我有一个简单的程序:在java中爪哇 - 字符串中的许多比较txt文件
- 解析txt文件
- 在Unix时间戳
- 转换数据保留值只有一行每个第二
下面是代码。 的txt文件都是这种格式:
21/03/2013 4时18分23秒6890 6830 6850 6770 6830 6400 6630 6710 6770 6850 35024 34976 21/03/2013 4时18分23秒6910 6800 6850 6770 6820 6410 6590 6710 6780 6820 35056 34976 21/03/2013 04:18:24 6890 6820 6860 6770 6830 6400 6580 6720 6770 6830 34912 34880 21/03/2013 04:18:24 6860 6840 6840 6770 6830 6390 6660 6700 6740 6890 35008 34880
我的程序转换的代码是这样的:
放sensor.rat.128 1364278801 7100传感器= A
放sensor.rat.128 1364278801 6910传感器= B
放sensor.rat.128 1364278801 6890传感器= C
放传感器.rat.128 1364278801 6630传感器= d
该方案工作得非常好与txt文件,在这个意义上,它使只有一个值的线,用于每个第二,但如果在不同相同的第二值txt文件,它无法识别打他们。
所以问题是:我该如何让代码保存每秒钟和跨多个文件的值的唯一一个列表? 我希望你们所有人都明白。
import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
import java.io.*;
import java.text.SimpleDateFormat;
import java.util.Date;
import org.apache.commons.io.FileUtils;
public class Downsampler {
public static void main(String[] args) throws Exception{
/*
* Scans all the files in a specified folder
* Obtains Cell number from the file name
*/
String path = "/home/alessandro/Data128prova"; // name of path
File folder = new File(path);
for (File file : folder.listFiles()) {
Scanner s = new Scanner(file);
ArrayList<String> list = new ArrayList<String>();
while (s.hasNext()){
list.add(s.next());
}
s.close();
//Arraylist to save modified values
ArrayList<String> ds = new ArrayList<String>();
int i;
String app = "";
for(i=0; i<=list.size()-13; i=i+14){
//combining the first to values to obtain data
String str = list.get(i)+" "+list.get(i+1);
//------convert data in epoch time
Date dt= new java.text.SimpleDateFormat("dd/MM/yyyy HH:mm:ss").parse(str);
long epochlong = dt.getTime()/1000;
String epoch = Long.toString(epochlong);
//------end conversion data
if (!str.equalsIgnoreCase(app)){
//add all the other values to arraylist ds
ds.add(epoch);
int j;
for(j=1; j<14; j++){
ds.add(list.get(i+j));
}
}
app = str;
}
int k;
String metric = "sensor.rat.128.riprova"; //name of the metric
for (k=0; k<=ds.size()-13; k=k+14){
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+2)+" sensor=A");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+3)+" sensor=B");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+4)+" sensor=C");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+5)+" sensor=D");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+6)+" sensor=E");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+7)+" sensor=F");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+8)+" sensor=G");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+9)+" sensor=H");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+10)+" sensor=I");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+11)+" sensor=L");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+12)+" sensor=M");
System.out.println ("put "+metric+" "+ds.get(k)+" "+ds.get(k+13)+" sensor=N");
}
} //end of for
}
}
问号在哪里? –
@Heuster,我编辑了这个问题,谢谢 – alessandrob
您的数据是否按时间顺序写入文件文件,以便文件中较早的值永远不会写入相同文件中后面的值之后? – lreeder