上午我做這些布爾方法正確

Question

設計了一個名爲密碼2如下面的UML圖指定類：

• 密碼必須至少有七個字符
• 密碼只由字母和數字
• 密碼必須包含至少三個數字

+Password2 (ps: String)個構造帶密碼進入

+verifyPassword2():void調用其他配套方法

-verifyLength():boolean返回真/假基於密碼的長度

-verifyValidCharacters():boolean返回真/假基於有效/無效字符

-verifyNumberOfDigits(): boolean返回true/false基於位數

+printReport():void打印總結報告爲特殊cified下面

這是我到目前爲止有：

public class Password2 { 
private boolean rule1; 
private boolean rule2; 
private boolean rule3; 
private String ps; 

public Password2(){ 
} 

public Password2(String ps){ 
    this.ps = ps; 
} 

public void verifyPassword2(){ 
    //Verify the password is the right length 
    rule1 = verifyLength(); 
    rule2 = verifyValidCharacters(); 
    rule3 = verifyNumberOfDigits(); 
}  
private boolean verifyLength(){ 
     if (ps.length() >= 8){ 
      return true; 
     return false; 
    } 
    //Verify if the password have the right characters 
private boolean verifyValidCharacters(){ 
    if (ps.matches("^[A-Za-z1-9]$"){ 
     return true;} 
    return false;  
    } 
    //Verify if the password contains at least 2 digits 
private boolean verifyNumberOfDigits(){ 
    int digit = 0; 
    for (int i = 0; i = <ps.length(); i++){ 
     char letter = password.charAt(i); 
     if(Character.isDigit(letter)){ 
       digit++; 
        } 
     if (digits >= 2){ 
      return true;{ 
        } 
     return false 
     } 
    } 

public void printReport(){ 
    String Report; 
    Report = "\t+++++++++++++++++ Password Report ++++++++++++++++++++\n"; 
    Report += "\tLebron Broadnax\n"; 
    Report += "\tCS1302\n"; 
    Report += "\t6/11/201\n3"; 
    Report += "\t+++++++++++++++++++++++++++++++++++++++++++++++++++++"; 

} 
} 

方法不是另一種方法裏面這只是我怎麼貼吧

Answer 1

一個提示改善：

//Verify if the password have the right characters 
private boolean verifyValidCharacters() 
{ 
    if (ps.matches("^[A-Za-z1-9]$") 
     return true; 

    return false; 
} 

可更改爲：

private boolean verifyValidCharacters() 
{ 
    return ps.matches("^[A-Za-z1-9]$"); 
} 

既然你想要返回true如果匹配，並且false如果不匹配，這正是matches(regex)所做的。

同樣的邏輯可以適用於：

private boolean verifyNumberOfDigits() 
    { 
     int digit = 0; 

     for (int i = 0; i = <ps.length(); i++) 
     { 
      char letter = password.charAt(i); 

      if(Character.isDigit(letter)) 
       digit++; 
     }    

     return (digit >= 2);    
} 

而且，你的方法應該不是你verifyPassword2方法內部聲明。

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