如何在java中连接int值？
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How to concatenate int values in java?
提问by Shamli
I have the following values:
我有以下价值观：
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 aby 10000, bby 1000.....and eby 1 will not working since b=0and therefore i will lose it when i add the values up.
我如何连接这些值，以便我最终得到一个字符串10221；请注意，乘以a10000、乘以b1000.....再e乘以 1 将不起作用b=0，因此当我将这些值相加时我会丢失它。
回答by T .
Actually,
实际上，
int result = a * 10000 + b * 1000 + c * 100 + d * 10 + e;
String s = Integer.toString(result);
willwork.
会工作。
Note: this will only work when ais greater than 0 and all of b, c, dand eare in [0, 9]. For example, if bis 15, Michael's method will get you the result you probably want.
注意：这仅在a大于 0 且所有b、c、d和e都在 [0, 9] 中时才有效。例如，如果b是 15，Michael 的方法将为您提供您可能想要的结果。
回答by Daniel A. White
If you multiply bby 1000, you will not lose any of the values. See below for the math.
如果乘以b1000，则不会丢失任何值。请参阅下面的数学。
10000
0
200
20
1
=====
10221
回答by Bj?rn
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.
正如其他答案向您展示的那样，将值保留为 int 是首选。
回答by Grundlefleck
I would suggest converting them to Strings.
我建议将它们转换为字符串。
StringBuilder concatenated = new StringBuilder();
concatenated.append(a);
concatenated.append(b);
/// etc...
concatenated.append(e);
Then converting back to an Integer:
然后转换回整数：
Integer.valueOf(concatenated.toString());
回答by Shervin Asgari
Use StringBuilder
使用 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by Jon Skeet
Others have pointed out that multiplying bby 1000 shouldn't cause a problem - but if awere zero, you'd end up losing it. (You'd get a 4 digit string instead of 5.)
其他人指出乘以b1000 不应该引起问题 - 但如果a为零，你最终会失去它。（你会得到一个 4 位数的字符串，而不是 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.)
这是另一种（通用）方法 - 假设所有值都在 0-9 范围内。（如果结果不是真的，你很可能应该输入一些代码来抛出异常，但为了简单起见，我在这里把它省略了。）
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:
在这种情况下，您可以使用以下命令调用它：
String result = concatenateDigits(a, b, c, d, e);
回答by Michael Borgwardt
The easiest (but somewhat dirty) way:
最简单（但有点脏）的方法：
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.
编辑：我不推荐这个并且同意 Jon 的评论。添加那些额外的空字符串可能是简短和清晰之间的最佳折衷。
回答by polygenelubricants
Michael Borgwardt's solution is the best for 5 digits, but if you have variable number of digits, you can use something like this:
Michael Borgwardt 的解决方案最适合 5 位数，但如果您的位数可变，则可以使用如下方法：
public static String concatenateDigits(int... digits) {
StringBuilder sb = new StringBuilder(digits.length);
for (int digit : digits) {
sb.append(digit);
}
return sb.toString();
}
回答by cHao
People were fretting over what happens when a == 0. Easy fix for that...have a digit before it. :)
人们担心当 a == 0 时会发生什么。很容易解决这个问题......在它之前有一个数字。:)
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.
最大的缺点：它创建了两个字符串。如果这是一个大问题，String.format 可以提供帮助。
int sum = a*10000 + b*1000 + c*100 + d*10 + e;
System.out.println(String.format("%05d", sum));
回答by Ali
You can Use
您可以使用
String x = a+"" +b +""+ c+""+d+""+ e;
int result = Integer.parseInt(x);
