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[英]Accept empty string or string with length of 0 as input using Scanner in Java
[英]Using a scanner to accept String input and storing in a String Array
有誰可以幫助我嗎。 我做了很多搜索,但無法在任何地方找到解決方案。 我是Java的初學者,目前在大學休息期間練習一些代碼。
我正在嘗試制作電話簿程序。 目前我正在嘗試添加一個新的聯系人,下面是我的代碼,但我不知道如何將數據存儲在一個數組中,有人可以給我一些指示。
import java.util.Scanner;
public class addContact {
public static void main(String [] args){
//declare arrays
String [] contactName = new String [12];
String [] contactPhone = new String [12];
String [] contactAdd1 = new String [12];
String [] contactAdd2 = new String [12];
//inputs
String name = "";
String phone = "";
String add1 = "";
String add2 = "";
//method of taken input
Scanner input = new Scanner(System.in);
//while name field is empty display prompt etc.
while (name.equals(""))
{
System.out.println("Enter contacts name: ");
name = input.nextLine();
name += contactName[];
}
while (add1.equals(""))
{
System.out.println("Enter contacts addressline1:");
add1 = input.nextLine();
add1 += contactAdd1[];
}
while (add2.equals(""))
{
System.out.println("Enter contacts addressline2:");
add2 = input.nextLine();
add2 += contactAdd2[];
}
while (phone.equals(""))
{
System.out.println("Enter contact phone number: ");
phone = input.nextLine();
phone += contactPhone[];
}
}
}
一個清潔的方法是創建一個Person
,其中包含對象contactName
, contactPhone
等。然后,使用一個ArrayList
而不是一個數組中添加新的對象。 創建一個接受每個`Person的所有字段的循環:
while (!done) {
Person person = new Person();
String name = input.nextLine();
person.setContactName(name);
...
myPersonList.add(person);
}
使用該列表將不再需要數組邊界檢查。
這段代碼的一個問題是:
name += contactName[];
該指令不會在數組中插入任何內容。 相反,它將連接變量名稱的當前值與contactName數組的字符串表示形式。
而是使用這個:
contactName[index] = name;
此指令將變量名稱存儲在索引index
的contactName數組中。
你遇到的第二個問題是你沒有變量index
。
你可以做的是一個循環,有12次迭代來填充你的所有數組。 ( index
將是你的迭代變量)
//go through this code I have made several changes in it//
import java.util.Scanner;
public class addContact {
public static void main(String [] args){
//declare arrays
String [] contactName = new String [12];
String [] contactPhone = new String [12];
String [] contactAdd1 = new String [12];
String [] contactAdd2 = new String [12];
int i=0;
String name = "0";
String phone = "0";
String add1 = "0";
String add2 = "0";
//method of taken input
Scanner input = new Scanner(System.in);
//while name field is empty display prompt etc.
while (i<11)
{
i++;
System.out.println("Enter contacts name: "+ i);
name = input.nextLine();
name += contactName[i];
}
while (i<12)
{
i++;
System.out.println("Enter contacts addressline1:");
add1 = input.nextLine();
add1 += contactAdd1[i];
}
while (i<12)
{
i++;
System.out.println("Enter contacts addressline2:");
add2 = input.nextLine();
add2 += contactAdd2[i];
}
while (i<12)
{
i++;
System.out.println("Enter contact phone number: ");
phone = input.nextLine();
phone += contactPhone[i];
}
}
}
這會更好嗎?
import java.util.Scanner;
public class Work {
public static void main(String[] args){
System.out.println("Please enter the following information");
String name = "0";
String num = "0";
String address = "0";
int i = 0;
Scanner input = new Scanner(System.in);
//The Arrays
String [] contactName = new String [7];
String [] contactNum = new String [7];
String [] contactAdd = new String [7];
//I set these as the Array titles
contactName[0] = "Name";
contactNum[0] = "Phone Number";
contactAdd[0] = "Address";
//This asks for the information and builds an Array for each
//i -= i resets i back to 0 so the arrays are not 7,14,21+
while (i < 6){
i++;
System.out.println("Enter contact name." + i);
name = input.nextLine();
contactName[i] = name;
}
i -= i;
while (i < 6){
i++;
System.out.println("Enter contact number." + i);
num = input.nextLine();
contactNum[i] = num;
}
i -= i;
while (i < 6){
i++;
System.out.println("Enter contact address." + i);
num = input.nextLine();
contactAdd[i] = num;
}
//Now lets print out the Arrays
i -= i;
while(i < 6){
i++;
System.out.print( i + " " + contactName[i] + " / " );
}
//These are set to print the array on one line so println will skip a line
System.out.println();
i -= i;
i -= 1;
while(i < 6){
i++;
System.out.print( i + " " + contactNum[i] + " / " );
}
System.out.println();
i -= i;
i -= 1;
while(i < 6){
i++;
System.out.print( i + " " + contactAdd[i] + " / " );
}
System.out.println();
System.out.println("End of program");
}
}
如果我錯了,請糾正我
public static void main(String[] args) {
Scanner na = new Scanner(System.in);
System.out.println("Please enter the number of contacts: ");
int num = na.nextInt();
String[] contactName = new String[num];
String[] contactPhone = new String[num];
String[] contactAdd1 = new String[num];
String[] contactAdd2 = new String[num];
Scanner input = new Scanner(System.in);
for (int i = 0; i < num; i++) {
System.out.println("Enter contacts name: " + (i+1));
contactName[i] = input.nextLine();
System.out.println("Enter contacts addressline1: " + (i+1));
contactAdd1[i] = input.nextLine();
System.out.println("Enter contacts addressline2: " + (i+1));
contactAdd2[i] = input.nextLine();
System.out.println("Enter contact phone number: " + (i+1));
contactPhone[i] = input.nextLine();
}
for (int i = 0; i < num; i++) {
System.out.println("Contact Name No." + (i+1) + " is "+contactName[i]);
System.out.println("First Contacts Address No." + (i+1) + " is "+contactAdd1[i]);
System.out.println("Second Contacts Address No." + (i+1) + " is "+contactAdd2[i]);
System.out.println("Contact Phone Number No." + (i+1) + " is "+contactPhone[i]);
}
}
`
到目前為止,java中沒有使用指針。 您可以從類中創建一個對象,並使用彼此鏈接的不同類,並使用主類中每個類的函數。
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