我将解释逻辑：我正在读取一个xml文件，其中包含许多soap格式的请求和响应，然后我将请求和响应存储在两个哈希Map中。在第一个哈希Map中，我将事务id（unique）存储为键，将值存储为请求时间til name。在第二个哈希Map中，我将事务id（unique）存储为键，将值存储为响应时间。在两个hashMap中，键是相同的，但是值是不同的，通过使用for循环迭代两个循环，我需要得到响应时间和请求时间之间的时间差，例如：requesttime:2020-01-30t11：07:08.351z和响应time:2020-01-30t11：07:10.152赫兹
public class MapTimeDiff {
public static void main(String[] args) throws ParseException {
File file =new File("C:\\Users\\gsanaulla\\Documents\\My Received Files\\ecarewsframework.xml");
Scanner in = null;
String tilname = null;
String transactionId = null;
String requesttime = null;
String responsetime = null;
Date dateOne = null;
Date dateTwo = null;
double timeDiff;
DateFormat df = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSS");
Map<String,ArrayList<String>> request=new HashMap<String,ArrayList<String>>();
ArrayList<String> req=new ArrayList<String>();
Map<String,ArrayList<String>> response=new HashMap<String,ArrayList<String>>();
ArrayList<String> res=new ArrayList<String>();
try {
in = new Scanner(file);
while(in.hasNext())
{
String line=in.nextLine();
if(line.contains("</S:Envelope>")) {
System.out.println(line);
tilname=line.split("StartRecord><")[1].split("><")[0].split(":")[1];
System.out.println("tilname :: "+tilname);
transactionId = line.split("transactionId>")[1].split("<")[0];
System.out.println("transactio id :: "+transactionId);
requesttime=line.split("sourceTimestamp>")[1].split("<")[0];
System.out.println("request time is :: "+requesttime);
dateOne = df.parse(requesttime);
}
req.add(tilname);
req.add(dateOne.toString());
System.out.println("req is==== " +req);
request.put(transactionId,req);
System.out.println("request is==== " +request.get(transactionId));
if(line.contains("</SOAP-ENV:Envelope>")) {
//System.out.println(line);
if(line.contains("transactionId"))
{
responsetime=line.split("sourceTimestamp>")[1].split("<")[0];
transactionId = line.split("transactionId>")[1].split("<")[0];
System.out.println("responsetime :: "+responsetime);
System.out.println("transaction id "+transactionId);
dateTwo = df.parse(responsetime);
}
res.add(dateTwo.toString());
System.out.println("res is===== "+res);
response.put(transactionId,res);
System.out.println("response is===== "+response.get(transactionId));
for (Entry<String, ArrayList<String>> entry : request.entrySet()) {
for (Entry<String, ArrayList<String>> entry1 : response.entrySet()) {
System.out.println("Key = " + entry.getKey() +
", Value = " + entry.getValue());
System.out.println("Key = " + entry1.getKey() +
", Value = " + entry1.getValue());
if(request.keySet().equals(response.keySet())) {
timeDiff = (dateTwo.getTime() - dateOne.getTime());
}
}
}
}
}
}
catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
1条答案
按热度按时间06odsfpq1#
我不确定我是否正确理解了你的问题，但也许你可以做如下类似的事情：
如您所见，transactionid 3没有响应，因此将忽略此项。如果键的列表中的元素大小不同（transactionid 2），多余的元素也将被忽略。