轉換爲int時出現Java字符串錯誤

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我想使用Integer.parseInt（）將表示電話號碼的字符串轉換爲我可以使用的int值。轉換爲int時出現Java字符串錯誤

這是我得到的錯誤：

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

這是我的代碼：

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，當我嘗試剛剛替補它的工作原理使用的Integer.parseInt一個硬編碼字符串（）。這導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。我是新來的java，所以我不知道爲什麼會這樣，或者我做錯了什麼。

來源

2017-03-05 Samier Mahagna