在java中将String格式化为Integer

Question

我面临以下问题:

与临时值相比,我将获得类似或更长的值:

 public class NumberFormat {
     public static void main(String arg[]){
     Integer numValue = null;
     String temp="5474151538110135";
     numValue=Integer
    .parseInt(temp.trim());
     System.out.println("--> "+numValue);

}
}

请提供解决方案.

Exception in thread "main" java.lang.NumberFormatException: For input string: "5474151538110135"
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:60)
    at java.lang.Integer.parseInt(Integer.java:473)
    at java.lang.Integer.parseInt(Integer.java:511)
    at com.filetransfer.August.NumberFormat.main(NumberFormat.java:10)

Answer 1

5474151538110135大于Integer.MAX_VALUE.使用Long.parseLong相反或BigInteger输入数字可能会显着增长

Long numValue = Long.parseLong(temp.trim());