Develop Reference

How can i split on a string

I have a .txt file that I browse through a bufferReader and I need to extract the last character from this String, I leave the line below
<path
action="m"
text-mod="true"
mods="true"
kind="file">branches/RO/2021Align01/CO/DGSIG-DAO/src/main/java/eu/ca/co/vo/CsoorspeWsVo.java</path>
I have the following code that takes my entire line and sets it in a list, but I just need it Cs00rspeWsVo
while ((line = bufferdReader.readLine()) != null) {
Excel4 excel4 = new Excel4();
if (line.contains("</path>")) {
int index1 = line.indexOf(">");
int index2 = line.lastIndexOf("<");
line = line.substring(index1, index2);
excel4.setName(line);
listExcel4.add(excel4);
}
}
and I only want to extract Cs00rspeWsVo from here.
can anyone help me? thanks

You can use Regex groups to get it for example
public static void main(String []args){
String input = "<path\n" +
" action=\"m\"\n" +
" text-mod=\"true\"\n" +
" mods=\"true\"\n" +
" kind=\"file\">branches/RO/2021Align01/CO/DGSIG-DAO/src/main/java/eu/ca/co/vo/CsoorspeWsVo.java</path>\n";
Pattern pattern = Pattern.compile("kind=\"file\">.+/(.+\\..+)</path>");
Matcher matcher = pattern.matcher(input);
if (matcher.find()) {
String fileName = matcher.group(1);
System.out.println(fileName);
}
}
Output will be -> CsoorspeWsVo.java
and if you want the fill path change the regex to
Pattern pattern = Pattern.compile("kind=\"file\">(.+)</path>");
The output will be:
branches/RO/2021Align01/CO/DGSIG-DAO/src/main/java/eu/ca/co/vo/CsoorspeWsVo.java
And you can get name and extension in two groups for example
Pattern pattern = Pattern.compile("kind=\"file\">.+/(.+)\\.(.+)</path>");
And inside the if
String fileName = matcher.group(1);
String fileExtension = matcher.group(2);

Get text in the URL with dynamic date - Regex Java

I need to get the text between the URL which has a date in Java
Input 1:
/test1/raw/2019-06-11/testcustomer/usr/pqr/DATA/mn/export/
Output: testcustomer
Only /raw/ remains, date will change and testcustomer will change
Input 2:
/test3/raw/2018-09-01/newcustomer/usr/pqr/DATA/mn/export/
Output: newcustomer
String url = "/test3/raw/2018-09-01/newcustomer/usr/pqr/DATA/mn/export/";
String customer = getCustomer(url);
public String getCustomer (String _url){
String source = "default";
String regex = basePath + "/raw/\\d{4}-\\d{2}-\\d{2}/usr*";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(_url);
if (m.find()) {
source = m.group(1);
} else {
logger.error("Cant get customer with regex " + regex);
}
return source;
}
It's returning 'default' :(

Your regex /raw/\\d{4}-\\d{2}-\\d{2}/usr* is missing the part for the value you want, you need a regex that find the date, and keep what's next :
/\w*/raw/[0-9-]+/(\w+)/.* or (?<=raw\/\d{4}-\d{2}-\d{2}\/)(\w+) will be good
Pattern p = Pattern.compile("/\\w*/raw/[0-9-]+/(\\w+)/.*");
Matcher m = p.matcher(str);
if (m.find()) {
String value = m.group(1);
System.out.println(value);
}
Or if it's always the 4th part, use split()
String value = str.split("/")[4];
System.out.println(value);
And here a >> code demo

Here, we can likely use raw followed by the date as a left boundary, then we would collect our desired output in a capturing group, we would add an slash and consume the rest of our string, with an expression similar to:
.+raw\/[0-9]{4}-[0-9]{2}-[0-9]{2}\/(.+?)\/.+
Demo
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = ".+raw\\/[0-9]{4}-[0-9]{2}-[0-9]{2}\\/(.+?)\\/.+";
final String string = "/test1/raw/2019-06-11/testcustomer/usr/pqr/DATA/mn/export/\n"
+ "/test3/raw/2018-09-01/newcustomer/usr/pqr/DATA/mn/export/";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
RegEx
If this expression wasn't desired or you wish to modify it, please visit regex101.com.
RegEx Circuit
jex.im visualizes regular expressions:

How to recover integers?

I get a string and I have to retrieve the values
Je pense que nous devons utiliser le ".slit"
if (stringReceived.contains("ID")&& stringReceived.contains("Value")) {
here is my character string:
I/RECEIVER: [1/1/0 3
I/RECEIVER: :32:11]
I/RECEIVER: Timestam
I/RECEIVER: p=946697
I/RECEIVER: 531 ID=4
I/RECEIVER: 3 Value=
I/RECEIVER: 18
I receive the value 1 byte by 1 byte.
I would like to recover the value of Timestamp, Id and Value..

You can also use regex for that. Something like:
String example="[11/2/19 9:48:25] Timestamp=1549878505 ID=4 Value=2475";
Pattern pattern=Pattern.compile(".*Timestamp=(\\d+).*ID=(\\d+).*Value=(\\d+)");
Matcher matcher = pattern.matcher(example);
while(matcher.find()) {
System.out.println("Timestamp is:" + matcher.group(1));
System.out.println("Id is:" + matcher.group(2));
System.out.println("Value is:" + matcher.group(3));
}
If the order of tokens can be different (for example ID can come before Timestamp) you can also do it. But since it looks like log which is probably structured I doubt you will need to.

