java - Android EditText 检查帮助

Question

我有一个 editText 字段，我想检查它是否符合某些条件，基本上它必须包含 2 个整数和 3 个字符串，如何在添加元素之前检查它。

 EditText.getText().toString();

我可以添加一张这样的支票吗

public boolean checkString(String StringPassed) {

   String s = StringPassed;
if(s.length == 5){
  boolean hasString = false;
  boolean hasInt = false;
  String letters = s.substring(0, 2);
  Pattern p = Pattern.compile("[a-zA-Z]");
  Matcher m = p.matcher(letters);
  if (m.matches()) {
    hasString = true;
  }
  String numbers=s.substring(2,5);
  try {
    int num = Integer.parseInt(numbers);
    String n = num + "";
    if (num > 0 && n.length() == 3)
        hasInt = true;
  } catch (Exception e) {
  }
  if (hasInt && hasString) {
   return true;
  }
 }else{
  return false;
 }
return false;
}

然后我有一个方法会说

public void addString() {

String StringPassed =  EditTextName.getString().toString();
checkString(String StringPassed);

if (StringPassed() == false) {
Display Toast;
}
else {
      add;
     }
}

Answer 1

    String s = edittext.gettext();
    if(s.length == 5){
      boolean hasString = false;
      boolean hasInt = false;
      String letters = s.substring(0, 2);
      Pattern p = Pattern.compile("^[a-zA-Z]+$");
      Matcher m = p.matcher(letters);
      if (m.matches()) {
        hasString = true;
      }
      String numbers=s.substring(2,5);
      try {
        int num = Integer.parseInt(numbers);
        String n = num + "";
        if (num > 0 && n.length() == 3)
            hasInt = true;
      } catch (Exception e) {
      }
      if (hasInt && hasString) {
        //success
      }
     }else{
      // incorrect string
     }

您还可以在 edittext 上添加一个 TextWatcher 来监听文本输入更改并自动调用您的方法