转换为int时出现Java字符串错误

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我想使用Integer.parseInt（）将表示电话号码的字符串转换为我可以使用的int值。转换为int时出现Java字符串错误

这是我得到的错误：

Enter a 10-digit telephone number with optional space or dash after the first and the second block of three digits > 
5558759142 
Processing encrypted number "5558759142" 
Exception in thread "main" java.lang.NumberFormatException: For input string: "5558759142" 
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) 
    at java.lang.Integer.parseInt(Integer.java:583) 
    at java.lang.Integer.parseInt(Integer.java:615) 
    at TelephoneNumberValidator.main(TelephoneNumberValidator.java:32) 
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) 
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) 
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) 
    at java.lang.reflect.Method.invoke(Method.java:498) 
    at com.intellij.rt.execution.application.AppMain.main(AppMain.java:147) 

Process finished with exit code 1

这是我的代码：

import java.util.*; 
import java.lang.Integer; 

public class TelephoneNumberValidator { 
    public static void main(String[] args) { 
     final String VALID_PHONE_PATTERN = "[0-9]{3}[-| ]?[0-9]{3}[-| ]?[0-9]{4}"; 
     final int NUMBERS_FOLLOWING_AREA_CODE = 7; 
     final int DECRYPTED_AREA_CODE = 212; 
     Scanner scanKeyboard = new Scanner(System.in); 
     String userInput; 

     do { 
      System.out.println("Enter a 10-digit telephone number with optional space or dash after the first and the second block of three digits >"); 
      userInput = scanKeyboard.nextLine(); 
      //check to see if the phone number is correct 
      if (!userInput.matches(VALID_PHONE_PATTERN)) 
      { 
       System.out.println("The given input \" " + userInput + " \" is not valid"); 
      } 
     }while (!userInput.matches(VALID_PHONE_PATTERN)); 

     //extract int from string 
     String phoneJustNumbers = userInput; 
     phoneJustNumbers = phoneJustNumbers.replaceAll("-",""); // replaces hyphens 
     phoneJustNumbers = phoneJustNumbers.replaceAll(" ", ""); 

     System.out.println("Processing encrypted number \"" + phoneJustNumbers + "\""); 
     //check the first 3 digits (aka area code) to see if it matches 212 
     int areaCode = Integer.parseInt(phoneJustNumbers);//gets area code 
     int placeholderForAreaCode = 111; //set 111 so that each number in area code can be accounted for 
     int placeholderForEntirePhoneNum = 1111111111; //set each digit to 1 so we can shift later 
     for (int j = 0; j <= 9; j++) 
     { 
      if ((areaCode + (placeholderForAreaCode * j) == DECRYPTED_AREA_CODE)) // we are shifting each digit by one 
                         // to see if it matches the decrypted area code 
      { 
       System.out.println("The telephone number is \"" + (Integer.parseInt(phoneJustNumbers) * placeholderForEntirePhoneNum * j) 
         + "\" with a shift value of " + j); 
      } 
      else 
      { 
       System.out.println("The telephone number \"" + userInput + "\" cannot be decrypted area code."); 
      } 
     } 

    } 
}

所以，当我尝试刚刚替补它的工作原理使用的Integer.parseInt一个硬编码字符串（）。这导致我相信我生成清理我的字符串的方式不知何故阻止了它的工作。我是新来的java，所以我不知道为什么会这样，或者我做错了什么。

来源

2017-03-05 Samier Mahagna