我想把第1行和第5行的用户名和日期相同但在第一行包含它的时间，在第5行它包含时间 我想阅读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，则将其作为单行打印在其他文本文件或哈希映射中
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
以上数据通过以下代码生成以上数据
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内容，但有些内容正在单行中正确显示，如第7行和第8行 有些是同一个日期的不同行，我希望它像我上面提供的例子一样在同一行打印， 无论是进出时间都记录在单行中的记录应该保留它...... PLZ可以帮助任何人......！
答案 0 :(得分：1)
考虑到您有lstFile中文件中所有行的列表。
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(","));
}
答案 1 :(得分：0)
这是基本的数据解析。在伪代码中，我就是这样做的
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.
答案 2 :(得分：0)
以下是一些示例代码，可帮助您入门。
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，以便您可以将它们分解。如果可能的话，尝试最初标准化日志，使它们符合类似的模式，之后不需要按摩数据。
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，您仍然需要清除“名称”和“日期”变量以查找错误，但希望这会有所帮助。