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.
Related
Want to capture the string after the last slash and before either a (; sid=) word or a (?) character.
sample data:
sessionId=30a793b1-ed7e-464a-a630; Url=https://www.example.com/mybook/order/newbooking/itemSummary; sid=KJ4dgQGdhg7dDn1h0TLsqhsdfhsfhjhsdjfhjshdjfhjsfddscg139bjXZQdkbHpzf9l6wy1GdK5XZp; targetUrl=https://www.example.com/mybook/order/newbooking/page1?id=122;
sessionId=sfdsdfsd-ba57-4e21-a39f-34; Url=https://www.example.com/mybook/order/newbooking/itemList?id=76734¶=jhjdfhj&type=new&ordertype=kjkf&memberid=273647632&iSearch=true; sid=Q4hWgR1GpQb8xWTLpQB2yyyzmYRgXgFlJLGTc0QJyZbW targetUrl=https://www.example.com/ mybook/order/newbooking/page1?id=123;
sessionId=0e1acab1-45b8-sdf3454fds-afc1-sdf435sdfds; Url=https://www.example.com/mybook/order/newbooking/; sid=hkm2gRSL2t5ScKSJKSJn3vg2sfdsfdsfdsfdsfdfdsfdsfdsfvJZkDD3ng0kYTjhNQw8mFZMn; targetUrl=https://www.example.com/mybook/order/newbooking/page1?id=343;
Expecting the below output:
1. itemSummary
2. itemList
3. ''(empty string)
Have build the below regex to capture it but its 100% accurate. It is capturing some additional part.
Regex
Url=.*\/(.*)(; sid|\?)
Could you please help me to improve the regex to get desired output?
Thanks in advance!
You may use this regex in Java with a greedy match after Url=:
\bUrl=\S+/([^?;/]+)(?=; sid|\?)
RegEx Demo
RegEx Demo:
\b: Word boundary
Url=: Match text Url=
\S+/: Match 1+ non-whitespace characters followed by a /
([^?;/]+): Match 1+ of a character that not ? and ; and /
(?=; sid|\?): Lookahead to assert that we have ; sid or ? ahead
Alternative solution:
Used regex:
"^Url=.*/(\\w+|)$"
Regex in test bench and context:
public static void main(String[] args) {
String input1 = "sessionId=30a793b1-ed7e-464a-a630; "
+ "Url=https://www.example.com/mybook/order/newbooking/itemSummary; "
+ "sid=KJ4dgQGdhg7dDn1h0TLsqhsdfhsfhjhsdjfhjshdjfhjsfddscg139bjXZQdkbHpzf9l6wy1GdK5XZp; "
+ "targetUrl=https://www.example.com/mybook/order/newbooking/page1?id=122;";
String input2 = "sessionId=sfdsdfsd-ba57-4e21-a39f-34; "
+ "Url=https://www.example.com/mybook/order/newbooking/itemList?id=76734¶=jhjdfhj&type=new&ordertype=kjkf&memberid=273647632&iSearch=true; "
+ "sid=Q4hWgR1GpQb8xWTLpQB2yyyzmYRgXgFlJLGTc0QJyZbW "
+ "targetUrl=https://www.example.com/mybook/order/newbooking/page1?id=123;";
String input3 = "sessionId=0e1acab1-45b8-sdf3454fds-afc1-sdf435sdfds; "
+ "Url=https://www.example.com/mybook/order/newbooking/; "
+ "sid=hkm2gRSL2t5ScKSJKSJn3vg2sfdsfdsfdsfdsfdfdsfdsfdsfvJZkDD3ng0kYTjhNQw8mFZMn; "
+ "targetUrl=https://www.example.com/mybook/order/newbooking/page1?id=343;";
List<String> inputList = Arrays.asList(input1, input2, input3);
// Pre-compiled Patterns should not be in loops - that is why they are placed outside the loops
Pattern replaceWithNewLinePattern = Pattern.compile(";?\\s|\\?");
Pattern extractWordFromUrlPattern = Pattern.compile("^Url=.*/(\\w+|)$", Pattern.MULTILINE);
int count = 0;
for(String input : inputList) {
String inputWithNewLines = replaceWithNewLinePattern.matcher(input).replaceAll("\n");
// System.out.println(inputWithNewLines); // Check the change...
