toast.makeText（getApplicationContext（）的比赛！
                                    Toast.LENGTH_LONG）.show（）;
                  editPassword.setText（NULL）;
                }
                其他
                {
                  Toast.makeText（getApplicationContext（），不匹配！
                            Toast.LENGTH_LONG）.show（）;
                }
            }
        }）;
    }
 

解决方案

添加此行的按钮移到你的的onClick 法

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

present您的EditText值是空字符串方式，

I am trying to compare EditText value with TextView, but always getting "Does not match !"

Is this the wrong way to compare two values ?

For an example : I have stored win in TextView and entering same win in EditText but getting Does Not Match ......

    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();
                }
            }
        });        
    }

解决方案

Add this line in your onClick method of the Buttton

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

Present your edittext value is empty string means ""

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