Question

我有一个Email EditText，我想通过电子邮件验证来检查它。

这是我的电子邮件验证码

public final static boolean isValidEmail(CharSequence target) {
    if (target == null) {
        return false;
    } else {
        return android.util.Patterns.EMAIL_ADDRESS.matcher(target).matches();
    }
}

public void showAlertValidation() {
    AlertDialog.Builder alertDialog = new AlertDialog.Builder(RegisterActivity.this);
    alertDialog.setTitle("Failed");
    alertDialog.setMessage("Invalid Email");
    alertDialog.setNegativeButton("Close", new DialogInterface.OnClickListener() {
        public void onClick(DialogInterface dialog, int which) {
            dialog.cancel();
        }
    });
    alertDialog.show();
}

这是我的EditText验证码

editTextEmail= (EditText) findViewById(R.id.editTextEmail);
email = editTextEmail.getText().toString();
if(email.length() == 0) {
    editTextEmail.setError("Email required!");
    if (isValidEmail(email)) {
          Toast.makeText(getApplicationContext(),"valid email address",Toast.LENGTH_SHORT).show();
     }
     else{
          showAlertValidation();
     }
}

问题是EditText的结果。当EditText的值为null时，它将运行showAlertValidation(); 但是如果EditText的值是＆＃34;电子邮件＆＃34;或者＆＃34;发送电子邮件@ example＆＃34;或＆＃34; email@example.com" ;,它没有运行showAlertValidation(); 我的代码出了什么问题？

Answer 1

工作!!!!

我们现在有简单的电子邮件模式匹配器

 private static boolean isValidEmail(String email) {
        return !TextUtils.isEmpty(email) && android.util.Patterns.EMAIL_ADDRESS.matcher(email).matches();
    }

Answer 2

您错误地将EditText文本长度与0进行比较，并且只有在它为真的情况下，才会执行验证逻辑。

这是正确的代码：

 @Override
        public void onClick(View v) {
            String email = editTextEmail.getText().toString();
            if(email.length() != 0) {
                if (isValidEmail(email)) {
                    Toast.makeText(getApplicationContext(), "Valid email address!", Toast.LENGTH_SHORT).show();
                }
                else{
                    editTextEmail.setError("Email required!");
                    showAlertValidation();
                }
            }
            else{
                editTextEmail.setError("Email required!");
            }
        }

Answer 3

最好创建一个验证电子邮件的类。这样您就可以在任何需要的地方重复使用它。您甚至不需要将edittext的文本放在单独的字符串中。制作以下课程：

public class EmailValidator {
private Pattern pattern;
private Matcher matcher;
private static EmailValidator sInstance;


public static EmailValidator getInstance() {
    if (sInstance == null) {
        sInstance = new EmailValidator();
    }
    return sInstance;
}
private static final String EMAIL_PATTERN =
        "^[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*@"
                + "[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})$";

public EmailValidator() {
    pattern = Pattern.compile(EMAIL_PATTERN);
}

public boolean validate(final String hex) {

    matcher = pattern.matcher(hex);
    return matcher.matches();

}}

现在使用此类检查edittext的电子邮件验证，如下所示：

if(!EmailValidator.getInstance().validate(editTextEmail.getText().toString().trim())){
                    editTextEmail.setError("Invalid email address");
                }

希望这有帮助。

Answer 4

我认为这是你犯了错误的地方（EditText验证码）：

editTextEmail= (EditText) findViewById(R.id.editTextEmail);
email = editTextEmail.getText().toString();
if(email.length() == 0) {
    editTextEmail.setError("Email required!");

}
else if (isValidEmail(email)) {
              Toast.makeText(getApplicationContext(),"valid email address",Toast.LENGTH_SHORT).show();
         }
         else{
              showAlertValidation();
         }

Answer 5

在使用String.trim（）;

检查电子邮件字符串的长度之前，您应该清空空格“”

if(email.trim().length() == 0) {//code here}

Answer 6

问题是你的条件

if(email.length() == 0) 
    {
        editTextEmail.setError("Email required!");
        if (isValidEmail(email)) 
            {
                Toast.makeText(getApplicationContext(),"valid email address",Toast.LENGTH_SHORT).show();
            }
        else
            {
                showAlertValidation();
            }
    }

因为它只在您的editText为空时验证您的电子邮件地址。你能做的是：

email.addTextChangedListener(new TextWatcher()
        {
            @Override
            public void beforeTextChanged(CharSequence s, int start, int count, int after)
                {

                }

            @Override
            public void onTextChanged(CharSequence s, int start, int before, int count)
                {

                }

            @Override
            public void afterTextChanged(Editable s)
                {
                    if(s.length() > 0) 
                        {
                            if (isValidEmail(email)) 
                                {
                                    editTextEmail.setError(null);
                                    Toast.makeText(getApplicationContext(),"valid email address",Toast.LENGTH_SHORT).show();
                                }
                            else
                               {
                                    Toast.makeText(getApplicationContext(),"valid email address Required",Toast.LENGTH_SHORT).show();
                               }
                        }
                }
        });

EditText上的电子邮件验证 - Android

6 个答案: