How could I get the first and the second text in "" from the string?
I could do it with indexOf but this is really boring ((
For example I have a String for parse like: "aaa":"bbbbb"perhapsSomeOtherText
And I d like to get aaa and bbbbb with the help of Regex pattern - this will help me to use it in switch statement and will greatly simplify my app/
If all that you have is colon delimited string just split it:
String str = ...; // colon delimited
String[] parts = str.split(":");
Note, that split() receives regex and compilies it every time. To improve performance of your code you can use Pattern as following:
private static Pattern pColonSplitter = Pattern.compile(":");
// now somewhere in your code:
String[] parts = pColonSplitter.split(str);
If however you want to use pattern for matching and extraction of string fragments in more complicated cases, do it like following:
Pattert p = Patter.compile("(\\w+):(\\w+):");
Matcher m = p.matcher(str);
if (m.find()) {
String a = m.group(1);
String b = m.group(2);
}
Pay attention on brackets that define captured group.
Something like this?
Pattern pattern = Pattern.compile("\"([^\"]*)\"");
Matcher matcher = pattern.matcher("\"aaa\":\"bbbbb\"perhapsSomeOtherText");
while (matcher.find()) {
System.out.println(matcher.group(1));
}
Output
aaa
bbbbb
String str = "\"aaa\":\"bbbbb\"perhapsSomeOtherText";
Pattern p = Pattern.compile("\"\\w+\""); // word between ""
Matcher m = p.matcher(str);
while(m.find()){
System.out.println(m.group().replace("\"", ""));
}
output:
aaa
bbbbb
there are several ways to do this
Use StringTokenizer or Scanner with UseDelimiter method
Related
I have been struggling to find the matched string(s) with Java Regular expression for the syntax {//<some string>/<some String>}
My regular expression should return with these matched cases: {//data/process_id}
Below is the String which i want to find matched syntax:
#process_id={//data/process_id}##history_id={//data/history_id}##Pdataxml={//data/dataxml}##Prules =_UNESCAPEXMLVALUE({//data/rules})##submitted_by={//data/submitted_by}##table_definition={//data/table_definition}
I have tried with below regx pattern but it did not work:
[a-zA-Z_/\\[\\]\\(\\)0-9|]+
Can someone please help me to solve this issue?
You can use the following regex:
\{\/\/[^\/{}\s]*\/[^\/{}\s]*\}
Demo on regex101
code:
String input = "#process_id={//data/process_id}##history_id={//data/history_id}##Pdataxml={//data/dataxml}##Prules =_UNESCAPEXMLVALUE({//data/rules})##submitted_by={//data/submitted_by}##table_definition={//data/table_definition}";
List<String> allMatches = new ArrayList<String>();
Matcher m = Pattern.compile("\\{\\/\\/[^\\/{}\\s]*\\/[^\\/{}\\s]*\\}").matcher(input);
while (m.find()) {
allMatches.add(m.group());
}
System.out.println(allMatches);
output:
[{//data/process_id}, {//data/history_id}, {//data/dataxml}, {//data/rules}, {//data/submitted_by}, {//data/table_definition}]
Try this regex with a Matcher:
"\\{//([^/]+)/([^/}]+)}"
The parts are captured in groups 1 and 2.
Like this:
Matcher m = Pattern.compile("\\{//([^/]+)/([^/}]+)}").matcher(str);
while (m.find()) {
String part1 = m.group(1);
String part2 = m.group(2);
// do something with the parts
}
To just grab the whole thing, which would be got from m.group(), use this regex:
"(?<=\\{)//[^/]+/[^/}]+(?=})"
I have string as follows
"ValueFilter("val1") AND ColumnFilter("val2") AND ValueFilter("val3")"
I have stored the following regex in a array. Using for loop I tried to match the pattern
"ValueFilter\\((.*?)\\)","ColumnFilter\\((.*?)\\)"
what I will do is I will replace the value in the bracket and copy it to a new string.
When I run this above regex against the string in the first loop i have XFilter so it will match both occurrence. But I want to do this in order.
Here is the i thing i want to achieve
first i want to match ValueFilter first then ColumnFilter then again ValueFilter. How can I achieve this?