First [11/2/19 9:48:25] seems unnecessary so let's remove it by jumping right into "Timestamp".
Using indexOf(), we can find where Timestamp starts.
// "Timestamp=1549878505 ID=4 Value=2475"
line = line.substring(line.indexOf("Timestamp"));
Since each string is separated by space, we can split it.
// ["Timestamp=1549878505", "ID=4" ,"Value=2475"]
line.split(" ");
Now for each tokens, we can substring it using index of '=' and parse it into string.
for(String token: line.split(" ")) {
int v = Integer.parseInt(token.substring(token.indexOf('=') + 1));
System.out.println(v);
}
Hope that helps :)

String text = "Timestamp=1549878505 ID=4 Value=2475";
Pattern p = Pattern.compile("ID=(\\d)");
Matcher m = p.matcher(text);
if (m.find()) {
System.out.println(m.group(1));
}
output
4

A simple regex is also an option:
private int fromString(String data, String key) {
Pattern pattern = Pattern.compile(key + "=(\\d*)");
Matcher matcher = pattern.matcher(data);
if (matcher.find()) {
return Integer.parseInt(matcher.group(1));
}
return -1;
}
private void test(String data, String key) {
System.out.println(key + " = " + fromString(data, key));
}
private void test() {
String test = "[11/2/19 9:48:25] Timestamp=1549878505 ID=4 Value=2475";
test(test, "Timestamp");
test(test, "ID");
test(test, "Value");
}
prints:
Timestamp = 1549878505
ID = 4
Value = 2475

You can try that:
String txt= "[11/2/19 9:48:25] Timestamp=1549878505 ID=4 Value=2475";
String re1= ".*?\\d+.*?\\d+.*?\\d+.*?\\d+.*?\\d+.*?\\d+.*?(\\d+).*?(\\d+).*?(\\d+)";
Pattern p = Pattern.compile(re1,Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
Matcher m = p.matcher(txt);
if (m.find())
{
String int1=m.group(1);
String int2=m.group(2);
String int3=m.group(3);
System.out.print("("+int1+")"+"("+int2+")"+"("+int3+")"+"\n");
}

Use below code, You will find your timestamp at index 0, id at 1 and value at 2 in List.
Pattern pattern = Pattern.compile("=\\d+");
Matcher matcher = pattern.matcher(stringToMatch);
final List<String> matches = new ArrayList<>();
while (matcher.find()) {
String ans = matcher.group(0);
matches.add(ans.substring(1, ans.length()));
}
Explaining the regex
= matches the character = literally
\d* matches a digit (equal to [0-9])
* Quantifier — Matches between zero and unlimited times, as many times as possible

Check if id in string and get value if so

I am trying to get a regex to match, then get the value with it. For example, I want to check for 1234 as an id and if present, get the status (which is 0 in this case). Basically its id:status. Here is what I am trying:
String topicStatus = "1234:0,567:1,89:2";
String someId = "1234";
String regex = "\\b"+someId+":[0-2]\\b";
if (topicStatus.matches(regex)) {
//How to get status?
}
Not only do I not know how to get the status without splitting and looping through, I don't know why it doesn't match the regex.
Any help would be appreciated. Thanks.

Use the Pattern class
String topicStatus = "1234:0,567:1,89:2";
String someId = "1234";
String regex = "\\b"+someId+":[0-2]\\b";
Pattern MY_PATTERN = Pattern.compile(regex);
Matcher m = MY_PATTERN.matcher(topicStatus);
while (m.find()) {
String s = m.group(1);
System.out.println(s);
}
The key here is to surround the position you want [0-2] in parenthesis which means it will be saved as the first group. You then access it through group(1)

I made some assumptions that your pairs we're always comma separate and then delimited by a colon. Using that I just used split.
String[] idsToCheck = topicStatus.split(",");
for(String idPair : idsToCheck)
{
String[] idPairArray = idPair.split(":");
if(idPairArray[0].equals(someId))
{
System.out.println("id : " + idPairArray[0]);
System.out.println("status: " + idPairArray[1]);
}
}

Use RegEx to extract number from coordinates

I am a beginner of Java Programming language.
When I input (1,2) into the console (brackets included), how can I write the code to extract the first and the second number using RegEx?
If there is no such expression to extract the first/second number within the brackets, I will have to change the way of inputing coordinates to x,y without the brackets and that should be a lot easier to extract numbers to be used.

Try this code:
public static void main(String[] args) {
String searchString = "(7,32)";
Pattern compile1 = Pattern.compile("\\(\\d+,");
Pattern compile2 = Pattern.compile(",\\d+\\)");
Matcher matcher1 = compile1.matcher(searchString);
Matcher matcher2 = compile2.matcher(searchString);
while (matcher1.find() && matcher2.find()) {
String group1 = matcher1.group();
String group2 = matcher2.group();
System.out.println("value 1: " + group1.substring(1, group1.length() - 1 ) + " value 2: " + group2.substring(1, group2.length() - 1 ));
}
}
Not that I think regex is the best to use here. If you know the input will be in the form of: (number, number), I would first get rid of brackets:
stringWithoutBrackets = searchString.substring(1, searchString.length()-1)
and than tokenize it with split
String[] coordiantes = stringWithoutBrackets.split(",");
Looked through Regex API and you can also do something like this:
public static void main(String[] args) {
String searchString = "(7,32)";
Pattern compile1 = Pattern.compile("(?<=\\()\\d+(?=,)");
Pattern compile2 = Pattern.compile("(?<=,)\\d+(?=\\))");
Matcher matcher1 = compile1.matcher(searchString);
Matcher matcher2 = compile2.matcher(searchString);
while (matcher1.find() && matcher2.find()) {
String group1 = matcher1.group();
String group2 = matcher2.group();
System.out.println("value 1: " + group1 + " value 2: " + group2);
}
}
The main change is that I used (?<==\)), (?=,), (?<=,), (?=\)), to search for brackets and commas but not caputre them. But I really think its an overkill for this task.