我想獲得下一個遞增值9（當然是10），但它返回一個ASCII碼

Question

我正在處理屬性文件。關於它，我有一個關鍵是「user_email」和值，我把它設置爲toomeuser0@gmail.com。

現在，在我的代碼中，我希望電子郵件的值能夠在我的程序運行時進行迭代，因此toomeuser0@gmail.com將會是toomeuser1@gmail.com等等，並且從那裏讀取屬性中的電子郵件值文件並在我的程序中調用它。

public String getEmailFromProperty(){ 

     String new_email = ""; 

     try{ 

      new ConfigReader(); 
      FileOutputStream fos = new FileOutputStream(property_file); 
      StringBuilder str = new StringBuilder(property.getProperty("user_email")); 

      char charWithNumber = str.charAt(9); // get the number character of the email (starts with 0) 

      int toInteger = (int) charWithNumber; // convert that character number to an Integer value 

      int numericValue = Character.getNumericValue(toInteger+1); // the number character in email is converted to int then incremented by 1 on each run 

      int toAscii = numericValue + 48; // the number character (which is now an Int) is added by 48 

      char toChar = (char) toAscii; // it is converted back to a character in order for it to be passed as a parameter to setCharAt() method 

      str.setCharAt(9, toChar); // attached the newly incremented number character to the email @ 9th index 

      new_email = str.toString(); // converted the StringBuilder variable str to an ordinary String in order to call toString() method 

      property.setProperty("user_email", new_email); // now, I wrote the new email to the property file using the "user_email" key 


      property.store(fos, null); 
      fos.close(); 


     } 
     catch(Exception e){ 
      System.out.println("Error is " + e.getMessage()); 

     } 

     return new_email; 


    } 

我知道這對你來說有點麻煩。但是，當電子郵件號字符達到值9，然後它增加，我預計它是10.但是，它返回'/'字符。我想運行後toomeuser10@gmail.com不toomeuser/@gmail.com

Answer 1

你感到困惑char和String。字符是單個字符，並且將永遠不能表示'10'，因爲此編號使用兩個chars：1和0。只有String可以代表10號正確，就像這樣："10"

因此，這些線路應改爲：

int numericValue = Character.getNumericValue(toInteger+1); // <-- This breaks at '9' 
int toAscii = numericValue + 48; // <-- This breaks after '9' 
char toChar = (char) toAscii; // <-- this should be a String 
str.setCharAt(9, toChar); // This cannote be used because we may need more than 1 char 

要像這樣（未經）：

int numericValue = Character.getNumericValue(toInteger) + 1; // Note the parenthesis 
String asString = String.valueOf(numericValue); 
String prefix = str.subString(9); 
String suffix = str.subString(10, str.length()); 
str = prefix + asString + suffix; 

Answer 2

，如果您喜歡支持大於9的數字，我會建議使用正則表達式功能來查找和替換數字：

public String incrementNumberInEmail(String email) 
{ 
    Matcher matcher = Pattern.compile("\\d+").matcher(email); 
    if(matcher.find()) { 
     StringBuffer buffer = new StringBuffer(); 
     int next = Integer.valueOf(matcher.group()) + 1; 
     matcher.appendReplacement(buffer, String.valueOf(next)); 
     matcher.appendTail(buffer); 
     return buffer.toString(); 
    } 
    else throw new IllegalArgumentException("Email does not contain number: " 
              + email); 
} 

@Test 
public void test() 
{ 
    assertThat(incrementNumberInEmail("foo42@fubar.com"), is("foo43@fubar.com")); 
} 

Answer 3

而不是使用下面的代碼

int numericValue = Character.getNumericValue(toInteger+1); // the number character in email is converted to int then incremented by 1 on each run 

    int toAscii = numericValue + 48; // the number character (which is now an Int) is added by 48 

    char toChar = (char) toAscii; // it is converted back to a character in order for it to be passed as a parameter to setCharAt() method 

    str.setCharAt(9, toChar); // attached the newly incremented number character to the email @ 9th index 

使用此

toInteger = toInteger+1; 
toString = toInteger + ""; 
str.setCharAt(9,toString); 

由於現時10沒有ASCII碼和ASCII有可供1至9 ASCII代碼，所以在這種情況下，10將像1和O不同的字符，所以改爲使用int轉換爲int（在這種情況下，您不需要添加48，因爲我們直接處理字符串）

您可以參考以下鏈接： http://ascii.cl/