How to use regular expression in Android

2019-01-12 05:36发布

I have a numberDecimal EditText which I want to validate using a regular expression. In validation what I want is:

  1. Before the decimal point, the maximum digit I want to enter is three and the digit should not start with zero like 2,23,342, etc.

  2. After the decimal point, the maximum digit I want to enter is one like .1, .3, .6, etc.

So the number that I allow the user to enter is like 2.1, 32.5, 444.8, 564.9, etc.

But in my code, what happens is:

  1. It allows the user to enter more than a three digit number before the decimal point like 3456, 4444, 5555 and after that it doesn't allow me to enter a decimal point after that.

  2. It allows me to enter 0 before the decimal point as the start of the digit.

So why does this happen, is anything wrong in the regular expression I have used? If anyone knows, please help me to solve this.

Code I have used:

weightEditText.addTextChangedListener(new TextWatcher() 
{           
    @Override
    public void onTextChanged(CharSequence s, int start, int before, int count) {               
    }           
    @Override
    public void beforeTextChanged(CharSequence s, int start, int count, int after) {            
    }           
    @Override
    public void afterTextChanged(Editable s) 
    {
        Pattern mPattern = Pattern.compile("^([1-9][0-9]{0,2})?(\\.[0-9]?)?$");

        Matcher matcher = mPattern.matcher(s.toString());               
        if(!matcher.find())
        {
            weightEditText.setText(); // Don't know what to place                   
        }
    }
});

5条回答
男人必须洒脱
2楼-- · 2019-01-12 06:01

Maybe my implementation will help anyone:

    etRepeaterCallsign.addTextChangedListener(new TextWatcher() {
        private final Pattern sPattern
                = Pattern.compile("^([A-Z]{0,2})?(\\d)?([A-Z-]{0,5})"); // ^([1-9][0-9]{0,2})?(\\.[0-9]?)?$

        private CharSequence mText;

        private boolean isValid(CharSequence s) {
            return sPattern.matcher(s).matches();
        }

        @Override
        public void beforeTextChanged(CharSequence r, int start, int count,
                                      int after) {
            mText = r.toString();
        }

        @Override
        public void onTextChanged(CharSequence s, int start, int before, int count) {
            bIsEdit = true;
        }
        @Override
        public void afterTextChanged(Editable s) {
            etRepeaterCallsign.removeTextChangedListener(this);

            int iCursorPosition = etRepeaterCallsign.getSelectionStart();
            etRepeaterCallsign.setText("");
            if (isValid(s))
                etRepeaterCallsign.append(s);
            else
                etRepeaterCallsign.append(mText);

            etRepeaterCallsign.setSelection(iCursorPosition);

            etRepeaterCallsign.addTextChangedListener(this);

        }
    });
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兄弟一词,经得起流年.
3楼-- · 2019-01-12 06:01

You could try something like this: ^[1-9][0-9]{0,2}(\\.\\d)?$. This should match any number which does not start with 0 (^[1-9]) and is followed by at most two numbers ([0-9]{0,2}). The regex also allows for an optional decimal value ((\\.\\d)?$).

That being said, ideally you should parse the string value to a double or float and make what ever validations you need to make using the actual numerical value.

To parse your string into a double value, you will need to use the Double.parseDouble(String s) like so: double myValue = Double.parseDouble(EditText.getText());

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干净又极端
4楼-- · 2019-01-12 06:03

I tried your pattern using my code as following. It works perfectly as 2.1, 32.5, 444.8, 564.9 etc.

My code:

public class WebPush extends Activity {  
    EditText editTxt;
    private TextView regresult;  

    protected void onCreate(Bundle savedInstanceState) {  
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main); 

        editTxt = (EditText) findViewById(R.id.editID);
        regresult = (TextView) findViewById(R.id.txtID);

        String urName = editTxt.getText().toString();
        editTxt.addTextChangedListener(new TextWatcher() {

            @Override
            public void onTextChanged(CharSequence arg0, int arg1, int arg2, int arg3) {
            }

            @Override
            public void beforeTextChanged(CharSequence arg0, int arg1, int arg2,int arg3) {
            }       

            @Override
            public void afterTextChanged(Editable s) {
                if (editTxt.getText().toString().matches("(^([0-9]{0,3})?)(\\.[0-9]{0,1})?$"))
                {
                    regresult.setText("");
                }
                else
                {
                    regresult.setText("invalid number");
                }
            }
        });
    }
}
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Lonely孤独者°
5楼-- · 2019-01-12 06:12

There's never any point in examining dest alone in an InputFilter; that's what's already present in the field. Change the regular expression match to be against source and it would be appropriate if you only wanted to check that certain characters were accepted into the field. However, you want to check field formatting, not just filter the input on a character-by-character basis. This is much more complex.

Every time the user makes a change to the contents of tempEditText, the system calls your filter's filter method before the change is actually made. It passes the current field contents and the proposed change (which can be insert/append, delete, or replace). The change is represented by a source CharSequence source (the characters—if any—to be added to the field), range start and end indexes within the source (the range is not necessarily all of source), a Spanned dest (the current field contents before the change) and range dstart and dend indexes within dest that are proposed to be replaced by the indicated source range.

The job of filter is to modify the change (if necessary) and return a CharSequence to use (in its entirety) in place of source (or null to go ahead and use source). Rather than checking dest as you are now doing, you will need to check whether the change will result in an acceptable field. To do this, you will need more complex logic. (Note, in particular, that the new character(s) may be intended for insert somewhere other than at the end; also, filter will be called when the user is deleting characters as well as adding them.)

It may be easier to implement a TextWatcher. In it's beforeTextChanged method, you can record the current contents and in it's afterTextChanged method, you can check (using a regular expression) whether the contents are acceptable and, if not, restore the before-the-change contents. (Make sure, though, that the text before the change was acceptable. If it isn't, substitute something acceptable—like clearing the field. Otherwise your code will go into an infinite loop because the TextWatcher is going to be invoked again when you correct the field contents.)

You also have an error in your regular expression: it allows a leading zero. Here's an improved version that fixes this problem (and removes one set of unnecessary parentheses):

"^([1-9][0-9]{0,2})?(\\.[0-9]?)?$"

(As an aside: you can use \\d instead of [0-9].)

EDIT

Here's my edit of your edit:

weightEditText.addTextChangedListener(new TextWatcher() 
{           
    private static final Pattern sPattern
        = Pattern.compile("^([1-9][0-9]{0,2})?(\\.[0-9]?)?$");

    private CharSequence mText;

    private boolean isValid(CharSequence s) {
        return sPattern.matcher(s).matches();
    }

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

    @Override
    public void beforeTextChanged(CharSequence s, int start, int count,
            int after){
        mText = isValid(s) ? new CharSequence(s) : "";
    }           

    @Override
    public void afterTextChanged(Editable s) 
    {
        if (!isValid(s))
        {
            weightEditText.setText(mText);
        }
        mText = null;
    }
});
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6楼-- · 2019-01-12 06:12

You could just parse the number and check that it is < 1000 and that 10*number is an integer while num is not. It would probably be more readable too.

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