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Min & Max Input values on Android Studio [duplicate]

This question already has answers here:
Is there a way to define a min and max value for EditText in Android?
(29 answers)
Closed 11 months ago.
I am creating an app that calculates the the users marks. How do I go forward, I'm trying to limit the user's input value to between min=0 and max=100?
I've got 3 inputs that I want to limit.
EditText a = (EditText) findViewById(R.id.Assignment1);
EditText b = (EditText) findViewById(R.id.Assignment2);
EditText c = (EditText) findViewById(R.id.Assignment3);
Tried using the following but only works when I execute my function.
public void YearMark(View v)
{
EditText a = (EditText) findViewById(R.id.Assignment1);
a.setFilters(new InputFilter[]{ new InputFilterMinMax("0", "100")});

As #JAY_Panchal mentioned, you can use an InputFilter:
public class RangeInputFilter implements InputFilter {
private final int min, max;
public RangeInputFilter(int min, int max) {
boolean rightOrder = min <= max;
this.min = rightOrder ? min : max;
this.max = rightOrder ? max : min;
}
#Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dStart, int dEnd) {
try {
String sourceStr = source.toString();
String destStr = dest.toString();
String result = destStr.substring(0, dStart) + sourceStr.substring(start, end) + destStr.substring(dEnd);
int input = Integer.parseInt(result);
if (min <= input && input <= max) return null;
} catch (NumberFormatException ignored) { }
return "";
}
}
When a user changes the content of an EditText, you can use an InputFilter to decide what actually changes in the EditText.
Returning null means that the input is accepted.
Returning something else (in this case "") means that source should be replaced with that (in this case, we want dest to stay as it is and thus have an empty source).
To assign your filter:
RangeInputFilter filter = new RangeInputFilter(0, 100);
InputFilter[] filters = new InputFilter[] {filter};
editText1.setFilters(filters);
editText2.setFilters(filters);
editText3.setFilters(filters);

Add this textwatcher for all three edittext a,b and c:
a.addTextChangedListener(new TextWatcher() {
#Override
public void afterTextChanged(Editable s) {
}
#Override
public void beforeTextChanged(CharSequence s, int start,
int count, int after) {
}
#Override
public void onTextChanged(CharSequence s, int start,
int before, int count) {
String strTest = editText1235.getText().toString().trim();
if (!strTest.equals("")) {
float no = Float.parseFloat(strTest);
if (no > 100) {
a.setText(null);
Toast.makeText(this, "enter a value less than 100", Toast.LENGTH_SHORT).show();
}
}
}
});
Also set input type of these text fields as number so that user cannot enter a negative sign,ie, any value less than 0.
android:inputType="number"
While submitting the details also check if the field is not empty like below:
if (a.getText().toString().trim().length() <=0 ||b.getText().toString().trim().length() <=0 ||c.getText().toString().trim().length() <= 0){
Toast.makeText(this, "Value cannot be empty", Toast.LENGTH_SHORT).show();
}

I don't know your usecase, but you might also want to use a SeekBar.
A SeekBar is a slider - you can select a number by sliding. It is also very customizable.
The main advantage is that the user does not need to open the keyboard.

