[英]Get last number from string
I want to get the last character of a string and if it's a even number to do different actions based on that. 我想获取字符串的最后一个字符,如果它是偶数,则可以基于该字符执行不同的操作。 Simple example:
简单的例子:
Server_3
First I want to get the last value 3 首先我要获得最后的值3
String str = "Server_3";
str.charAt(str.length()-1)
The problem is next how to construct the logic to have different cased based on that is this a even number or not. 接下来的问题是如何基于是否为偶数来构造具有不同大小写的逻辑。 Can you help me to complete this?
你能帮我完成这个吗?
The correct code is: 正确的代码是:
if (Integer.valueOf(str.charAt(str.length()-1) + "") % 2 == 0) {
// Even logic
} else {
// Odd logic
}
Or.... to reduce method calls and the get rid of the funky string concatenation: 或者....以减少方法调用并摆脱时髦的字符串连接:
if ((str.charAt(str.length()-1) - '0') % 2) {
// Even logic
} else {
// Odd logic
}
Convert to integer and check for even 转换为整数并检查偶数
if(Integer.valueOf(str.charAt((str.length()-1) + "") % 2 ==0) {
// do your work here
}
使用模( %
)运算符: i % 2
如果i
为偶数,则i % 2
为0,如果i
为奇数,则为1。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.