使用 .next() 或 .nextLine() 的 java 字符串变量
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java string variable using .next() or .nextLine()
提问by wesleylim1993
the following is my sources code:
以下是我的源代码：
package functiontest;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;
import java.util.Scanner;
public class FunctionTest {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int option;
String cname, cpassword="testpassword",chp="010-000000";
// cname="clerk test";
System.out.println("1. add");
System.out.println("2. delete");
option = scan.nextInt();
switch(option)
{
case 1:
System.out.print("Enter clerk name:\t");
cname = scan.nextLine();
File cfile = new File("clerk/"+cname+".txt");
FileWriter cwrite;
try
{
cwrite = new FileWriter(cfile);
BufferedWriter cbuf = new BufferedWriter(cwrite);
cbuf.write("clerk name:\t" +cname);
cbuf.write("clerk password:\t"+cpassword);
cbuf.write("clerk handphone number:\t"+chp);
cbuf.flush();
cbuf.close();
System.out.println("The clerk profile create completely");
}catch(IOException e)
{
System.out.println("Error! clerk profile cannot create");
}
;break;
case 2:
String dclerk;
// dclerk = "clerk test";
System.out.print("\nPlease enter clerk name for delete:\t");
dclerk = scan.next();
File dcfile = new File("clerk/"+dclerk+".txt");
if(!dcfile.exists())
{
System.out.println("Error! the clerk profile not exist");
}
try
{
dcfile.delete();
System.out.println(dclerk + "'s prifle successful delete");
}catch(Exception e)
{
System.out.println("Something wrong! " + dclerk +" profile cannot delete");
};break;
}
}
}
i cannot the enter the variable name cname
我不能输入变量名 cname
cname = scan.nextLine()
when i run the program it show the result as below:
当我运行程序时，它显示的结果如下：
run:
1. add
2. delete
1
Enter clerk name: The clerk profile create completely
when i use .next():
当我使用 .next():
cname = scan.next()
it cannot read the
它无法读取
cname
with spaces, example
带空格，示例
clerk test
it will read
它会读
clerk
only how should i do?
只是我该怎么办？
回答by Dawood ibn Kareem
Your problem is that the line
你的问题是这条线
option = scan.nextInt();
does not read the new line character after the integer. So that new line is finally read when you do nextLine(), later on.
不读取整数后的换行符。因此，当您这样做时nextLine()，最终会读取该新行，稍后。
To fix this, you should add an extra
要解决此问题，您应该添加一个额外的
scan.nextLine()
after the call to nextInt(), then use
调用后nextInt()，然后使用
cname = scan.nextLine();
when you want to read the clerk's name.
当你想读店员的名字时。