how to set decimal number using inputfilter android studio

I'm making an application that can control weight (kilogram).
I have an edittext that can accept enter decimal numbers using:
et_beratbadan.setInputType(InputType.TYPE_CLASS_NUMBER | InputType.TYPE_NUMBER_FLAG_DECIMAL | InputType.TYPE_NUMBER_FLAG_SIGNED);
my edittext already able to limit entering numbers, 3 digits before zero and 1 digit after zero (120.1) using:
et_beratbadan.setFilters(new InputFilter[]{new DecimalDigitsInputFilter(3, 1)});
what i want is, my edittext able to limit entering numbers, 3 digits before zero and 1 digit after zero, not accept minus values and start from one.
so it would allow : 123.4, 12.3, 1.2
not allow : 0, -12.3, 0.1
i already try code from https://stackoverflow.com/a/38793820 to make my edittext do not accept minus value. and some other code, that i lost the link for it.
and i have a class for implement inputfilter:
public class DecimalDigitsInputFilter implements InputFilter {
private int mDigitsBeforeZero;
private int mDigitsAfterZero;
private Pattern mPattern;
private static final int DIGITS_BEFORE_ZERO_DEFAULT = 100;
private static final int DIGITS_AFTER_ZERO_DEFAULT = 100;
public DecimalDigitsInputFilter(Integer digitsBeforeZero, Integer digitsAfterZero) {
this.mDigitsBeforeZero = (digitsBeforeZero != null ? digitsBeforeZero : DIGITS_BEFORE_ZERO_DEFAULT);
this.mDigitsAfterZero = (digitsAfterZero != null ? digitsAfterZero : DIGITS_AFTER_ZERO_DEFAULT);
mPattern = Pattern.compile("-?[0-9]{0," + (mDigitsBeforeZero) + "}+((\\.[0-9]{0," + (mDigitsAfterZero)
+ "})?)||(\\.)?");
}
#Override
public CharSequence filter(CharSequence source, int start, int end, Spanned dest, int dstart, int dend) {
String replacement = source.subSequence(start, end).toString();
String newVal = dest.subSequence(0, dstart).toString() + replacement
+ dest.subSequence(dend, dest.length()).toString();
Matcher matcher = mPattern.matcher(newVal);
if (matcher.matches())
return null;
if (TextUtils.isEmpty(source))
return dest.subSequence(dstart, dend);
else
return "";
}
}
sorry I'm still learning English and java. thanks!

Forming a pattern of bits from a integer

I need help to design java code for generating bit array for any given integer in following manner:
23 should produce output as 1101011 (min length array)
explaination :
positions are given as 1 -2 4 -8 16 -32 ....
So 1101011 can be evaluated as:
1*1 + 1*-2 + 0*4+ 1*-8 + 0*16 +1*-32 + 1*64 = 23

This is the so-called negabinary representation of numbers (described first by Vittorio Grünwald in 1885). They can be encoded in a fashion very similar to the usual binary representation, just working with -2 instead of 2 as base (Java code inspired by C# code on https://en.wikipedia.org/wiki/Negative_base ):
class EncodeNegaBinary {
public static void main(String[] args) {
int n=0,input=0;
String result="";
final String[] BITS = { "0","1" };
if (args.length != 1) {
System.err.println("Please enter an integer to be converted");
return;
} else {
input = n = Integer.parseInt(args[0]);
}
while (n != 0) {
int r = n%-2;
n /= -2;
if (r == -1) {
r=1;
n++;
}
result = BITS[r] + result;
}
System.out.printf( "%d -> %s\n", input, result);
}
}

Since it is not usual int to binary conversion, at each step we need to consider two cases as at each position there can be only two choices 0 or 1. This is done recursively in the below program:
public class ModifiedIntToBinaryConversion{
public static int calcBinaryString(int reqSum, int currSum, int add, String bs) {
if (reqSum == currSum) { // base condtion 1
System.out.println("The string is \n" + bs);
return currSum;
}
if (add + currSum > reqSum) { // base condtion 2
return 0;
}
int newAdd = add * -2;
// System.out.println("new add is "+ newAdd +" currSum is "+ currSum);
int s1 = calcBinaryString(reqSum, currSum + add, newAdd, bs + "1");
if (s1 == reqSum)
return s1;
int s2 = calcBinaryString(reqSum, currSum, newAdd, bs + "0");
return s2;
}
public static void calcBinaryString(int sum) {
int s1 = calcBinaryString(sum, 0, 1, "");
if(s1 != sum) {
System.out.println("The binary equivalent couldn't be found");
}
}
public static void main(String[] args) {
calcBinaryString(23);
}
}
Now base condition 1 is clear as I am just checking whether required sum and calculated sum are equal.
For base condition 2, I will accept it's result of debugging and a bit of thought as I was getting Stackoverflow errors. Once the calculated sum becomes greater than the required sum and then we take the next -ve number so that it become less than req. sum. But then the next +ve number will be greater than the -ve number we just considered and thus the chances are very less that the calculated sum will ever be equal to req. sum.