Edit : Added Code
String expr = "\"ValueFilter(\"val1\") AND ColumnFilter(\"val2\") AND ValueFilter(\"val3\")\"";
String patterns = {"ValueFilter\\((.*?)\\)", "ColumnFilter\\((.*?)\\)"}
for (String pattern : patterns) {
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(expr);
while (m.find()) {
//do something
}
}
Expected Output
ValueFilter("val1")
ColumnFilter("val2")
ValueFilter("val3")
You can use this regex [XY]Filter\((.*?)\) with pattern and you have to loop throw the matches using :
String str = "\"XFilter(\"val1\") AND YFilter(\"val2\") AND XFilter(\"val3\")\"";
String regex = "[XY]Filter\\((.*?)\\)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
System.out.println(matcher.group());
}
Note you can i use [XY] which mean to match both X or Y,
Output
XFilter("val1")
YFilter("val2")
XFilter("val3")
regex demo
If you want to get only the value you can get the group 1 like matcher.group(1) instead, the output should be :
"val1"
"val2"
"val3"
Edit
what if I have filtername as "ValueFilter" and "ColumnFilter" instead
of X and Y
In this case you can use (Value|Column) instead of [XY] which mean match ValueFilter or ColumnFilter, the regex should look like :
String str = "\"ValueFilter(\"val1\") AND ColumnFilter(\"val2\") AND ValueFilter(\"val3\")\"";
String regex = "(Value|Column)Filter\\((.*?)\\)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
System.out.println(matcher.group());
}
Output
ValueFilter("val1")
ColumnFilter("val2")
ValueFilter("val3")
Check code demo
Hi I am trying to build one regex to extract 4 digit number from given string using java. I tried it in following ways:
String mydata = "get the 0025 data from string";
Pattern pattern = Pattern.compile("^[0-9]+$");
//Pattern pattern = Pattern.compile("^[0-90-90-90-9]+$");
//Pattern pattern = Pattern.compile("^[\\d]+$");
//Pattern pattern = Pattern.compile("^[\\d\\d\\d\\d]+$");
Matcher matcher = pattern.matcher(mydata);
String val = "";
if (matcher.find()) {
System.out.println(matcher.group(1));
val = matcher.group(1);
}
But it's not working properly. How to do this. Need some help. Thank you.
Change you pattern to:
Pattern pattern = Pattern.compile("(\\d{4})");
\d is for a digit and the number in {} is the number of digits you want to have.
If you want to end up with 0025,
String mydata = "get the 0025 data from string";
mydata = mydata.replaceAll("\\D", ""); // Replace all non-digits
Pattern pattern = Pattern.compile("\\b[0-9]+\\b");
This should do it for you.^$ will compare with the whole string.It will match string with only numbers.
Remove the anchors.. put paranthesis if you want them in group 1:
Pattern pattern = Pattern.compile("([0-9]+)"); //"[0-9]{4}" for 4 digit number
And extract out matcher.group(1)
Many better answers, but if you still have to use in the same way.
String mydata = "get the 0025 data from string";
Pattern pattern = Pattern.compile("(?<![-.])\\b[0-9]+\\b(?!\\.[0-9])");
Matcher matcher = pattern.matcher(mydata);
String val = "";
if (matcher.find()) {
System.out.println(matcher.group(0));
val = matcher.group(0);
}
changed matcher.group(1); to matcher.group(0);
You can go with \d{4} or [0-9]{4} but note that by specifying the ^ at the beginning of regex and $ at the end you're limiting yourself to strings that contain only 4 digits.
My recomendation: Learn some regex basics.
Scanner sc=new Scanner(System.in);
HashMap<String,String> a=new HashMap<>();
ArrayList<String> b=new ArrayList<>();
String s=sc.nextLine();
Pattern p=Pattern.compile("\\d{4}");
Matcher m=p.matcher(s);
while(m.find())
{
String x="";
x=x+(m.group(0));
a.put(x,"0");
b.add(x);
}
System.out.println(a.size());
System.out.println(b);
You can find all matched digit patterns and unique patterns (for unique use Set<String> k=b.keySet();)
If you want to match any number of digits then use pattern like the following:
^\D*(\d+)\D*$
And for exactly 4 digits go for
^\D*(\d{4})\D*$
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 want to break a string like :
String s = "xyz213123kop234430099kpf4532";
into tokens where each token starts with an alphabet and ends with a number. So the above string can be broken down into 3 tokens :
xyz213123
kop234430099
kpf4532
This string s could be very big but the pattern will remain the same, i.e each token will start with 3 alphabets and end with a number.
How do I split them ?
Try this:
\w+?\d+
Java Matcher:
Pattern pattern = Pattern.compile("\\w+?\\d+"); //compiles the pattern we want to use
Matcher matcher = pattern.matcher("xyz213123kop234430099kpf4532"); //we create the matcher on certain string using our pattern
while(matcher.find()) //while the matcher can find the next match
{
System.out.println(matcher.group()); //print it
}
And then you could use Regex.Matches C#:
foreach(Match m in Regex.Matches("xyz213123kop234430099kpf4532", #"\w+?\d+"))
{
Console.WriteLine(m.Value);
}
And for the future this:
RegExr
Do it like this,
String s = "xyz213123kop234430099kpf4532";
Pattern p = Pattern.compile("\\w+?\\d+");
Matcher match = p.matcher(s);
while(match.find()){
System.out.println(match.group());
}
OUTPUT
xyz213123
kop234430099
kpf4532
You can start from such regexp: (\w+?\d+)
http://regexr.com?36utt