Matcher matcher = extractWordFromUrlPattern.matcher(inputWithNewLines);
while (matcher.find()) {
System.out.printf( "%d. '%s'%n", ++count, matcher.group(1));
}
}
}
Output:
1. 'itemSummary'
2. 'itemList'
3. ''
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
I have a long string let's say
I like this #computer and I want to buy it from #XXXMall.
I know the regular expression pattern is
Pattern tagMatcher = Pattern.compile("[#]+[A-Za-z0-9-_]+\\b");
Now i want to get all the hashtags in an array. How can i use this expression to get array of all hash tags from string something like
ArrayList hashtags = getArray(pattern, str)
You can write like?
private static List<String> getArray(Pattern tagMatcher, String str) {
Matcher m = tagMatcher.matcher(str);
List<String> l = new ArrayList<String>();
while(m.find()) {
String s = m.group(); //will give you "#computer"
s = s.substring(1); // will give you just "computer"
l.add(s);
}
return l;
}
Also you can use \\w- instead of A-Za-z0-9-_ making the regex [#]+[\\w]+\\b
This link would surely be helpful for achieving what you want.
It says:
The find() method searches for occurrences of the regular expressions
in the text passed to the Pattern.matcher(text) method, when the
Matcher was created. If multiple matches can be found in the text, the
find() method will find the first, and then for each subsequent call
to find() it will move to the next match.
The methods start() and end() will give the indexes into the text
where the found match starts and ends.
Example:
String text =
"This is the text which is to be searched " +
"for occurrences of the word 'is'.";
String patternString = "is";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(text);
int count = 0;
while(matcher.find()) {
count++;
System.out.println("found: " + count + " : "
+ matcher.start() + " - " + matcher.end());
}
You got the hint now.
Here is one way, using Matcher
Pattern tagMatcher = Pattern.compile("#+[-\\w]+\\b");
Matcher m = tagMatcher.matcher(stringToMatch);
ArrayList<String> hashtags = new ArrayList<>();
while (m.find()) {
hashtags.add(m.group());
}
I took the liberty of simplifying your regex. # does not need to be in a character class. [A-Za-z0-9_] is the same as \w, so [A-Za-z0-9-_] is the same as [-\w]
You can use :
String val="I like this #computer and I want to buy it from #XXXMall.";
String REGEX = "(?<=#)[A-Za-z0-9-_]+";
List<String> list = new ArrayList<String>();
Pattern pattern = Pattern.compile(REGEX);
Matcher matcher = pattern.matcher(val);
while(matcher.find()){
list.add(matcher.group());
}
(?<=#) Positive Lookbehind - Assert that the character # literally be matched.
you can use the following code for getting the names
String saa = "#{akka}nikhil#{kumar}aaaaa";
Pattern regex = Pattern.compile("#\\{(.*?)\\}");
Matcher m = regex.matcher(saa);
while(m.find()) {
String s = m.group(1);
System.out.println(s);
}
It will print
akka
kumar
I have a program where I want to filter Strings with a set number of "+"'s at the beginning.
For example:
+++Adam is working very well.
++Adam is working well.
+Adam is doing OK.
How do I only pick up each particular case (i.e. only one plus sign, only two plus signs, only three plus signs)? I usually get a return of anything beginning with a +.
I have the following regex patterns compiled, but I either get only one return (usually the two ++) or all of them:
public static String regexpluschar = "^\\Q+\\E{1}[\\w <]";
public static String regexpluspluschar = "^\\Q+\\E{2}[\\w <]";
public static String regexpluspluspluschar = "^\\Q+\\E{3}[\\w <]";
Pattern plusplusplus = Pattern.compile(regexpluspluspluschar);
Pattern plusplus = Pattern.compile(regexpluspluschar);
Pattern plus = Pattern.compile(regexpluschar);
I then try to find using a Matcher class - I've used .find() and .matches() but don't get the result I'm after (java+regex newbie alert here).
