Java 如何检查字符串输入的格式
我不确定我是否在错误的区域检查,或者我要检查的代码是否错误。我正在尝试检查999-A格式的员工Id是否有效。如有任何帮助,将不胜感激Java 如何检查字符串输入的格式,java,string,Java,String,我不确定我是否在错误的区域检查,或者我要检查的代码是否错误。我正在尝试检查999-A格式的员工Id是否有效。如有任何帮助,将不胜感激 public class Employee { private String name; private String employeeNumber; private String hireDate; // Constructor public Employee(String name, String employeeNumber,
public class Employee {
private String name;
private String employeeNumber;
private String hireDate;
// Constructor
public Employee(String name, String employeeNumber, String hireDate){
this.name = name;
if(isValidEmployeeNumber(employeeNumber) == true){
this.employeeNumber = employeeNumber;
} else{
}
this.hireDate = hireDate;
}
// Getters and Setters
public String getName(){
return name;
}
public String getEmployeeNumber(){
return employeeNumber;
}
public String getHireDate(){
return hireDate;
}
public void setName(String name){
this.name = name;
}
public void setEmployeeNumber(String employeeNumber){
this.employeeNumber = employeeNumber;
}
public void setHireDate(String hireDate){
this.hireDate = hireDate;
}
//999-X
private boolean isValidEmployeeNumber(String employeeNumber){
if(Character.getNumericValue(employeeNumber.charAt(4)) > 10 &
Character.getNumericValue(employeeNumber.charAt(4)) < 22)
{
/*
int cutNum = Integer.parseInt(employeeNumber.substring(0,1));
if(cutNum <= 0 && cutNum <= 9){
cutNum = Integer.parseInt(employeeNumber.substring(1, 2));
if(cutNum <= 0 && cutNum <= 9){
cutNum = Integer.parseInt(employeeNumber.substring(2, 3));
if(cutNum <= 0 && cutNum <= 9){
setEmployeeNumber(employeeNumber);
return true;
}
}
}*/
return true;
}
return false;
}
}
公共类员工{
私有字符串名称;
私有字符串employeeNumber;
私人租赁;
//建造师
公共雇员(字符串名称、字符串employeeNumber、字符串hireDate){
this.name=名称;
如果(isValidEmployeeNumber(employeeNumber)==true){
this.employeeNumber=employeeNumber;
}否则{
}
this.hireDate=hireDate;
}
//接球手和接球手
公共字符串getName(){
返回名称;
}
公共字符串getEmployeeNumber(){
返回员工编号;
}
公共字符串getHireDate(){
回租;
}
公共void集合名(字符串名){
this.name=名称;
}
public void setEmployeeNumber(字符串employeeNumber){
this.employeeNumber=employeeNumber;
}
public void setHireDate(字符串hireDate){
this.hireDate=hireDate;
}
//999-X
私有布尔值isValidEmployeeNumber(字符串employeeNumber){
如果(Character.getNumericValue(employeeNumber.charAt(4))>10&
Character.getNumericValue(employeeNumber.charAt(4))<22)
{
/*
int-cutNum=Integer.parseInt(employeeNumber.substring(0,1));
if(cutNum\d{1}只是表示\d的一种更复杂的方式。如果070-N有效,但70-N无效
"\d{3}-\D"
可能就是你要找的,或者
"\d\d\d-\D"
但是\D匹配每个非数字,因此它也会匹配“097-?”。在Java源代码中,您还需要屏蔽反斜杠:
"097-X".matches ("\\d\\d\\d-[A-Z]")
注意,中间的减去不需要掩蔽,只有在集合中,如果要从字面上理解:
"097-X".matches (".*[,\\-/].")
在许多语言中,可以通过将其放在第一位或最后来避免:
"097-X".matches (".*[-,/].")
"097-X".matches (".*[,/-].")
正则表达式非常适合此任务:
private boolean isValidEmployeeNumber(String employeeNumber) {
return employeeNumber.matches("[0-9]{3}-[A-Z]")
}
要分解正则表达式,请执行以下操作:
[0-9]{3}
一个数字(0-9)3次
-
仪表板
[A-Z]
A-Z一次
其他任何东西都是无效的。注意\D
对任何非数字都是有效的,我认为这不是您想要的