Develop Reference

Add zeros to the left of Number

I know that I can add left zeros to String but what about Long?
I need to put left zeros until the Long size is 10 digits. For example, if it's 8 digits (12345678), it should add 2 left zeros (0012345678)
I want to add this in the getValue() method.
public Long getValue() {
// Should always be 10 digits, If it's 8, add zeros
return value;
}
I'm using spring. This issue is that the database that cuts the left zeros. Maybe is there a annotation to avoid extra code?

This is not possible. A Long does not contain data about the String representation of its value. In fact, the Long is actually stored in binary, not decimal, and the long object is unaware of this.
If you want to convert it to a String with leading zeroes, String.format("%017d" , number); will pad it to make sure it has 10 digits.

In java, a long (wrapped in a Long) will always be stored on 8 bytes,
There is no way to "add" extra zeros as they're already existing.
Either your database must change its type to String and add padding zeros when you store your Long object either change your inner code to String and add padding zeros when you pull the data from your Long in db.

You cannot because a long does not have a leading zero.
A string of characters like 0012345678 is not an integer, 12345678 is.
but there are two way in java to add leading zeroes
WAY 1: format() method
int number = 9;
String str = String.format("%04d", 9); // 0009
System.out.printf("original number %d, numeric string with padding : %s", 9, str);
WAY 2 : DecimalFormat
DecimalFormat df = new DecimalFormat("0000");
String c = df.format(9); // 0009
String a = df.format(99); // 0099
String b = df.format(999); // 0999
but in both case you get string instead of Long
for more reading

Try this one,
int number = 12345678;
String str = String.format("%10d", number);
System.out.println("original number %d, numeric string with padding : %s", number, str);

Java - Construct a signed numeric String and convert it to an integer

Can I somehow prepend a minus sign to a numeric String and convert it into an int?
In example:
If I have 2 Strings :
String x="-";
String y="2";
how can i get them converted to an Int which value is -2?

You will first have to concatenate both Strings since - is not a valid integer character an sich. It is however acceptable when it's used together with an integer value to denote a negative value.
Therefore this will print -2 the way you want it:
String x = "-";
String y = "2";
int i = Integer.parseInt(x + y);
System.out.println(i);
Note that the x + y is used to concatenate 2 Strings and not an arithmetic operation.

Integer.valueOf("-") will throw a NumberFormatException because "-" by itself isn't a number. If you did "-1", however, you would receive the expected value of -1.
If you're trying to get a character code, use the following:
(int) "-".charAt(0);
charAt() returns a char value at a specific index, which is a two-byte unicode value that is, for all intensive purposes, an integer.

(JAVA) how to use charAt() to find last digit number (or rightmost) of any real number?

Hello Im new to programming
and as title I was wonder how you can find last digit of any given number?
for example when entering in 5.51123123 it will display 3
All I know is I should use charAt
Should I use while loop?
thanks in advance

You would want to do something like this:
double number = 5.51123123;
String numString = number + "";
System.out.println(numString.charAt(numString.length()-1));
When you do the number + "", Java "coerces" the type of number from a double to a string and allows you to perform string functions on it.
The numString.length()-1 is because numString.length() returns the count of all the characters in the string BUT charAt() indexes into the string and its indexing begins at 0, so you need to do the -1 or you'll get a StringIndexOutOfBoundsException.

You can use below function simply and you can customize data types accordingly,
private int getLastDigit(double val){
String tmp = String.valueOf(val);
return Integer.parseInt(tmp.substring(tmp.length()-1));
}

You can't use charAt on a floating point variable (or any other data type other than Strings). (Unless you define a charAt method for your own classes...)
You can, however, convert that floating point number to a string and check the charAt(str.length()-1).

Convert your float variable to string and use charAt(strVar.length - 1).
double a=5.51123123;
String strVar=String.valueOf(a);
System.out.print(strVar.charAt(strVar.length -1);

Double doubleNo = 5.51123123;
String stringNo = doubleNo.toString();
System.out.println(stringNo.charAt(stringNo.length()-1));

Is it possible to get only the first character of a String?

I have a for loop in Java.
for (Legform ld : data)
{
System.out.println(ld.getSymbol());
}
The output of the above for loop is
Pad
CaD
CaD
CaD
Now my question is it possible to get only the first characer of the string instead of the whole thing Pad or CaD
For example if it's Pad I need only the first letter, that is P
For example if it's CaD I need only the first letter, that is C
Is this possible?

Use ld.charAt(0). It will return the first char of the String.
With ld.substring(0, 1), you can get the first character as String.

String has a charAt method that returns the character at the specified position. Like arrays and Lists, String is 0-indexed, i.e. the first character is at index 0 and the last character is at index length() - 1.
So, assuming getSymbol() returns a String, to print the first character, you could do:
System.out.println(ld.getSymbol().charAt(0)); // char at index 0

The string has a substring method that returns the string at the specified position.
String name="123456789";
System.out.println(name.substring(0,1));

Here I am taking Mobile No From EditText It may start from +91 or 0 but i am getting actual 10 digits.
Hope this will help you.
String mob=edit_mobile.getText().toString();
if (mob.length() >= 10) {
if (mob.contains("+91")) {
mob= mob.substring(3, 13);
}
if (mob.substring(0, 1).contains("0")) {
mob= mob.substring(1, 11);
}
if (mob.contains("+")) {
mob= mob.replace("+", "");
}
mob= mob.substring(0, 10);
Log.i("mob", mob);
}

Answering for C++ 14,
Yes, you can get the first character of a string simply by the following code snippet.
string s = "Happynewyear";
cout << s[0];
if you want to store the first character in a separate string,
string s = "Happynewyear";
string c = "";
c.push_back(s[0]);
cout << c;

Java strings are simply an array of char. So, char c = s[0] where s is string.

Java: string tokenizer and assign to 2 variables?

Let's say I have a time hh:mm (eg. 11:22) and I want to use a string tokenizer to split. However, after it's split I am able to get for example: 11 and next line 22. But how do I assign 11 to a variable name "hour" and another variable name "min"?
Also another question. How do I round up a number? Even if it's 2.1 I want it to round up to 3?

Have a look at Split a string using String.split()
Spmething like
String s[] = "11:22".split(":");;
String s1 = s[0];
String s2 = s[1];
And ceil for rounding up
Find ceiling value of a number using Math.ceil

Rounding a number up isn't too hard. First you need to determine whether it's a whole number or not, by comparing it cast as both an int and a double. If they don't match, the number is not whole, so you can add 1 to the int value to round it up.
// num is type double, but will work with floats too
if ((int)num != (double)num) {
int roundedNum = (int)num + 1;
}