问题
I have the following values:
int a=1;
int b=0;
int c=2;
int d=2;
int e=1;
How do i concatenate these values so that i end up with a String that is 10221;
please note that multiplying a by 10000, b by 1000.....and e by 1 will not working since b=0 and therefore i will lose it when i add the values up.
回答1:
The easiest (but somewhat dirty) way:
String result = "" + a + b + c + d + e
Edit: I don't recommend this and agree with Jon's comment. Adding those extra empty strings is probably the best compromise between shortness and clarity.
回答2:
Michael Borgwardt's solution is the best for 5 digits, but if you have variable number of digits, you can use something like this:
public static String concatenateDigits(int... digits) {
StringBuilder sb = new StringBuilder(digits.length);
for (int digit : digits) {
sb.append(digit);
}
return sb.toString();
}
回答3:
This worked for me.
int i = 14;
int j = 26;
int k = Integer.valueOf(String.valueOf(i) + String.valueOf(j));
System.out.println(k);
It turned out as 1426
回答4:
just to not forget the format method
String s = String.format("%s%s%s%s%s", a, b, c, d, e);
(%1.1s%1.1s%1.1s%1.1s%1.1s if you only want the first digit of each number...)
回答5:
Actually,
int result = a * 10000 + b * 1000 + c * 100 + d * 10 + e;
String s = Integer.toString(result);
will work.
Note: this will only work when a is greater than 0 and all of b, c, d and e are in [0, 9]. For example, if b is 15, Michael's method will get you the result you probably want.
回答6:
StringBuffer sb = new StringBuffer();
sb.append(a).append(b).append(c)...
Keeping the values as an int is preferred thou, as the other answers show you.
回答7:
If you multiply b by 1000, you will not lose any of the values. See below for the math.
10000
0
200
20
1
=====
10221
回答8:
Others have pointed out that multiplying b by 1000 shouldn't cause a problem - but if a were zero, you'd end up losing it. (You'd get a 4 digit string instead of 5.)
Here's an alternative (general purpose) approach - which assumes that all the values are in the range 0-9. (You should quite possibly put in some code to throw an exception if that turns out not to be true, but I've left it out here for simplicity.)
public static String concatenateDigits(int... digits)
{
char[] chars = new char[digits.length];
for (int i = 0; i < digits.length; i++)
{
chars[i] = (char)(digits[i] + '0');
}
return new String(chars);
}
In this case you'd call it with:
String result = concatenateDigits(a, b, c, d, e);
回答9:
For fun... how NOT to do it ;-)
String s = Arrays.asList(a,b,c,d,e).toString().replaceAll("[\\[\\], ]", "");
Not that anyone would really think of doing it this way in this case - but this illustrates why it's important to give access to certain object members, otherwise API users end up parsing the string representation of your object, and then you're stuck not being able to modify it, or risk breaking their code if you do.
回答10:
Using Java 8 and higher, you can use the StringJoiner, a very clean and more flexible way (especially if you have a list as input instead of known set of variables a-e):
int a = 1;
int b = 0;
int c = 2;
int d = 2;
int e = 1;
List<Integer> values = Arrays.asList(a, b, c, d, e);
String result = values.stream().map(i -> i.toString()).collect(Collectors.joining());
System.out.println(result);
If you need a separator use:
String result = values.stream().map(i -> i.toString()).collect(Collectors.joining(","));
To get the following result:
1,0,2,2,1
Edit: as LuCio commented, the following code is shorter:
Stream.of(a, b, c, d, e).map(Object::toString).collect(Collectors.joining());
回答11:
How about not using strings at all...
This should work for any number of digits...
int[] nums = {1, 0, 2, 2, 1};
int retval = 0;
for (int digit : nums)
{
retval *= 10;
retval += digit;
}
System.out.println("Return value is: " + retval);
回答12:
I would suggest converting them to Strings.
StringBuilder concatenated = new StringBuilder();
concatenated.append(a);
concatenated.append(b);
/// etc...
concatenated.append(e);
Then converting back to an Integer:
Integer.valueOf(concatenated.toString());
回答13:
Use StringBuilder
StringBuilder sb = new StringBuilder(String.valueOf(a));
sb.append(String.valueOf(b));
sb.append(String.valueOf(c));
sb.append(String.valueOf(d));
sb.append(String.valueOf(e));
System.out.print(sb.toString());
回答14:
People were fretting over what happens when a == 0. Easy fix for that...have a digit before it. :)
int sum = 100000 + a*10000 + b*1000 + c*100 + d*10 + e;
System.out.println(String.valueOf(sum).substring(1));
Biggest drawback: it creates two strings. If that's a big deal, String.format could help.
int sum = a*10000 + b*1000 + c*100 + d*10 + e;
System.out.println(String.format("%05d", sum));
回答15:
You can Use
String x = a+"" +b +""+ c+""+d+""+ e;
int result = Integer.parseInt(x);
回答16:
Assuming you start with variables:
int i=12;
int j=12;
This will give output 1212:
System.out.print(i+""+j);
And this will give output 24:
System.out.print(i+j);
回答17:
Best solutions are already discussed. For the heck of it, you could do this as well: Given that you are always dealing with 5 digits,
(new java.util.Formatter().format("%d%d%d%d%d", a,b,c,d,e)).toString()
I am not claiming this is the best way; just adding an alternate way to look at similar situations. :)
回答18:
int number =0; int[] tab = {your numbers}.
for(int i=0; i<tab.length; i++){
number*=10;
number+=tab[i];
}
And you have your concatenated number.
回答19:
Couldn't you just make the numbers strings, concatenate them, and convert the strings to an integer value?
回答20:
public class joining {
public static void main(String[] args) {
int a=1;
int b=0;
int c=2;
int d=2;
int e=1;
String j = Long.toString(a);
String k = Long.toString(b);
String l = Long.toString(c);
String m = Long.toString(d);
String n = Long.toString(e);
/* String s1=Long.toString(a); // converting long to String
String s2=Long.toString(b);
String s3=s2+s1;
long c=Long.valueOf(s3).longValue(); // converting String to long
*/
System.out.println(j+k+l+m+n);
}
}
来源：https://stackoverflow.com/questions/2674707/how-to-concatenate-int-values-in-java