我已经开始在我的AS级计算课程中学习Java，并且已经完成了我们设置的第一个DIY任务。
我使用了do-while语句来查看用户的输入用户名是否在数组“names”中 - 如果不是，则请求重新输入用户名，直到插入正确的用户名为止。我还设置了一个布尔值，所以当输入正确的用户名时，它会取消do-while循环并继续使用代码 - 但它没有。
String[] names = {"mckeownl", "heardj", "williamsc"};
String[] attendance = {"yes", "no", "yes"};
int[] grade = {96, 66, 73};
boolean loggedin = false;
Scanner user_input = new Scanner(System.in);
String login;
login = user_input.next();
do { // beginning of while - login
System.out.println("Insert student's surname followed by the first letter");
System.out.print("of their first name (e.g John Smith = smithj): ");
if (Arrays.asList(names).contains(login)) {
System.out.println("Student selected: "+login+".");
loggedin = true;
}
else {
System.out.println("Incorrect student name! Please try again.");
loggedin = false;
}
} while ( ! loggedin);
if (login.equals(names[0])) {
System.out.println("Attend today: "+attendance[0]);
System.out.println("Grade: ");
}
else {
System.out.println("poo");
}
}
}
正确名称的输出是;
“插入学生姓氏后跟第一个字母 他们的名字（例如John Smith = smithj）：mckeownl 学生入选：mckeownl。“
为什么不是最终的if语句输出？
答案 0 :(得分：0)
你应该要求在循环中登录。
如果失败，您不应将loggedin变量设置为false。只需提供失败的文本，然后它将返回到顶部并再次请求登录。
您可以在方法调用中拥有多行。所以你可以：
System.out.println("Insert student's surname followed by the first letter " +
"of their first name (e.g John Smith = smithj): ");
而不是：
System.out.println("Insert student's surname followed by the first letter");
System.out.print("of their first name (e.g John Smith = smithj): ");
答案 1 :(得分：0)
将login = user_input.next();放入循环中..为了让用户再次loggedin ..
do { // beginning of while - login
System.out.println("Insert student's surname followed by the first letter");
System.out.print("of their first name (e.g John Smith = smithj): ")
login = user_input.next(); // <--- you should ask for login here
...
<强>更新
//for the final if statement
if (loggedin) { //just use this boolean variable since you used it as an indicator if it is valid name or not
System.out.println("Attend today: "+attendance[0]);
System.out.println("Grade: ");
}
else {
System.out.println("poo");
}
实际上，你不需要在循环之后放置条件，因为你已经在循环中过滤它，所以如果名称无效，它将不会退出循环，直到用户输入有效名称..你可以做在循环之后它就像这样
do{
..
}while(..)
System.out.println("Attend today: "+attendance[0]);
System.out.println("Grade: ");
答案 2 :(得分：0)
我从零开始，创造了这个，它使用相同的原则，但已经解决了我的答案。
有三个用户，每个用户都有自己的登录名，密码创建和密码输入（如果你有更好的方法，请说）。
package codeyom9a;
import java.util.Scanner;
public class Codeyom9a {
public static void main(String[] args) {
String[] names = {"Luke", "Jack", "Brad" };
String[] surnames = {"Mckeown", "Heard", "Reed" };
Scanner user_input = new Scanner(System.in);
boolean firstloggedin;
boolean passloggedin;
String firstlogin;
do { //login first name
firstloggedin = false;
System.out.print("Enter your first name: ");
firstlogin = user_input.next();
if (firstlogin.equals(names[0]) || firstlogin.equals(names[1]) || firstlogin.equals(names[2])) {
firstloggedin = true;
}
else {
System.out.println("Please try again.");
}
} while (! firstloggedin);
boolean secondloggedin;
String secondlogin;
do { //login surname
secondloggedin = false;
System.out.print("Enter your surname: ");
secondlogin = user_input.next();
if (secondlogin.equals(surnames[0]) & firstlogin.equals(names[0])|| secondlogin.equals(surnames[1]) & firstlogin.equals(names[1]) || secondlogin.equals(surnames[2]) & firstlogin.equals(names[2])) {
secondloggedin = true;
}
else {
System.out.println("Please try again.");
}
} while (! secondloggedin);
if (secondlogin.equals(surnames[0]) & firstlogin.equals(names[0])) { //pass login user 1
String password1; //pass create user 1
System.out.print("Create a password (no spaces): ");
password1 = user_input.next();
boolean passloggedin1 = false;
do{
String passwordenter1; //pass enter user 1
System.out.print("Enter your password now: ");
passwordenter1 = user_input.next();
if (passwordenter1.equals(password1)) {
passloggedin1 = true;
System.out.println("Correct! You have now logged in.");
}
else {
System.out.println("Incorrect password!");
}
} while (! passloggedin1);
} //end user 1
if (secondlogin.equals(surnames[1]) & firstlogin.equals(names[1])) { //pass login user 2
String password2; //pass create user 2
System.out.print("Create a password (no spaces): ");
password2 = user_input.next();
boolean passloggedin2 = false;
do{
String passwordenter2; //pass enter user 2
System.out.print("Enter your password now: ");
passwordenter2 = user_input.next();
if (passwordenter2.equals(password2)) {
passloggedin2 = true;
System.out.println("Correct! You have now logged in.");
}
else {
System.out.println("Incorrect password!");
}
} while (! passloggedin2);
} //end user 2
if (secondlogin.equals(surnames[2]) & firstlogin.equals(names[2])) { //pass login user 3
String password3; //pass create user 3
System.out.print("Create a password (no spaces): ");
password3 = user_input.next();
boolean passloggedin3 = false;
do{
String passwordenter3; //pass enter user 3
System.out.print("Enter your password now: ");
passwordenter3 = user_input.next();
if (passwordenter3.equals(password3)) {
passloggedin3 = true;
System.out.println("Correct! You have now logged in.");
}
else {
System.out.println("Incorrect password!");
}
} while (! passloggedin3);
} //end user 3
}
}
如果您输入＆＃34; Luke＆＃34;，＆＃34; Jack＆＃34;或＆＃34; Brad＆＃34;，则它会请求姓氏（在同一个索引中） ＆＃39;姓氏＆＃39;阵列）。如果两者都正确，则请求创建密码，然后要求用户输入创建的密码。
关于我的第一个代码，我不知道为什么这个有用而另一个没有，任何想法为什么？