Question

我有多个字符串，格式如下：
12/18/2009 02:08:26在南大厅录入Doe，John（卡＃111）[In]

从这些字符串中我需要知道该人的日期，时间，姓名和卡号。录取的词可以省略，卡号的最后一位后面的任何内容都可以忽略我有一种感觉，我想为此使用StringTokenizer，但我并不积极。
有什么建议吗？

Answer 1

当你有一个共同的分隔符时，String Tokenizer很棒，但在这种情况下我会选择正则表达式。

Answer 2

我会选择java.util.Scanner ...这段代码会让你开始......你应该使用扫描仪方法的Pattern形式，而不是我使用的String形式。

import java.util.Scanner;

public class Main
{
    public static void main(String[] args)
        throws Exception
    {
        final String  str;
        final Scanner scanner;
        final String  date;
        final String  time;
        final String  word;
        final String  lastName;
        final String  firstName;

        str       = "12/18/2009 02:08:26 Admitted Doe, John (Card #111) at South Lobby [In]";
        scanner   = new Scanner(str);
        date      = scanner.next("\\d+/\\d+/\\d+");
        time      = scanner.next("\\d+:\\d+:\\d+");
        word      = scanner.next();
        lastName  = scanner.next();
        firstName = scanner.next();
        System.out.println("date : " + date);
        System.out.println("time : " + time);
        System.out.println("word : " + word);
        System.out.println("last : " + lastName);
        System.out.println("first: " + firstName);
    }
}

Answer 3

您的记录格式非常简单，我只需使用String的split方法来获取日期和时间。正如评论中所指出的那样，拥有可以包含空格的名称会使事情变得复杂，以至于按空格分割记录将不适用于每个字段。我使用正则表达式来获取其他三条信息。

public static void main(String[] args) {
    String record1 = "12/18/2009 02:08:26 Admitted Doe, John (Card #111) at South Lobby [In]";
    String record2 = "12/18/2009 02:08:26 Admitted Van Halen, Eddie (Card #222) at South Lobby [In]";
    String record3 = "12/18/2009 02:08:26 Admitted Thoreau, Henry David (Card #333) at South Lobby [In]";

    summary(record1);
    summary(record2);
    summary(record3);
}

public static void summary(String record) {
    String[] tokens = record.split(" ");

    String date = tokens[0];
    String time = tokens[1];

    String regEx = "Admitted (.*), (.*) \\(Card #(.*)\\)";
    Pattern pattern = Pattern.compile(regEx);
    Matcher matcher = pattern.matcher(record);
    matcher.find();

    String lastName = matcher.group(1);
    String firstName = matcher.group(2);
    String cardNumber = matcher.group(3);

    System.out.println("\nDate: " + date);
    System.out.println("Time: " + time);
    System.out.println("First Name: " + firstName);
    System.out.println("Last Name: " + lastName);
    System.out.println("Card Number: " + cardNumber);
}

正则表达式"Admitted (.*), (.*) \\(Card #(.*)\\)"使用分组括号来存储您尝试提取的信息。必须转义记录中存在的括号。

运行上面的代码给出了以下输出：

Date: 12/18/2009
Time: 02:08:26
First Name: John
Last Name: Doe
Card Number: 111

Date: 12/18/2009
Time: 02:08:26
First Name: Eddie
Last Name: Van Halen
Card Number: 222

Date: 12/18/2009
Time: 02:08:26
First Name: Henry David
Last Name: Thoreau
Card Number: 333

Answer 4

在解析此行时要记住以下几点：

姓氏可以包含空格，因此您应该寻找，
名字可以有空格，所以请查找（

由于这个原因，我会使用TofuBeer的答案并调整下一个名字。由于额外的空间，字符串拆分会变得混乱。

Answer 5

最短的正则表达式解决方案（使用类型转换）：

String stringToParse = "12/18/2009 02:08:26 Admitted Doe, John (Card #111) at South Lobby [In] ";
Pattern pattern = Pattern.compile("((\\d{2}/){2}\\d{4}\\s(\\d{2}:){2}\\d{2})\\s(\\w+)\\s((.*)),\\s((.*))\\s.*#(\\d+)");
Matcher matcher = pattern.matcher(stringToParse);
matcher.find();

String firstName = matcher.group(6);
String lastName = matcher.group(5);
int cardNumber = Integer.parseInt(matcher.group(7));

DateFormat df = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss");
Date date = df.parse(matcher.group(1));

Answer 6

相信你的勇气...... :) 使用StringTokenizer：

import java.io.*;
import java.util.StringTokenizer;
public class Test {
  public Test() {
  }

public void execute(String str) {
    String date, time, firstName, lastName, cardNo;
    StringTokenizer st = new StringTokenizer(str, " ");
    date = st.nextToken();
    time = st.nextToken();
    st.nextToken(); //Admitted
    lastName = st.nextToken(",").trim();
    firstName = st.nextToken(",(").trim();
    st.nextToken("#"); //Card
    cardNo = st.nextToken(")#");
    System.out.println("date = " + date +"\ntime = " + time +"\nfirstName = " + firstName +"\nlastName = "+ lastName +"\ncardNo = " +cardNo);
  }

public static void main(String args[]) {
    Test t = new Test();
    String record1 = "12/18/2009 02:08:26 Admitted Doe, John (Card #111) at South Lobby [In]";
    String record2 = "12/18/2009 02:08:26 Admitted Van Halen, Eddie (Card #222) at South Lobby [In]";
    String record3 = "12/18/2009 02:08:26 Admitted Thoreau, Henry David (Card #333) at South Lobby [In]";
    t.execute(record1);
    t.execute(record2);
    t.execute(record3);
  }
}

在Java中打破一个字符串

6 个答案: