[英]How to determine how many characters are in a string and if the characters are digits or not Java
I am doing a customer information scanner project for school, currently I am working on the phone number portion.我正在为学校做一个客户信息扫描仪项目,目前我正在处理电话号码部分。 I need to have the customer enter their phone number and I need to detect as many errors as possible.我需要让客户输入他们的电话号码,并且我需要检测尽可能多的错误。 This includes: make sure it's all digits, make sure its ten digits, then to divide into - <3 digit String for Area Code> <3 digit String for Central Office > <4 digit String for Station Code> and finally add dashes between the area code, central office, and station code.这包括:确保都是数字,确保它的十位数字,然后分为 - <区号的3位字符串> <中心局的3位字符串> <站点代码的4位字符串>最后在之间添加破折号区号、中心局和车站代码。 Are there any easy ways to do this, here is my current code.有什么简单的方法可以做到这一点,这是我当前的代码。 Thanks in advance提前致谢
System.out.println(DATA_DIV);
System.out.println("\nCustomer Phone Number Information");
System.out.println("-----------------------------------\n");
System.out.println("Enter the Phone Number:");
phoneNumber = uIn.next();
if (phoneNumber.matches("\\d+")) {
}
else {
Garbage = uIn.next();
System.out.println("\n\tError Data Type: you entered ( " + Garbage + " ) for Phone Number");
System.out.println("Phone Number must be made up of numbers only");
System.out.println();
System.out.println("Re-Enter the Phone Number :");
phoneNumber = uIn.next();
}
I would suggest you get familiar with regular expressions as they come really handy in situations like this one.我建议您熟悉正则表达式,因为它们在这种情况下非常方便。 The regex I would suggest you use: ^(?\d{3})(?\d{3})(?\d{4})$我建议您使用的正则表达式: ^(?\d{3})(?\d{3})(?\d{4})$
^ - means start of the sequence ^ - 表示序列的开始
(?<area/office/code>) - that's how we basically create a group and give it a name so we can call it by it later when we need it. (?<area/office/code>) - 这就是我们基本上创建一个组并为其命名的方式,以便我们以后需要时可以通过它调用它。
\d - digit \d - 数字
{3} - the exact number of the preceding element we're looking for {3} - 我们正在寻找的前一个元素的确切编号
$ - end of sequence $ - 序列结束
After that you should use the Pattern and Matcher classes in order to get the right sequences, there are handy methods like find(groupName) in the Matcher class.之后,您应该使用 Pattern 和 Matcher 类来获得正确的序列,在 Matcher class 中有一些方便的方法,例如 find(groupName)。 I think splitting is quite straightforward after this.我认为在此之后拆分非常简单。
Personally, I don't like using regex to solve this kind of problem.就个人而言,我不喜欢使用正则表达式来解决这类问题。 The alternative to using regex as mentioned in the comments is to build a scanner and parser from a Backus-Naur description for a phone number.如评论中提到的,使用正则表达式的替代方法是从电话号码的 Backus-Naur 描述构建扫描仪和解析器。 The description appears below:描述如下:
<phone number> ::= <area code>-<central office>-<station code>
<area code> ::= <digit><digit><digit>
<central office> ::= <digit><digit><digit>
<station code> ::= <digit><digit><digit><digit>
<digit> ::= 0|1|2|3|4|5|6|7|8|9
The scanner and parser is as follows:扫描器和解析器如下:
import java.util.Scanner;
public class PhoneNumberParser {
PhoneNumber number;
int look; // Current char read
int index; // Index into input buffer
String input; // Raw phone number input
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
PhoneNumberParser parser = new PhoneNumberParser();
try {
PhoneNumber number = parser.parse(scanner.nextLine());
System.out.println(number);
}catch(IllegalArgumentException ex) {
System.err.println(ex);
}
}
public PhoneNumber parse(String s) throws IllegalArgumentException{
number = new PhoneNumber();
input = s;
index = 0;
//if( s.length() != 10 )
// throw new IllegalArgumentException("Bad phone number. Expected a ten digit number.");
scan();
number.areaCode = areaCode();
scan();
if( look != '-')
throw new IllegalArgumentException("Bad phone number. Expected a '-' after the area code.");
scan();
number.centralOffice = centralOffice();
scan();
if( look != '-')
throw new IllegalArgumentException("Bad phone number. Expected a '-' after the central office part.");
scan();
number.stationCode = stationCode();
return number;
}
private void scan() {
try {
look = input.charAt(index++);
}catch(Exception ex) {
look = -1;
}
}
private int areaCode() {
StringBuffer sb = new StringBuffer();
sb.append(digit());
scan();
sb.append(digit());
scan();
sb.append(digit());
return Integer.parseInt(sb.toString());
}
private int centralOffice() {
StringBuffer sb = new StringBuffer();
sb.append(digit());
scan();
sb.append(digit());
scan();
sb.append(digit());
return Integer.parseInt(sb.toString());
}
private int stationCode() {
StringBuffer sb = new StringBuffer();
sb.append(digit());
scan();
sb.append(digit());
scan();
sb.append(digit());
scan();
sb.append(digit());
return Integer.parseInt(sb.toString());
}
private int digit() {
if( ! Character.isDigit((char) look) )
throw new IllegalArgumentException("Bad phone number. Expected a digit, but found a '" + (char)look + "' character.");
int digit = (char)look - '0';
return digit;
}
class PhoneNumber {
int areaCode;
int centralOffice;
int stationCode;
public String toString() {
return "Area code: " + areaCode + " Central office: " + centralOffice + " Station code: " + stationCode;
}
}
}
Output when supplied with the string 999-888-1234 is: Output 与字符串 999-888-1234 一起提供时是:
Area code: 999 Central office: 888 Station code: 1234
As you can see, the scanner parser gives better error reporting for bad numbers than regexp and also separates out the number into its constituent parts for further processing.如您所见,扫描仪解析器对错误数字的错误报告比正则表达式更好,并且还将数字分成其组成部分以供进一步处理。 For example, it may be important to just display the area code portion of the number.例如,仅显示号码的区号部分可能很重要。
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