We Keep Coding

Efficient way of adding zeros to a String number?

I would like to turn all values into the following number format, x.xxx. I am doing this fine by using a while loop and concatenating 0 to string values that do not have the correct length. However, I would like to know if there is a more efficient way of doing this. To be more clear of what I am trying to do, here are some examples.
Ex1--Change a string value such as 2.5 to 2.500
Ex2--Change a string value such as 2.51 to 2.510.
This is how I am currently doing this.
while (str.length() < 5) {
str= str+ "0";
}

You can use DecimalFormat
DecimalFormat decimalFormat = new DecimalFormat("0.000");
System.out.println(decimalFormat.format(2.5));

Wow,that was an amazing one Reimeus!
Anywy,how is this ? :P
int totalLenght = str.length();
if(totalLenght>5){
str = str.substring(0,5);
}else{
int remLength = 5 - totalLenght;
for(int i=0;i<remLength;i++){
str = str +"0";
}
}

Java: Get Float and String value from a String

I have a string, such as "4.25GB"
I'd like to get the floating part "4.25"
And get the string part "GB"
How to get the two values respectively in Java.
Thanks.

Try
String s = "4.25GB"
Float value = Float.valueOf(s.replaceAll("[^\\d.]", "")); // remove all non-numeric symbols
String f = s.replaceAll("[0-9]",""); // remove all numbers

To get Number Part: String numberPart = "4.25GB".replaceAll("[^0-9.]", "");
To get String part: String stringPart = "4.25GB".replaceAll("[^A-Za-z]", "");

Use String.replaceAll to first replace all non-digits and dot with "" to get the number then otherwise

You can write a function that will be similar to C# int.TryParse method, and use it in loop on your string, it will only work if you alwayes have a (NUM)(STRING) formation :
boolean tryParse(String value)
{
try
{
Integer.parseInt(value);
return true;
} catch(NumberFormatException e)
{
return false;
}
}

Use split/ substring concept. divide the string like below:
String Str = new String("4.25GB");
for (String retval: Str.split("G")){
System.out.println(retval);
}
//or u can use
String[] r = s.split("(?=\\p{Upper})");

You could use public String substring(int beginIndex, int endIndex)
String start = "4.25GB";
String numbers = start.substring(0,4);
String letters = start.substring(4,6);
Read more about substrings and how to use them here

Tested, works:
String str = "4.25GB" ;
String parts[] = str.split("(?i)(?<=\\d)(?=[a-z])|(?<=[a-z])(?=\\d)");
float number = Float.parseFloat(parts[0]) ;
String string = parts[1] ;
System.out.println(number); //4.25
System.out.println(string); //GB

You can use regular expression like this :
String s = "4.25GB";
String num = s.replaceAll("[^0-9.]", "");
System.out.println(num);
String str = s.replaceAll("[0-9.]", "");
System.out.println(str);
wish help you.

That depends on what "such as" means. Are all the strings in the format "x.xxGB"? If that's the case, then you can use substring(), as you know the exact number of 'float' chars and 'suffix' chars.
String theStr = "x.xxGB";
String numStr = theStr.substring(0, 4); // grab first 4 chars: "x.xx"
float numFloat = Float.parseFloat(numStr);
String suffix = theStr.substring(5); // or .substring(5, 7); if you know the exact length
If it's more variable than that, it gets more complicated. If you don't know the length of the leading number string, you'd have to check the first part as a valid float, with perhaps the easiest way to be gathering characters as the start and checking each succession as a valid float, with all the rest being considered a suffix. Maybe something like this (pseudocode-ish):
String theStr = "324.994SUFFIX"; // SomeArbitraryNumberAndSuffixString
String currNumStr = "";
Boolean bHaveFloat = true;
for (int i = 1; i < theStr.length(); i++){
String testStr = theStr.substring(0, i);
try{
float f = Float.parseFloat(testStr);
} catch (NumberFormatException nfe){
// handle the exception, printStackTrace, etc...
// failed so No longer have Valid String...
break;
}
currNumStr = testStr;
}
// currNumStr now has the valid numberString

convert string to int in java

i have a string that has a int value in it. i just want to extract the int value from the string and print.
String str="No. of Days : 365";
String daysWithSplChar = str.replaceAll("[a-z][A-Z]","").trim();
char[] ch = daysWithSplChar.toCharArray();
StringBuffer stb = new StringBuffer();
for(char c : ch)
{
if(c >= '0' && c <= '9')
{
stb.append(c);
}
}
int days = Integer.ParseInt(stb.toString());
is there any better way than this. please let me know.

try String.replaceAll
String str = "No. of Days : 365";
str = str.replaceAll(".*?(\\d+).*", "$1");
System.out.println(str);
you will get
365

