java - 检查字符串是否包含数值和字母值

Question

我正在尝试检测字符串是否具有 NumericValue 和 AlphaValue。如果它只是数字，则 AmountBoolean = true，如果它是数字和 Alpha，则 AmountBoolean = false。我找不到让这个工作的方法。看来我找到了一种方法，但在运行时崩溃，在 logcat 中声明 onClick if 语句是一个问题，但我不知道它为什么或如何。

用于帮助解决此问题的链接是:

How to check if a string contains only digits in Java

Java - See if a string contains any characters in it

Tell if string contains a-z chars

主要 Activity ；

public class MainActivity extends AppCompatActivity {

Button Enter;
EditText eName,eDate,eAmount;
ListView lDebtList;
TextView tName,tDate,tAmount;

Boolean AmountBoolean = false;
String currentDate,name,amount,date;
Toast toastName,toastDate,toastAmount;

ArrayList<String> AmountlistItems = new ArrayList<String>();
ArrayList<String> DateListItems = new ArrayList<String>();
ArrayList<String> NameListItems = new ArrayList<String>();

ArrayAdapter<String> AmountAdapter;
ArrayAdapter<String> DateAdapter;
ArrayAdapter<String> NameAdapter;

String numRegex   = ".*[0-9].*";
String alphaRegex = ".*[A-Z].*";
@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    Enter = findViewById(R.id.Enter);
    eAmount = findViewById(R.id.eAmount);
    eDate = findViewById(R.id.eDate);
    eName = findViewById(R.id.eName);
    lDebtList = findViewById(R.id.lDebtList);
    tAmount = findViewById(R.id.tAmount);
    tDate = findViewById(R.id.tDate);
    tName = findViewById(R.id.tName);

    eAmount.clearFocus();
    eDate.clearFocus();
    eName.clearFocus();

    currentDate = new SimpleDateFormat("MM-dd-yyyy", Locale.getDefault()).format(new Date());

    tAmount.setText("Amount:");
    tDate.setText("Date Owed");
    tName.setText("Name:");

    eAmount.setHint("$$$");
    eDate.setHint(currentDate);
    eName.setHint("John Doe");


    amount = eAmount.getText().toString();
    date = eDate.getText().toString();
    name = eName.getText().toString();



    if (amount.contains(numRegex) && !amount.contains(alphaRegex)) {
        AmountBoolean = true;
    }
     if(amount.matches(numRegex) && amount.contains(alphaRegex)){
        AmountBoolean = false;
    }


    AmountAdapter = new ArrayAdapter<String>(MainActivity.this,android.R.layout.simple_list_item_1, AmountlistItems);
    lDebtList.setAdapter(AmountAdapter);
}

public void onEnter(View view){
    if(AmountBoolean){
        //AmountAdapter.add(amount);
        AmountAdapter.add(eAmount.getText().toString());
        AmountAdapter.notifyDataSetChanged();
    }

    if(!AmountBoolean){
        toastAmount = Toast.makeText(this, "Please correct Amount Owed", Toast.LENGTH_SHORT);
        toastAmount.show();

    }


}


}  

感谢任何和所有的帮助，此外，如果有更简单的方法来实现这一点，请不要犹豫，展示如何。我知道之前有人问过这个问题，但我花了几个小时尝试各种方法，但都没有奏效，我正在求助于此。

Answer 1

使用下面的正则表达式:

只检查数字

^[0-9]+$

只检查字母

^[a-zA-Z]+$

检查字符串是否包含字母

.*([a-zA-Z]).*

检查字符串是否为字母数字

^[a-zA-Z0-9]+$