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Comparing integers and determining similarity

OK so I am writing a program that compares the number entered by a user to a computer generated number. Each number is 3 digits. What I'd like to know how to do is compare these integers. So for example if the last two digits of the three digit numbers are the same it will run a block of code. I'm not completely sure how to go about this. Any youtube videos or any links would be much appreciated. I would rather understand than get an answer to my code.

The fastet way would be: Convert the Integer to a String and then split the string at each char. Then you have an Char Array with each digit and you can compare them like whatever you want.
int number = 324;
String number_string = String.valueOf(number); //converts integer to string
char[] digits = number_string.toCharArray(); //converts string to char array
if(digits[2] == 4){ //checks if 3rd digit of the char array is 4
//do something
}

You can get last two digit by modulo(%) operator.when you modulo three digit number by 100 it return last two digit and compare these digit in if condition like,
int num=154; //number of three digit
int rem=num%100;//modulo by 100 return remainder
if(rem==54){ //checking condition
//code
}
else{
//code
}

I suppose that the number entered is a String so first you want to convert it to an integer.
String value = .... <whatever read from the user>
int number = Integer.parseInt(value)
Then, if you like to get information on some of the digits, use modulus, ie:
number%10
will give you the most right integer (ie 127%10 is 7)
number%100
will give you 2 digits (ie 626%100 is 26)

Treat octal numbers as decimals

I'm learning JAVA and recently I had the same problem with a few training tasks.
I have a some numbers and some of them are starting with 0. I found out that these numbers are octal which means it won't be the number I wanted or it gives me an error (because of the "8" or the "9" because they are not octal digits) after I read it as an int or long...
Until now I only had to work with two digit numbers like 14 or 05.
I treated them as Strings and converted them into numbers with a function that checks all of the String numbers and convert them to numbers like this
String numStr = "02";
if(numStr.startsWith("0")) {
int num = getNumericValue(numStr.charAt(1));
} else {
int num = Integer.parseInt(numStr);
}
Now I have an unkown lot of number with an unknown number of digits (so maybe more than 2). I know that if I want I can use a loop and .substring(), but there must be an easier way.
Is there any way to simply ignore the zeros somehow?
Edit:
Until now I always edited the numbers I had to work with to be Strings because I couldn't find an easier way to solve the problem. When I had 0010 27 09 I had to declare it like:
String[] numbers = {"0010", "27", "09"};
Because if I declare it like this:
int[] numbers = {0010, 27, 09};
numbers[0] will be 8 instead of 10 and numbers[2] will give me an error
Actually I don't want to work with Strings. What I actually want is to read numbers starting with zero as numbers (eg.: int or long) but I want them to be decimal. The problem is that I have a lot of number from a source. I copied them into the code and edited it to be a declaration of an array. But I don't want to edit them to be Strings just to delete the zeros and make them numbers again.

I'm not quite sure what you want to achieve. Do you want to be able to read an Integer, given as String in a 8-based format (Case 1)? Or do you want to read such a String and interpret it as 10-based though it is 8-based (Case 2)?
Or do you simply want to know how to create such an Integer without manually converting it (Case 3)?
Case 1:
String input = "0235";
// Cut the indicator 0
input = input.substring(1);
// Interpret the string as 8-based integer.
Integer number = Integer.parseInt(input, 8);
Case 2:
String input = "0235";
// Cut the indicator 0
input = input.substring(1);
// Interpret the string as 10-based integer (default).
Integer number = Integer.parseInt(input);
Case 3:
// Java interprets this as octal number
int octal = 0235;
// Java interprets this as hexadecimal number
int hexa = 0x235
// Java interprets this as decimal number
int decimal = 235
You can expand Case 1 to a intelligent method by reacting to the indicator:
public Integer convert(final String input) {
String hexaIndicator = input.substring(0, 2);
if (hexaIndicator.equals("0x")) {
return Integer.parseInt(input.substring(2), 16);
} else {
String octaIndicator = input.substring(0, 1);
if (octaIndicator.equals("0")) {
return Integer.parseInt(input.substring(1), 8);
} else {
return Integer.parseInt(input);
}
}
}

Parsing string to double/float throwing errors

I am extracting couple of values like 1234, 2456.00 etc from UI as string. When I try to parse this string to float, 1234 is becoming 1234.0 and when I tried to parse as double its throwing error. How can I solve this?
I am using selenium web driver and java. Below are few things I tried.
double Val=Double.parseDouble("SOQ");
double Val=(long)Double.parseDouble("SOQ");``

