无法在循环中输入第一个字符串，在java中它总是从第二个输入开始

我创建了一个Employee类，在该类中我只传递值并打印数组。你可以看到：

package practicequestion;

import java.util.Scanner;
import practicequestion.employeeq.Employee;

public class EmployeeMain {
    public static void main(String[] args) {
        Employee[] obj = new Employee[3];
        Scanner sc = new Scanner(System.in);
        System.out.println("Enter the number of inputs you want ");
        int n = sc.nextInt();
        for (int i = 0; i < n; i++) {

            System.out.println("Enter FirstName");
            String firstname = sc.nextLine();
            System.out.println("Enter LastName");
            String lastname = sc.nextLine();
            System.out.println("Enter Email");
            String email = sc.nextLine();
            System.out.println("Enter Salary");
            int salary = sc.nextInt();
            System.out.println("Enter Mobile Number");
            long mobileno = sc.nextLong();
            System.out.println("Enter Experience");
            int empexp = sc.nextInt();
            System.out.println("Enter Employee Id");
            int empid = sc.nextInt();
            if (i < n) {
                obj[i] = new Employee(firstname, lastname, email, salary, empid, empexp, mobileno);
            }
        }
        for (int i = 0; i < 3; i++) {
            System.out.println("Employee " + i +  " values are: ");
            obj[i].Display();
        }
        if (sc != null) {
            sc.close();   
        }
    }

}

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有人能解决这个问题吗？

这里已经回答了这个问题：

基本上，问题在于Scanner.nextInt（）在进入循环之前调用它。 Scanner.nextLine（）只读取用户输入的下一个整数，而不是整行，因此当您在直接接受输入整数的行的剩余部分（和空部分）后调用Scanner.nextLine（）时

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要解决此问题，您必须调用Scanner.nextLine（）一次，或者使用Scanner.nextLine（）获取整数，然后将其强制转换为整数。

尝试此方法。我将记录我在哪里做的更改

Scanner sc = new Scanner(System.in);
System.out.println("Enter the number of inputs you want ");
int n = sc.nextInt();
// allocate the Employee array after you get the number of inputs.
Employee[] obj = new Employee[n];

for (int i = 0; i < n; i++) {

    // get the newline out of the buffer here. That also
    // takes care of the `nextInt` problem at the end of this loop
    sc.nextLine();
    
    System.out.println("Enter FirstName");
    String firstname = sc.nextLine();
    System.out.println("Enter LastName");
    String lastname = sc.nextLine();
    System.out.println("Enter Email");
    String email = sc.nextLine();
    System.out.println("Enter Salary");
    int salary = sc.nextInt();
    System.out.println("Enter Mobile Number");
    long mobileno = sc.nextLong();
    System.out.println("Enter Experience");
    int empexp = sc.nextInt();
    System.out.println("Enter Employee Id");
    int empid = sc.nextInt();
    // you don't need to verify `i` here.  The for loop does that
    // for you
    obj[i] = new Employee(firstname, lastname, email, salary,
            empid, empexp, mobileno);
    
}

// use the number of inputs `n` to limit the loop
for (int i = 0; i < n; i++) {
    System.out.println("Employee " + i + " values are: ");
    obj[i].Display();
}

Scanner sc=新扫描仪（System.in）；
System.out.println（“输入所需的输入数量”）；
int n=sc.nextInt（）；
//在获得输入数后分配Employee数组。
员工[]obj=新员工[n]；
对于（int i=0；i

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关闭扫描仪通常不是一个好的做法。

谢谢你的建议。现在我也可以输入，但无法打印名字。请指导我怎么做？

Scanner sc = new Scanner(System.in);
System.out.println("Enter the number of inputs you want ");
int n = sc.nextInt();
// allocate the Employee array after you get the number of inputs.
Employee[] obj = new Employee[n];

for (int i = 0; i < n; i++) {

    // get the newline out of the buffer here. That also
    // takes care of the `nextInt` problem at the end of this loop
    sc.nextLine();
    
    System.out.println("Enter FirstName");
    String firstname = sc.nextLine();
    System.out.println("Enter LastName");
    String lastname = sc.nextLine();
    System.out.println("Enter Email");
    String email = sc.nextLine();
    System.out.println("Enter Salary");
    int salary = sc.nextInt();
    System.out.println("Enter Mobile Number");
    long mobileno = sc.nextLong();
    System.out.println("Enter Experience");
    int empexp = sc.nextInt();
    System.out.println("Enter Employee Id");
    int empid = sc.nextInt();
    // you don't need to verify `i` here.  The for loop does that
    // for you
    obj[i] = new Employee(firstname, lastname, email, salary,
            empid, empexp, mobileno);
    
}

// use the number of inputs `n` to limit the loop
for (int i = 0; i < n; i++) {
    System.out.println("Employee " + i + " values are: ");
    obj[i].Display();
}