Another way of using regex (other than the way suggested by #EvgeniyDorofeev) which is closer to what you did:
str.replaceAll("[^0-9]",""); // give you "365"
which means, replace everything that is not 0-9 with empty string (or, in another word, remove all non-digit characters)
This is meaning the same, just a matter of taste which one is more comfortable to you:
str.replaceAll("\\D",""); // give you "365"

Scanner in = new Scanner("Hello123").useDelimiter("[^0-9]+");
int integer = in.nextInt();
This will get you the integer

following code gives you integer value
String str = "No. of Days : 365";
str = str.replaceAll(".*?(\\d+)", "$1");
System.out.println(str);
Integer x = Integer.valueOf(str);//365 in integer type
System.out.println(x+1);//output 366

Remove trailing zero in Java

I have Strings (from DB), which may contain numeric values. If it contains numeric values, I'd like to remove trailing zeros such as:
10.0000
10.234000
str.replaceAll("\\.0*$", ""), works on the first one, but not the second one.
A lot of the answers point to use BigDecimal, but the String I get may not be numeric. So I think a better solution probably is through the Regex.

there are possibilities:
1000 -> 1000
10.000 -> 10 (without point in result)
10.0100 -> 10.01
10.1234 -> 10.1234
I am lazy and stupid, just
s = s.indexOf(".") < 0 ? s : s.replaceAll("0*$", "").replaceAll("\\.$", "");
Same solution using contains instead of indexOf as mentioned in some of the comments for easy understanding
s = s.contains(".") ? s.replaceAll("0*$","").replaceAll("\\.$","") : s

Use DecimalFormat, its cleanest way
String s = "10.1200";
DecimalFormat decimalFormat = new DecimalFormat("0.#####");
String result = decimalFormat.format(Double.valueOf(s));
System.out.println(result);

Kent's string manipulation answer magically works and also caters for precision loss, But here's a cleaner solution using BigDecimal
String value = "10.234000";
BigDecimal stripedVal = new BigDecimal(value).stripTrailingZeros();
You can then convert to other types
String stringValue = stripedVal.toPlainString();
double doubleValue = stripedVal.doubleValue();
long longValue = stripedVal.longValue();
If precision loss is an ultimate concern for you, then obtain the exact primitive value. This would throw ArithmeticException if there'll be any precision loss for the primitive. See below
int intValue = stripedVal.intValueExact();

String value = "10.010"
String s = new DecimalFormat("0.####").format(Double.parseDouble(value));
System.out.println(s);
Output:
10.01

I find all the other solution too complicated. Simply
s.replaceFirst("\\.0*$|(\\.\\d*?)0+$", "$1");
does the job. It tries the first alternative first, so that dot followed by all zeros gets replaced by nothing (as the group doesn't get set). Otherwise, if it finds a dot followed by some digits (as few as possible due to the lazy quantifier *?) followed by some zeros, the zeros get discarded as they're not included in the group. It works.
Warning
My code relies on my assumption that appending a unmatched group does nothing. This is true for the Oracle implementation, but not for others, including Android, which seem to append the string "null". I'd call the such implementations broken as it just may no sense, but they're correct according to the Javadoc.

The following works for all the following examples:
"1" -> "1"
"1.0" -> "1"
"1.01500" -> "1.015"
"1.103" -> "1.103"
s = s.replaceAll("()\\.0+$|(\\..+?)0+$", "$2");

What about replacing
(\d*\.\d*)0*$
by
\1
?

You could replace with:
String result = (str.indexOf(".")>=0?str.replaceAll("\\.?0+$",""):str);
To keep the Regex as simple as possible. (And account for inputs like 1000 as pointed out in comments)

My implementation with possibility to select numbers of digits after divider:
public static String removeTrailingZero(String number, int minPrecise, char divider) {
int dividerIndex = number.indexOf(divider);
if (dividerIndex == -1) {
return number;
}
int removeCount = 0;
for (int i = dividerIndex + 1; i < number.length(); i++) {
if (number.charAt(i) == '0') {
removeCount++;
} else {
removeCount = 0;
}
}
int fracLen = number.length() - dividerIndex - 1;
if (fracLen - removeCount < minPrecise) {
removeCount = fracLen - minPrecise;
}
if (removeCount < 0) {
return number;
}
String result = number.substring(0, number.length() - removeCount);
if (result.endsWith(String.valueOf(divider))) {
return result.substring(0, result.length() - 1);
}
return result;
}

In addition to Kent's answer.
Be careful with regex in Kotlin. You have to manually write Regex() constructor instead of a simple string!
s = if (s.contains("."))
s.replace(Regex("0*\$"),"").replace(Regex("\\.\$"),"")
else s

Try to use this code:
DecimalFormat df = new DecimalFormat("#0.#####");
String value1 = df.format(101.00000);
String value2 = df.format(102.02000);
String value3 = df.format(103.20000);
String value4 = df.format(104.30020);
Output:
101
102.02
103.2
104.3002

Separate out the fraction part first. Then you can use the below logic.
BigDecimal value = BigDecimal.valueOf(345000);
BigDecimal div = new BigDecimal(10).pow(Integer.numberOfTrailingZeros(value.intValue()));
System.out.println(value.divide(div).intValue());

Converting a string to an integer on Android

How do I convert a string into an integer?
I have a textbox I have the user enter a number into:
EditText et = (EditText) findViewById(R.id.entry1);
String hello = et.getText().toString();
And the value is assigned to the string hello.
I want to convert it to a integer so I can get the number they typed; it will be used later on in code.
Is there a way to get the EditText to a integer? That would skip the middle man. If not, string to integer will be just fine.