integer to word conversion in java using map continue question

This is a probable answer of my question in stack overflow.Integer to word conversion
At first I have started with dictionary. Then I came to know it is obsolete. So now I use Map instead of dictionary. My code is work well for number till Millions. But the approach I take here is a naive approach. The main problem of this code is
First: Huge numbers of variable use
2nd: Redundant code block as per program requirement
3rd: Multiple if else statement
I am thinking about this problems
Solution for 2nd problem: using user define function or macros to eliminate redundant code block
Solution for 3rd problem: Using switch case
My code:
public class IntegerEnglish {
public static void main(String args[]){
Scanner in=new Scanner(System.in);
System.out.println("Enter the integer");
int input_number=in.nextInt();
Map<Integer,String> numbers_converter = new HashMap<Integer,String>();
Map<Integer,String> number_place = new HashMap<Integer,String>();
Map<Integer,String> number_2nd = new HashMap<Integer,String>();
numbers_converter.put(0,"Zero");
numbers_converter.put(1,"One");
numbers_converter.put(2,"Two");
numbers_converter.put(3,"Three");
numbers_converter.put(4,"Four");
numbers_converter.put(5,"Five");
numbers_converter.put(6,"Six");
numbers_converter.put(7,"Seven");
numbers_converter.put(8,"Eight");
numbers_converter.put(9,"Nine");
numbers_converter.put(10,"Ten");
numbers_converter.put(11,"Eleven");
numbers_converter.put(12,"Twelve");
numbers_converter.put(13,"Thirteen");
numbers_converter.put(14,"Fourteen ");
numbers_converter.put(15,"Fifteen");
numbers_converter.put(16,"Sixteen");
numbers_converter.put(17,"Seventeen");
numbers_converter.put(18,"Eighteen");
numbers_converter.put(19,"Nineteen");
number_place.put(3,"Hundred");
number_place.put(4,"Thousand");
number_place.put(7,"Million");
number_place.put(11,"Billion");
number_2nd.put(2,"Twenty");
number_2nd.put(3,"Thirty");
number_2nd.put(4,"Forty");
number_2nd.put(5,"Fifty");
number_2nd.put(6,"Sixty");
number_2nd.put(7,"Seventy");
number_2nd.put(8,"Eighty");
number_2nd.put(9,"Ninty");
if(input_number== 0){
System.out.println("zero");
}
else if(input_number>0 && input_number<19){
System.out.println(numbers_converter.get(input_number));
}
else if(input_number>19 && input_number<100){
int rem=input_number%10;
input_number=input_number/10;
System.out.print(number_2nd.get(input_number));
System.out.print(numbers_converter.get(rem));
}
else if(input_number==100){
System.out.println(number_place.get(3));
}
else if(input_number>100 && input_number<1000){
int reminder=input_number%100;
int r1=reminder%10;
int q1=reminder/10;
int quot=input_number/100;
System.out.print(numbers_converter.get(quot) + "hundred");
if(reminder>0 && reminder<20){
System.out.print(numbers_converter.get(reminder));
}
else{
System.out.println(number_2nd.get(q1) + numbers_converter.get(r1));
}
}
else if(input_number==1000){
System.out.println(number_place.get(4));
}
else if(input_number>1000 && input_number<10000){
int rem=input_number%100;
int rem_two=rem%10;
int quotient =rem/10;
input_number=input_number/100;
int thousand=input_number/10;
int hundred = input_number%10;
System.out.print(numbers_converter.get(thousand) + "thousand" + numbers_converter.get(hundred)+ " hundred");
if(rem >0 && rem<20){
System.out.print(numbers_converter.get(rem));
}
else if(rem >19 && rem <100){
System.out.print(number_2nd.get(quotient) + numbers_converter.get(rem_two));
}
}
else if(input_number>10000 && input_number<1000000000){
//Say number 418,229,356
int third_part=input_number%1000;//hold 356
input_number=input_number/1000;//hold 418,229
int sec_part=input_number%1000;//hold 229
input_number=input_number/1000;// hold 418
int rem_m=third_part%100;//hold 56
int rem_m1=rem_m%10;//hold 6
int rem_q=rem_m/10;// hold 5
int q_m=third_part/100;// hold 3
int sec_part_rem=sec_part%100;// hold 29
int sec_part_rem1=sec_part_rem%10;//9
int sec_part_q=sec_part_rem/10;//hold 2
int sec_q=sec_part/100;// hold 2
int input_q=input_number/100;// hold 4
int input_rem=input_number%100;//hold 18
int input_q_q=input_rem/10;//hold 1
int input_rem1=input_rem%10;// hold 8
System.out.print(numbers_converter.get(input_q) + " hundred ");
if(input_rem>0 && input_rem<20){
System.out.print(numbers_converter.get(input_rem)+ " Million ");
}
else{
System.out.print(number_2nd.get(input_q_q) + " " + numbers_converter.get(input_rem1) + " Million ");
}
System.out.print(numbers_converter.get(sec_q) + " hundred ");
if(sec_part_rem >0 && sec_part_rem<20){
System.out.println(numbers_converter.get(sec_part_rem) + " thousand ");
}
else{
System.out.print(number_2nd.get(sec_part_q) + " " + numbers_converter.get(sec_part_rem1) + " thousand ");
}
System.out.print(numbers_converter.get(q_m) + " hundred ");
if(rem_m>0 && rem_m<20){
System.out.print(numbers_converter.get(rem_m));
}
else{
System.out.print(number_2nd.get(rem_q) + " " + numbers_converter.get(rem_m1));
}
}
}
}
Redundant Code Blocks
int rem=input_number%100;
int rem_two=rem%10;
int quotient =rem/10;
input_number=input_number/100;
int thousand=input_number/10;
int hundred = input_number%10;
This type of code block used almost every where. Taking a number divide it with 100 or 1000 to find out the hundred position then then divide it with 10 to find out the tenth position of the number. Finally using %(modular division) to find out the ones position.
How could I include user define function and switch case to minimize the code block.

Instead of storing the results in variables, use a method call:
int remainder100(int aNumber) {
return aNumber % 100;
}
int remainder10(int aNumber) {
return aNumber % 10;
}
...etc.
System.out.println(numbers_converter.get(remainder100(input_number)));
About 3rd problem: I wouldn't use switch ... case, too many cases.
Instead, take advantage that numbering repeats itself every 3 digits. That means the pattern for thousands and millions is the same (and billions, trillions, etc).
To do that, use a loop like this:
ArrayList<String> partialResult = new ArrayList<String>();
int powersOf1000 = 0;
for (int kiloCounter = input_number; kiloCounter > 0; kiloCounter /= 1000) {
partialResult.add(getThousandsMilionsBillionsEtc(powersOf1000++);
partialResult.add(convertThreeDigits(kiloCounter % 1000));
}
Then you can print out the contents of partialResult in reverse order to get the final number.

I'd suggest you break your single main method down into a couple of classes. And if you haven't already create a few unit tests to allow you to easily test / refactor things. You'll find it quicker than starting the app and reading from stdin.
You'll find it easier to deal with the number as a string. Rather than dividing by 10 all the time you just take the last character of the string. You could have a class that does that bit for you, and a separate one that does the convert.
Here's what I came up with, but I'm sure it can be improved. It has a PoppableNumber class which allows the last character of the initial number to be easily retrieved. And the NumberToString class which has a static convert method to perform the conversion.
An example of a test would be
#Test
public void Convert102356Test() {
assertEquals("one hundred and two thousand three hundred and fifty six", NumberToString.convert(102356));
}
And here's the NumberToString class :
import java.util.HashMap;
import java.util.Map;
public class NumberToString {
// billion is enough for an int, obviously need more for long
private static String[] power3 = new String[] {"", "thousand", "million", "billion"};
private static Map<String,String> numbers_below_twenty = new HashMap<String,String>();
private static Map<String,String> number_tens = new HashMap<String,String>();
static {
numbers_below_twenty.put("0","");
numbers_below_twenty.put("1","one");
numbers_below_twenty.put("2","two");
numbers_below_twenty.put("3","three");
numbers_below_twenty.put("4","four");
numbers_below_twenty.put("5","five");
numbers_below_twenty.put("6","six");
numbers_below_twenty.put("7","seven");
numbers_below_twenty.put("8","eight");
numbers_below_twenty.put("9","nine");
numbers_below_twenty.put("10","ten");
numbers_below_twenty.put("11","eleven");
numbers_below_twenty.put("12","twelve");
numbers_below_twenty.put("13","thirteen");
numbers_below_twenty.put("14","fourteen ");
numbers_below_twenty.put("15","fifteen");
numbers_below_twenty.put("16","sixteen");
numbers_below_twenty.put("17","seventeen");
numbers_below_twenty.put("18","eighteen");
numbers_below_twenty.put("19","nineteen");
number_tens.put(null,"");
number_tens.put("","");
number_tens.put("0","");
number_tens.put("2","twenty");
number_tens.put("3","thirty");
number_tens.put("4","forty");
number_tens.put("5","fifty");
number_tens.put("6","sixty");
number_tens.put("7","seventy");
number_tens.put("8","eighty");
number_tens.put("9","ninty");
}
public static String convert(int value) {
if (value == 0) {
return "zero";
}
PoppableNumber number = new PoppableNumber(value);
String result = "";
int power3Count = 0;
while (number.hasMore()) {
String nextPart = convertUnitTenHundred(number.pop(), number.pop(), number.pop());
nextPart = join(nextPart, " ", power3[power3Count++], true);
result = join(nextPart, " ", result);
}
if (number.isNegative()) {
result = join("minus", " ", result);
}
return result;
}
public static String convertUnitTenHundred(String units, String tens, String hundreds) {
String tens_and_units_part = "";
if (numbers_below_twenty.containsKey(tens+units)) {
tens_and_units_part = numbers_below_twenty.get(tens+units);
}
else {
tens_and_units_part = join(number_tens.get(tens), " ", numbers_below_twenty.get(units));
}
String hundred_part = join(numbers_below_twenty.get(hundreds), " ", "hundred", true);
return join(hundred_part, " and ", tens_and_units_part);
}
public static String join(String part1, String sep, String part2) {
return join(part1, sep, part2, false);
}
public static String join(String part1, String sep, String part2, boolean part1Required) {
if (part1 == null || part1.length() == 0) {
return (part1Required) ? "" : part2;
}
if (part2.length() == 0) {
return part1;
}
return part1 + sep + part2;
}
/**
*
* Convert an int to a string, and allow the last character to be taken off the string using pop() method.
*
* e.g.
* 1432
* Will give 2, then 3, then 4, and finally 1 on subsequent calls to pop().
*
* If there is nothing left, pop() will just return an empty string.
*
*/
static class PoppableNumber {
private int original;
private String number;
private int start;
private int next;
PoppableNumber(int value) {
this.original = value;
this.number = String.valueOf(value);
this.next = number.length();
this.start = (value < 0) ? 1 : 0; // allow for minus sign.
}
boolean isNegative() {
return (original < 0);
}
boolean hasMore() {
return (next > start);
}
String pop() {
return hasMore() ? number.substring(--next, next+1) : "";
}
}
}

