Develop Reference

Setting minimum length for auto link

I have a big paragraph which may have numbers, email addresses and links. So I have to set setAutoLinkMask(Linkify.PHONE_NUMBERS | Linkify.EMAIL_ADDRESSES | Linkify.WEB_URLS) for my textview.
The content may contain digits of varying numbers. I want to set numbers having atleast 8 digits as phone number links.(For Eg : 12345678)
Is it possible to set minimum length for Linkify.PHONE_NUMBERS ?
Is there anyway to achieve this?

In case you can use Linkify.MatchFilter to specify minimum length or your some other requirements. There is not any direct way provided by Android.
Also somewhere in this SO post found some good examples.

use below pattern :
SpannableString buffer = new SpannableString(text);
Pattern pattern = Pattern.compile("^[0-9]\d{7,9}$");
Linkify.addLinks(buffer , pattern,"");

Yes, its possible. I researched this phenomenon :-)
To set the minimum length for a phone number, use this code:
private final Linkify.MatchFilter matchFilterForPhone = new Linkify.MatchFilter() {
#Override
public 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 >= 6) { // HERE: number 6 is minimum
return true;
}
}
}
return false;
}
};
To properly format and link phone numbers, use:
final SpannableString s = new SpannableString(myTekst);
Linkify.addLinks(s, android.util.Patterns.PHONE, "tel:", matchFilterForPhone, Linkify.sPhoneNumberTransformFilter);
Now place the formatted s in your TextView, and call:
findViewById(R.id.message).setLinkTextColor(Color.BLUE);
findViewById(R.id.message).setMovementMethod(LinkMovementMethod.getInstance());
That's all. Thanks for vote.

How to get last 4 digit of number in android?

I have an app which contain mobile number edit text in which user can edit mobile number and I have to send two request to server like:- mobile number and mssdn,mobile number(which is full lenghth ) and mssdn(which contain mobile number last 4 digit).How can I do that

Try this. Check for length greater than 4 before calling subString to avoid IndexOutOfBounds Exception.
EditText mEdtPhoneNumber = (EditText) findViewById(R.id.edtPhoneNumber);
String phoneNumber = mEdtPhoneNumber.getText().toString().trim();
String strLastFourDi = phoneNumber.length() >= 4 ? phoneNumber.substring(phoneNumber.length() - 4): "";
Also what is mssdn?? Is it msisdn??

Use the modulus (%) operator:
To get the last digit: use number % 10
To get the last 2 digits: use number % 100
and so on
For example:
42455%10000 = 2455

You could do something like this:
EditText phoneNumberEditText = (EditText) findViewById(R.id.phoneNumberEditText);
String phoneNumber = phoneNumberEditText.getText().toString();
String lastFourDigits = phoneNumber.substring(phoneNumber.length() - 4);

you should use regex because this will only give you result if the last four letters are actually numbers on the other hand the substring function simply give you last four letters no matter they are numbers or characters. e.g 4344sdsdss4 will give you dss4 which is clearly not a part of phone number
String str="4444ntrjntkr555566";
Pattern p = Pattern.compile("(\\d{4})$");
Matcher m = p.matcher(str);
if (m.find()) {
System.out.println(m.group(m.groupCount()));
}
this will produce 5566
Working
//d mean digits
{4} for fix length as 4
$ mean at the end

List<Integer> f(String str){
ArrayList<Integer> digits = new ArrayList<>();
if (null == str || str.length() < 4){
Log.i(LOG_TAG, "there are less than 4 digits");
return digits;
}
String digitsStr = str.substring(str.length() - 4);
for (char c : digitsStr.toCharArray()){
try {
Integer digit = Integer.parseInt(String.valueOf(c));
digits.add(digit);
} catch (Exception e){
continue;
}
}
return digits;
}

We can also use a new method introduced in kotlin: takeLast(n)
fun getLastDigit(data: String, n:Int): String {
return if(data.length > n){
data.takeLast(n)
}else {
""
}
}

Android - How to filter emoji (emoticons) from a string?

I'm working on an Android app, and I do not want people to use emoji in the input.
How can I remove emoji characters from a string?

Emojis can be found in the following ranges (source) :
U+2190 to U+21FF
U+2600 to U+26FF
U+2700 to U+27BF
U+3000 to U+303F
U+1F300 to U+1F64F
U+1F680 to U+1F6FF
You can use this line in your script to filter them all at once:
text.replace("/[\u2190-\u21FF]|[\u2600-\u26FF]|[\u2700-\u27BF]|[\u3000-\u303F]|[\u1F300-\u1F64F]|[\u1F680-\u1F6FF]/g", "");