I think you mixed it up a bit when trying to figure out how to parse the numbers. So here is an overview:
// lets say you have two Strings, one with a simple int number and one floating point number
String anIntegerString = "1234";
String aDoubleString = "1234.123";
// you can parse the String with the integer value as double
double integerStringAsDoubleValue = Double.parseDouble(anIntegerString);
System.out.println("integer String as double value = " + integerStringAsDoubleValue);
// or you can parse the integer String as an int (of course)
int integerStringAsIntValue = Integer.parseInt(anIntegerString);
System.out.println("integer String as int value = " + integerStringAsIntValue);
// if you have a String with some sort of floating point number, you can parse it as double
double doubleStringAsDoubleValue = Double.parseDouble(aDoubleString);
System.out.println("double String as double value = " + doubleStringAsDoubleValue);
// but you will not be able to parse an int as double
int doubleStringAsIntegerValue = Integer.parseInt(aDoubleString); // this throws a NumberFormatException because you are trying to force a double into an int - and java won't assume how to handle the digits after the .
System.out.println("double String as int value = " + doubleStringAsIntegerValue);
This code would print out:
integer String as double value = 1234.0
integer String as int value = 1234
double String as double value = 1234.123
Exception in thread "main" java.lang.NumberFormatException: For input string: "1234.123"
Java will stop "parsing" the number right when it hits the . because an integer can never have a . and the same goes for any other non-numeric vales like "ABC", "123$", "one" ... A human may be able to read "123$" as a number, but Java won't make any assumptions on how to interpret the "$".
Furthermore: for float or double you can either provide a normal integer number or anything with a . somewhere, but no other character besides . is allowed (not even , or ; and not even a WHITESPACE)
EDIT:
If you have a number with "zeros" at the end, it may look nice and understandable for a human, but a computer doesn't need them, since the number is still mathematically correct when omitting the zeros.
e.g. "123.00" is the same as 123 or 123.000000
It is only a question of formatting the output when printing or displaying the number again (in which case the number will be casted back into a string). You can do it like this:
String numericString = "2456.00 "; // your double as a string
double doubleValue = Double.parseDouble(numericString); // parse the number as a real double
// Do stuff with the double value
String printDouble = new DecimalFormat("#.00").format(doubleValue); // force the double to have at least 2 digits after the .
System.out.println(printDouble); // will print "2456.00"
You can find an overview on DecimalFormat here.
For example the # means "this is a digit, but leading zeros are omitted" and 0 means "this is a digit and will not be omitted, even if zero"
hope this helps

Your first problem is that "SOQ" is not a number.
Second, if you want create a number using a String, you can use parseDouble and give in a value that does not have a decimal point. Like so:
Double.parseDouble("1");
If you have a value saved as a long you do not have to do any conversions to save it as a double. This will compile and print 10.0:
long l = 10l;
double d = l;
System.out.println(d);
Finally, please read this Asking a good question

The problem is you cannot parse non-numeric input as a Double.
For example:
Double.parseDouble("my text");
Double.parseDouble("alphanumeric1234");
Double.parseDouble("SOQ");
will cause errors.
but the following is valid:
Double.parseDouble("34");
Double.parseDouble("1234.00");

The number you want to parse into Double contains "," and space so you need first to get rid of them before you do the parsing
String str = "1234, 2456.00".replace(",", "").replace(" ", "");
double Val=Double.parseDouble(str);

Multiplying ascii code and getting string back

I am trying to convert a string to ascii code and then multiplying that concatenation of the ascii codes by a number.
For example
String message = "Hello";
String result = "";
ArrayList arrayList = new ArrayList();
int temp;
for(int i = 0; i < message.length(); i++){
temp = (int) message.charAt(i);
result = result + String.value(temp).toString();
arrayList.add(String.valueOf(temp).toString());
}
I have tried two different ways, but there is always a catch with each one.
If I just concatenate all the ascii codes together into a string and I get 72101108108111 as my new string, the problem now is how can I get the original string back from this? This is because it is not obvious where each one character code starts and ends and the next one begins.
Another way I tried doing this was to use an array. I would receive |72|101|108|108|111| in an array. Obviously the codes are split here, but if I wanted to multiply this whole array (all the numbers as one number) by a number and then how would I get the array back together?
These are two different ways I have thought to solve this, but I have no idea how to get the string back out of these if I multiply the ascii by a number.

You don't need to modify the original string nor the ascii code string. Just have them both there, then whenever you need to get the numerical value of the string, just use X.valueOf(...)** method. Example,
final String message = "Hello";
final String result = "72101108108111";
long value = Long.valueOf(result);
If you do not want to store the two strings, then you should go with the array method. To get a numerical value, you simply concatenate all the strings in the array into one and use the X.valueOf(..) method.
And to get back the original string, use Integer.valueOf(...) on each string in the array, then cast each one to char.
System.out.println((char)Integer.valueOf("111").intValue());
** Note by X.valueOf(..), X doesn't have to be Long or Integer as I have shown. As you mentioned the value can get really large so BigInteger should be prefered above others

converts parts of a string to int in java

I was wondering how can I take some numbers in a string and convert them to an integer type? for example if a user entered 12:15pm how can I get 1 and 2 and make an int with value 12?

Given the example above, you could try something like this:
final int value = Integer.parseInt(input.substring(0, input.indexOf(':'))); //value = 12
Where input = 12:15pm in this case.
Generally speaking, just use a combination of String#indexOf(String), String#substring(int, int) and Integer.parseInt(String).

Read the String and Integer API's
You can use the String.split() to get the two numeric strings
You can use Integer.parseInt(...) to convert the String to an int.
Edit: Using the split() you can do something like:
String time = "12:34pm";
int hour = Integer.parseInt( time.split(":")[0] );