-2

我想取第 1 行和第 5 行，它们的用户名和日期相同，但在第 1 行它包含时间，在第 5 行它包含时间我想读取这两行并比较它以检查两行是否具有相同的用户名和日期，如果是这样，在其他文本文件或哈希图中将其打印为单行

像这样的例子：（"sangeetha-May 02, 2013 , -in-09:48:06:61 -out-08:08:19:27在 JAVA 中）

这是文本文件的内容：

line 1. "sangeetha-May 02, 2013 , -in-09:48:06:61
line 2. "lohith-May 01, 2013 , -out-09:10:41:61
line 3 . "sushma-May 02, 2013 , -in-09:48:06:61
line 4. "sangeetha-May 01, 2013 , -out-08:36:38:50
line 5. "sangeetha-May 02, 2013 , -out-08:08:19:27
line 6. "sushma-May 02, 2013 , -out-07:52:13:51
line 7. "sangeetha-Jan 01, 2013 , -in-09:27:17:52-out-06:47:48:00
line 8. "madhusudhan-Jan 01, 2013 , -in-09:38:59:31-out-07:41:06:40

以上数据是使用下面的代码生成的

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.util.HashSet;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Set;
import java.util.Map;
import java.util.TreeMap;

public class FlatFileParser 
{
    public static void main(String[] args)
    {    
        // The stream we're reading from
        BufferedReader in;
        BufferedWriter out1;
         BufferedWriter out2;
        // Return value of next call to next()
        String nextline;
        try 
        {
            if (args[0].equals("1"))
            {
                in = new BufferedReader(new FileReader(args[1]));
                nextline = in.readLine();
                while(nextline != null)
                {
                    nextline = nextline.replaceAll("\\<packet","\n<packet");
                    System.out.println(nextline);
                    nextline = in.readLine();
                }
                in.close();
            }
            else
            {
                in = new BufferedReader(new FileReader(args[1]));
                out1 = new BufferedWriter(new FileWriter("inValues.txt" , true));
                 out2 = new BufferedWriter(new FileWriter("outValues.txt"));
                nextline = in.readLine();
                HashMap<String,String> inout = new HashMap<String,String>();
                while(nextline != null)
                {
                    try
                    {
                        if (nextline.indexOf("timetracker")>0)
                        {
                            String from = "";
                            String indate = "";
                            if (nextline.indexOf("of in")>0)
                            {

                                int posfrom = nextline.indexOf("from");
                                int posnextAt = nextline.indexOf("@", posfrom);
                                int posts = nextline.indexOf("timestamp");
                                from = nextline.substring(posfrom+5,posnextAt);
                                indate = nextline.substring(posts+11, posts+23);
                                String dd = indate.split(" ")[1];
                                String key = dd+"-"+from+"-"+indate;
                                //String key = from+"-"+indate;
                                String intime = "-in-"+nextline.substring(posts+24, posts+35);
                                inout.put(key, intime);    

                            }
                            else if (nextline.indexOf("of out")>0)
                            {
                                int posfrom = nextline.indexOf("from");
                                int posnextAt = nextline.indexOf("@", posfrom);
                                int posts = nextline.indexOf("timestamp");
                                from = nextline.substring(posfrom+5,posnextAt);
                                indate = nextline.substring(posts+11, posts+23);
                                String dd = indate.split(" ")[1];
                                String key = dd+"-"+from+"-"+indate;
                                String outtime = "-out-"+nextline.substring(posts+24, posts+35);
                                if (inout.containsKey(key))
                                {
                                    String val = inout.get(key);
                                    if (!(val.indexOf("out")>0))
                                        inout.put(key, val+outtime);                    
                                }
                                else
                                    inout.put(key, outtime);
                            }
                        }
                    }
                    catch(Exception e)
                    {
                        System.err.println(nextline);
                        System.err.println(e.getMessage());
                    }
                    nextline = in.readLine();    
                }
                in.close();

                for(String key: inout.keySet())
                {
                    String val = inout.get(key);
                    out1.write(key+" , "+val+"\n");
                    System.out.println(key + val);
                }
                out1.close();
            }
        } 
        catch (IOException e)
        {
            throw new IllegalArgumentException(e);
        }
    }
}

描述：这些是员工的登录和注销时间，我正在从日志文件中读取这些，但有些正确地出现在单行中，如第 7 行和第 8 行，有些在同一日期出现在不同的行中，我想要它像我上面提供的示例一样打印在同一行中，并且在输入和输出时间都以单行形式出现的记录应该保持原样.... PLZ 任何人都可以帮助....！

4

3 回答 3

1

考虑到您在lstFile.

你可以这样做

String output="",line1,line2;
for(int i=0;i<lstFile.size();i++)
{

    line1=lstFile.get(i);
    if(line1.contains("in") && line1.contains("out"))continue;
    for(int j=i+1;j<lstFile.size();j++)
    {
        line2=lstFile.get(j);

        if(line2.contains("in") && line2.contains("out"))continue;

        if(line1.contains(getNameDate(line2)) && line2.contains("out") && line1.contains("in"))
        {
              output+=line1+line2.substring(line2.lastIndexOf(","),line2.length());
              output+=System.getProperty("line.separator");
              break;
        }
    }
}
//output now contains your desired result

以下方法将获取名称和日期

public String getNameDate(String input)
{
    return input.substring(0,input.lastIndexOf(","));
}
于 2013-07-11T04:44:13.523 回答
0

这是基本的数据解析。在伪代码中，我会这样做

Create a class that holds 4 valus, Employee, Date, InTime, OutTime
Instantiate a HashMap for all the final log lines
For each log line
    Parse the line using RegEx to find Employee, Date and in and/or out time
    Create the HashKey using Employee + Date
    See if the HashMap already contains such an object, else create one
    Populate with the in and/or out times found on the current line
Done, the HashMap now contains all parsed data.
于 2013-07-11T04:24:57.583 回答
0

这里有一些示例代码可以帮助您入门。

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;

public class ParseLogs {

    public static void main(String[] args) {
        BufferedReader br = null;
        try {
            String line;

            br = new BufferedReader(new FileReader("src/main/resources/log.txt"));
            while ((line = br.readLine()) != null) {
                String[] split = line.split(" ");

                if (split.length > 2) {
                    String name = split[2].split("-")[0];
                    name = name.replace("\"", "");
                    System.out.println(name);
                }

                if (split.length > 5) {
                    String date = split[6];
                    System.out.println(date);
                }
            }

        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            try {
                if (br != null) {
                    br.close();
                }
            } catch (IOException ex) {
                ex.printStackTrace();
            }
        }
    }
}

另一篇文章是正确的，您需要使用正则表达式并查看字符串行的模式，以便将它们分解。如果可能，请尝试在最初对日志进行标准化，以使它们符合类似的模式，并且之后您无需处理数据。

这是程序的输出

sangeetha
-in-09:48:06:61
lohith
-out-09:10:41:61
.
,
sangeetha
-out-08:36:38:50
sangeetha
-out-08:08:19:27
sushma
-out-07:52:13:51
sangeetha
-in-09:27:17:52-out-06:47:48:00
madhusudhan
-in-09:38:59:31-out-07:41:06:40

请注意，您仍然需要清理“名称”和“日期”变量以查找错误，但希望这会有所帮助。

于 2013-07-11T04:40:02.707 回答