Latest emoji data can be found here:
http://unicode.org/Public/emoji/
There is a folder named with emoji version.
As app developers a good idea is to use latest version available.
When You look inside a folder, You'll see text files in it.
You should check emoji-data.txt. It contains all standard emoji codes.
There are a lot of small symbol code ranges for emoji.
Best support will be to check all these in Your app.
Some people ask why there are 5 digit codes when we can only specify 4 after \u.
Well these are codes made from surrogate pairs. Usually 2 symbols are used to encode one emoji.
For example, we have a string.
String s = ...;
UTF-16 representation
byte[] utf16 = s.getBytes("UTF-16BE");
Iterate over UTF-16
for(int i = 0; i < utf16.length; i += 2) {
Get one char
char c = (char)((char)(utf16[i] & 0xff) << 8 | (char)(utf16[i + 1] & 0xff));
Now check for surrogate pairs. Emoji are located on the first plane, so check first part of pair in range 0xd800..0xd83f.
if(c >= 0xd800 && c <= 0xd83f) {
high = c;
continue;
}
For second part of surrogate pair range is 0xdc00..0xdfff. And we can now convert a pair to one 5 digit code.
else if(c >= 0xdc00 && c <= 0xdfff) {
low = c;
long unicode = (((long)high - 0xd800) * 0x400) + ((long)low - 0xdc00) + 0x10000;
}
All other symbols are not pairs so process them as is.
else {
long unicode = c;
}
Now use data from emoji-data.txt to check if it's emoji.
If it is, then skip it. If not then copy bytes to output byte array.
Finally byte array is converted to String by
String out = new String(outarray, Charset.forName("UTF-16BE"));

For those using Kotlin, Char.isSurrogate can help as well. Find and remove the indexes that are true from that.

Here is what I use to remove emojis. Note: This only works on API 24 and forwards
public String remove_Emojis_For_Devices_API_24_Onwards(String name)
{
// we will store all the non emoji characters in this array list
ArrayList<Character> nonEmoji = new ArrayList<>();
// this is where we will store the reasembled name
String newName = "";
//Character.UnicodeScript.of () was not added till API 24 so this is a 24 up solution
if (Build.VERSION.SDK_INT > 23) {
/* we are going to cycle through the word checking each character
to find its unicode script to compare it against known alphabets*/
for (int i = 0; i < name.length(); i++) {
// currently emojis don't have a devoted unicode script so they return UNKNOWN
if (!(Character.UnicodeScript.of(name.charAt(i)) + "").equals("UNKNOWN")) {
nonEmoji.add(name.charAt(i));//its not an emoji so we add it
}
}
// we then cycle through rebuilding the string
for (int i = 0; i < nonEmoji.size(); i++) {
newName += nonEmoji.get(i);
}
}
return newName;
}
so if we pass in a string:
remove_Emojis_For_Devices_API_24_Onwards("😊 test 😊 Indic:ढ Japanese:な 😊 Korean:ㅂ");
it returns: test Indic:ढ Japanese:な Korean:ㅂ
Emoji placement or count doesn't matter

Small Caps on TextViews, EditTexts, and Buttons in Android

Is there something I can do to make the text look in small caps/capital? As described here: http://en.wikipedia.org/wiki/Small_caps. I used a converter but some characters are missing.

EDIT 2015-08-02: As of API 21 (Lollipop) you can simply add:
android:fontFeatureSettings="smcp"
to your TextView declaration in XML, or at runtime, invoke:
textView.setFontFeatureSettings("smcp");
Of course, this only works for API 21 and up, so you'd still have to handle the old solution manually until you are only supporting Lollipop and above.
Being a bit of a typography geek at heart, this seemed like a really good question. I got to learn some more about Unicode today, as well as an answer for your question. :)
First, you'll need to have a font that includes "actual" small-caps characters. I'm assuming you know that since you're asking, but typically most professional fonts include these. Unfortunately most professional fonts are not licensed for distribution, so you may not be able to use them in your application. Anyway, in the event that you do find one (I used Chaparral Pro as an example here), this is how you can get small caps.
From this answer I found that the small caps characters (for A-Z) are located starting at Unicode-UF761. So I built a mapping of these characters:
private static char[] smallCaps = new char[]
{
'\uf761', //A
'\uf762',
'\uf763',
'\uf764',
'\uf765',
'\uf766',
'\uf767',
'\uf768',
'\uf769',
'\uf76A',
'\uf76B',
'\uf76C',
'\uf76D',
'\uf76E',
'\uf76F',
'\uf770',
'\uf771',
'\uf772',
'\uf773',
'\uf774',
'\uf775',
'\uf776',
'\uf777',
'\uf778',
'\uf779',
'\uf77A' //Z
};
Then added a helper method to convert an input string to one whose lowercase letters have been replaced by their Small Caps equivalents:
private static String getSmallCapsString (String input) {
char[] chars = input.toCharArray();
for(int i = 0; i < chars.length; i++) {
if(chars[i] >= 'a' && chars[i] <= 'z') {
chars[i] = smallCaps[chars[i] - 'a'];
}
}
return String.valueOf(chars);
}
Then just use that anywhere:
String regularCase = "The quick brown fox jumps over the lazy dog.";
textView.setText(getSmallCapsString(regularCase));
For which I got the following result:

Apologies for dragging up a very old question.
I liked #kcoppock's approach to this, but unfortunately the font I'm using is missing the small-cap characters. I suspect many others will find themselves in this situation.
That inspired me to write a little util method that will take a mixed-case string (e.g. Small Caps) and create a formatted spannable string that looks like Sᴍᴀʟʟ Cᴀᴘs but only uses the standard A-Z characters.
It works with any font that has the A-Z characters - nothing special required
It is easily useable in a TextView (or any other text-based view, for that matter)
It doesn't require any HTML
It doesn't require any editing of your original strings
I've posed the code here: https://gist.github.com/markormesher/3e912622d339af01d24e

Found an alternative here Is it possible to have multiple styles inside a TextView?
Basically you can use html tags formatting the size of the characters and give a small caps effect....

Just call this getSmallCaps(text) function:
public SpannableStringBuilder getSmallCaps(String text) {
text = text.toUpperCase();
text = text.trim();
SpannableStringBuilder spannableStringBuilder = new SpannableStringBuilder();
if (text.contains(" ")) {
String[] arr = text.split(" ");
for (int i = 0; i < arr.length; i++) {
spannableStringBuilder.append(getSpannableStringSmallCaps(arr[i]));
spannableStringBuilder.append(" ");
}
} else {
spannableStringBuilder=getSpannableStringSmallCaps(text);
}
return spannableStringBuilder;
}
public SpannableStringBuilder getSpannableStringSmallCaps(String text) {
SpannableStringBuilder spannableStringBuilder = new SpannableStringBuilder(
text);
spannableStringBuilder.setSpan(new AbsoluteSizeSpan(36), 0, 1, 0);
spannableStringBuilder.setSpan(new StyleSpan(Typeface.BOLD), 0, 1, 0);
spannableStringBuilder.setSpan(new StyleSpan(Typeface.BOLD), 1,
text.length(), 0);
return spannableStringBuilder;
}

This is not my code but its works perfectly.
public SpannableString getSmallCapsString(String input) {
// values needed to record start/end points of blocks of lowercase letters
char[] chars = input.toCharArray();
int currentBlock = 0;
int[] blockStarts = new int[chars.length];
int[] blockEnds = new int[chars.length];
boolean blockOpen = false;
// record where blocks of lowercase letters start/end
for (int i = 0; i < chars.length; ++i) {
char c = chars[i];
if (c >= 'a' && c <= 'z') {
if (!blockOpen) {
blockOpen = true;
blockStarts[currentBlock] = i;
}
// replace with uppercase letters
chars[i] = (char) (c - 'a' + '\u0041');
} else {
if (blockOpen) {
blockOpen = false;
blockEnds[currentBlock] = i;
++currentBlock;
}
}
}
// add the string end, in case the last character is a lowercase letter
blockEnds[currentBlock] = chars.length;
// shrink the blocks found above
SpannableString output = new SpannableString(String.valueOf(chars));
for (int i = 0; i < Math.min(blockStarts.length, blockEnds.length); ++i) {
output.setSpan(new RelativeSizeSpan(0.8f), blockStarts[i], blockEnds[i], Spannable.SPAN_EXCLUSIVE_INCLUSIVE);
}
return output;
}
Example:
SpannableString setStringObj = getSmallCapsStringTwo("Object"); tvObj.setText(setStringObj);

in XML
edit text has property :android:capitalize=""

Reformatting a String in Java

For my app I have created a QR Code, then took that bitmap and added text to the bitmap, however I need the text not to extend longer then the bitmap is. So what I want to do is create an Array of the text by taking 25 characters then find the last index of (" ") in that 25 character section. at that space I want to be able to replace that space that was located with \n to start a new line.
So the plan is if I have a String that looks like "Hello this is my name and I am longer than 25 charters and I have lots of spaces so that this example will work well."
I want it to out up this
Hello this is my name and
I am longer than 25
charters and I have lots
of spaces so that this
example will work well.
To make this I counted 25 characters then went back to the most resent space, at that point I hit enter, I want my app to do this for me.
I am not very good at English so if something doesn't make sense tell me and I will try to explain it. Thanks

I haven't tested this but you can try it and tweak as necessary
String fullText = "your text here";
String withBreaks = "";
while( fullText.length() > 25 ){
String line = fullText.substring(0,24);
int breakPoint = line.lastIndexOf( " ");
withBreaks += fullText.substring(0,breakPoint ) + "\n";
fullText = fullText.substring( breakPoint );
withBreaks += fullText;

char [] way (more C like):
public static String reduceLength(String s, int len){
char [] c = s.toCharArray();
int i=len, j=0, k;
while(true){
for(k=j; k<=i; k++){
if (k >= s.length()) return new String(c);
if (c[k] == ' ') j=k;
}
c[j] = '\n';
i= j+ len;
}
}
This isn't safe, just something i threw together.