Calling Zip Code -- Android

I have both the phone number and address (including the zipcode) in the same TextView. I want the phone number to be clickable to call; however, this is making the zip code also clickable to make a phone call. How do I make it no longer clickable without creating another TextView? Thanks!
tvInfo.setText(Html.fromHtml("John Smith<br>123 Fake Street<br>Faketown, FK 12345<br><b>(804) 932-3300</b><br>"));
Linkify.addLinks(tvInfo, Linkify.PHONE_NUMBERS);

android will regard digits with count >=5 as phone numbers. so I think there will be at least 2 solutions:
1)a simple work around : if you are sure the length of phone numbers is more than 5, for example, at least 6 digits, you could make some work around:
private final static int MY_PHONE_NUMBER_MINIMUM_DIGITS = 6;
Linkify.addLinks(main, Patterns.PHONE, "tel:", new Linkify.MatchFilter() {
public final boolean acceptMatch(CharSequence s, int start, int end) {
int digitCount = 0;
for (int i = start; i < end; i++) {
if (Character.isDigit(s.charAt(i))) {
digitCount++;
if (digitCount >= MY_PHONE_NUMBER_MINIMUM_DIGITS ) {
return true;
}
}
}
return false;
}
}, Linkify.sPhoneNumberTransformFilter);
this workaround is based on android source code of Linkify, in Linkify, the method:
gatherLinks(links, text, Patterns.PHONE,
new String[] { "tel:" },
sPhoneNumberMatchFilter, sPhoneNumberTransformFilter);
will be called, while sPhoneNumberMatchFilter will filter the digits less than 5:
public static final MatchFilter sPhoneNumberMatchFilter = new MatchFilter() {
public final boolean acceptMatch(CharSequence s, int start, int end) {
int digitCount = 0;
for (int i = start; i < end; i++) {
if (Character.isDigit(s.charAt(i))) {
digitCount++;
if (digitCount >= PHONE_NUMBER_MINIMUM_DIGITS/*=5*/) {
return true;
}
}
}
return false;
}
};
so we just replace the "PHONE_NUMBER_MINIMUM_DIGITS" with 6
2)a more complicated solution is, if your phone numbers are in a more specific format for example, must be something like "(xxx)xxx-xxxx", you could use your own pattern to replace the Patterns.PHONE, to extract and apply links of phone numbers more accurately