See the Integer class and the static parseInt() method:
http://developer.android.com/reference/java/lang/Integer.html
Integer.parseInt(et.getText().toString());
You will need to catch NumberFormatException though in case of problems whilst parsing, so:
int myNum = 0;
try {
myNum = Integer.parseInt(et.getText().toString());
} catch(NumberFormatException nfe) {
System.out.println("Could not parse " + nfe);
}

int in = Integer.valueOf(et.getText().toString());
//or
int in2 = new Integer(et.getText().toString());

Use regular expression:
String s="your1string2contain3with4number";
int i=Integer.parseInt(s.replaceAll("[\\D]", ""));
output:
i=1234;
If you need first number combination then you should try below code:
String s="abc123xyz456";
int i=NumberFormat.getInstance().parse(s).intValue();
output:
i=123;

Use regular expression:
int i=Integer.parseInt("hello123".replaceAll("[\\D]",""));
int j=Integer.parseInt("123hello".replaceAll("[\\D]",""));
int k=Integer.parseInt("1h2el3lo".replaceAll("[\\D]",""));
output:
i=123;
j=123;
k=123;

Use regular expression is best way to doing this as already mentioned by ashish sahu
public int getInt(String s){
return Integer.parseInt(s.replaceAll("[\\D]", ""));
}

Try this code it's really working.
int number = 0;
try {
number = Integer.parseInt(YourEditTextName.getText().toString());
} catch(NumberFormatException e) {
System.out.println("parse value is not valid : " + e);
}

Best way to convert your string into int is :
EditText et = (EditText) findViewById(R.id.entry1);
String hello = et.getText().toString();
int converted=Integer.parseInt(hello);

You should covert String to float. It is working.
float result = 0;
if (TextUtils.isEmpty(et.getText().toString()) {
return;
}
result = Float.parseFloat(et.getText().toString());
tv.setText(result);

You can use the following to parse a string to an integer:
int value=Integer.parseInt(textView.getText().toString());
(1) input: 12 then it will work.. because textview has taken this 12 number as "12" string.
(2) input: "abdul" then it will throw an exception that is NumberFormatException.
So to solve this we need to use try catch as I have mention below:
int tax_amount=20;
EditText edit=(EditText)findViewById(R.id.editText1);
try
{
int value=Integer.parseInt(edit.getText().toString());
value=value+tax_amount;
edit.setText(String.valueOf(value));// to convert integer to string
}catch(NumberFormatException ee){
Log.e(ee.toString());
}
You may also want to refer to the following link for more information:
http://developer.android.com/reference/java/lang/Integer.html

There are five ways to convert
The First Way :
String str = " 123" ;
int i = Integer.parse(str);
output : 123
The second way :
String str = "hello123world";
int i = Integer.parse(str.replaceAll("[\\D]" , "" ) );
output : 123
The Third Way :
String str"123";
int i = new Integer(str);
output "123
The Fourth Way :
String str"123";
int i = Integer.valueOf(Str);
output "123
The Fifth Way :
String str"123";
int i = Integer.decode(str);
output "123
There could be other ways
But that's what I remember now

You can also do it one line:
int hello = Integer.parseInt(((Button)findViewById(R.id.button1)).getText().toString().replaceAll("[\\D]", ""));
Reading from order of execution
grab the view using findViewById(R.id.button1)
use ((Button)______) to cast the View as a Button
Call .GetText() to get the text entry from Button
Call .toString() to convert the Character Varying to a String
Call .ReplaceAll() with "[\\D]" to replace all Non Digit Characters with "" (nothing)
Call Integer.parseInt() grab and return an integer out of the Digit-only string.

The much simpler method is to use the decode method of Integer so for example:
int helloInt = Integer.decode(hello);

Kotlin
There are available Extension methods to parse them into other primitive types.
"10".toInt()
"10".toLong()
"true".toBoolean()
"10.0".toFloat()
"10.0".toDouble()
"10".toByte()
"10".toShort()
Java
String num = "10";
Integer.parseInt(num );

I have simply change this
int j = Integer.parseInt(user.replaceAll("[\\D]", ""));
into
int j = 1;
AND IT SOLVE MY PROBLEM