Matcher matcherplusplusplus = plusplusplus.matcher(check);
Matcher matcherplusplus = plusplus.matcher(check);
Matcher matcherplus = plus.matcher(check);
//OK we have 3+'s
if ((matcherplusplusplus.find())==true){
System.out.println("Filtering 3 +s.");
System.out.println("filter is " + filter + " in the 3 + filter.");
String toChange = getItem(i);
setItemFiltered(i, toChange);
}
//OK - we have 2 +'s
if ((matcherplusplus.find())==true){
System.out.println("Filtering 2 +s.");
System.out.println("filter is " + filter + " in the 2 + filter.");
String toChange = getItem(i);
setItemFiltered(i, toChange);
}
//OK - we have 1 +'s
if ((matcherplus.find())==true){
System.out.println("Filtering 1 +.");
System.out.println("filter is " + filter + " in the 1 + filter.");
String toChange = getItem(i);
setItemFiltered(i, toChange);
}
For the very curious, the above if's are embedded in a for loop that cycles around some JTextFields. Full code at: http://pastebin.ca/2199327
Why not simpler :
public static String regexpluschar = "^\\+[\\w <]";
public static String regexpluspluschar = "^\\+{2}[\\w <]";
public static String regexpluspluspluschar = "^\\+{3}[\\w <]";
or even
public static String regexpluschar = "^\\+[^\\+]";
public static String regexpluspluschar = "^\\+{2}[^\\+]";
public static String regexpluspluspluschar = "^\\+{3}[^\\+]";
Edit : It's working on my test program, but I had to removed your specific code :
String toChange = getItem(i);
setItemFiltered(i, toChange);
proof : my output is :
Filtering 3 +s.
+++Adam is working very well. is in the 3 + filter.
Filtering 2 +s.
++Adam is working well. is in the 2 + filter.
Filtering 1 +.
+Adam is doing OK. is in the 1 + filter.
Your filter is working, but you specific code may not... (maybe have a look at setItemFiltered?)
I was thinking something like this would be easier:
public static void main(String[] args) {
Pattern pattern = Pattern.compile("^(\\+{1,3}).*");
Matcher matcher = pattern.matcher(<your text>);
if (matcher.matches()) {
String pluses = matcher.group(1);
switch (pluses.length()) {
}
}
}
And if you want to be sure that ++++This is insane does not match then change the pattern to
Pattern pattern = Pattern.compile("^(\\+{1,3})[^+].*");
I'm working on strings like "[ro.multiboot]: [1]". How do I just select 1(it can also be 0) out of this string?
I am looking for a regex in Java.
Usually, you would do something like (assuming 0 and 1 were the only options):
^.*\[([01])\].*$
If you only wanted the value for ro.multiboot, you could change it to something like:
^.*\[ro.multiboot\].*\[([01])\].*$
(depending on how complex any of the non-bracketed stuff is allowed to be).
These would both basically only extract the value between square brackets if it were zero or one, and capture it into a capture variable so you could use it.
Of course, regex is not a world-wide standard, nor are the environments in which you use it. That means it depends a lot on your actual environment how you will actually code this up.
For Java, the following sample program may help:
import java.util.regex.*;
class Test {
public static void main(String args[]) {
Pattern p = Pattern.compile("^.*\\[ro.multiboot\\].*\\[([01])\\].*$");
String str;
Matcher m;
str = "[ro.multiboot]: [0]";
m = p.matcher (str);
if (m.find()) {
System.out.println ("str0 has " + m.group(1));
}
str = "[ro.multiboot]: [1]";
m = p.matcher (str);
if (m.find()) {
System.out.println ("str1 has " + m.group(1));
}
str = "[ro.multiboot]: [2]";
m = p.matcher (str);
if (m.find()) {
System.out.println ("str2 has " + m.group(1));
}
}
}
This results in (as expected):
str0 has 0
str1 has 1
#paxdiablo's regexps are correct, but complete answer for "How do I just select 1(it can also be 0) out of this string?" is:
1. very simple solution
String input = "[ro.multiboot]: [1]";
String matched = input.replaceFirst( "^.*\\[ro.multiboot\\].*\\[([01])\\].*$", "$1" );
2. same functionality, more complicated but with better performance
String input = "[ro.multiboot]: [1]";
Pattern p = Pattern.compile( "^.*\\[ro.multiboot\\].*\\[([01])\\].*$" );
Matcher m = p.matcher( input );
String matched = null;
if ( m.matches() ) matched = m.group( 1 );
Performance is better because the pattern is compiled just once (for example when you are matching array os such Strings);
Notes:
in both examples the group is part of regexps between ( and ) (if not escaped)
in Java you have to use \\[, because \[ returns error - it is not correct escape sequence for String