I'm new to using Regex, I've been going through a rake of tutorials but I haven't found one that applies to what I want to do,
I want to search for something, but return everything following it but not the search string itself
e.g. "Some lame sentence that is awesome"
search for "sentence"
return "that is awesome"
Any help would be much appreciated
This is my regex so far
sentence(.*)
but it returns: sentence that is awesome
Pattern pattern = Pattern.compile("sentence(.*)");
Matcher matcher = pattern.matcher("some lame sentence that is awesome");
boolean found = false;
while (matcher.find())
{
System.out.println("I found the text: " + matcher.group().toString());
found = true;
}
if (!found)
{
System.out.println("I didn't find the text");
}
You can do this with "just the regular expression" as you asked for in a comment:
(?<=sentence).*
(?<=sentence) is a positive lookbehind assertion. This matches at a certain position in the string, namely at a position right after the text sentence without making that text itself part of the match. Consequently, (?<=sentence).* will match any text after sentence.
This is quite a nice feature of regex. However, in Java this will only work for finite-length subexpressions, i. e. (?<=sentence|word|(foo){1,4}) is legal, but (?<=sentence\s*) isn't.
Your regex "sentence(.*)" is right. To retrieve the contents of the group in parenthesis, you would call:
Pattern p = Pattern.compile( "sentence(.*)" );
Matcher m = p.matcher( "some lame sentence that is awesome" );
if ( m.find() ) {
String s = m.group(1); // " that is awesome"
}
Note the use of m.find() in this case (attempts to find anywhere on the string) and not m.matches() (would fail because of the prefix "some lame"; in this case the regex would need to be ".*sentence(.*)")
if Matcher is initialized with str, after the match, you can get the part after the match with
str.substring(matcher.end())
Sample Code:
final String str = "Some lame sentence that is awesome";
final Matcher matcher = Pattern.compile("sentence").matcher(str);
if(matcher.find()){
System.out.println(str.substring(matcher.end()).trim());
}
Output:
that is awesome
You need to use the group(int) of your matcher - group(0) is the entire match, and group(1) is the first group you marked. In the example you specify, group(1) is what comes after "sentence".
You just need to put "group(1)" instead of "group()" in the following line and the return will be the one you expected:
System.out.println("I found the text: " + matcher.group(**1**).toString());
Related
I'm new to using Regex, I've been going through a rake of tutorials but I haven't found one that applies to what I want to do,
I want to search for something, but return everything following it but not the search string itself
e.g. "Some lame sentence that is awesome"
search for "sentence"
return "that is awesome"
Any help would be much appreciated
This is my regex so far
sentence(.*)
but it returns: sentence that is awesome
Pattern pattern = Pattern.compile("sentence(.*)");
Matcher matcher = pattern.matcher("some lame sentence that is awesome");
boolean found = false;
while (matcher.find())
{
System.out.println("I found the text: " + matcher.group().toString());
found = true;
}
if (!found)
{
System.out.println("I didn't find the text");
}
You can do this with "just the regular expression" as you asked for in a comment:
(?<=sentence).*
(?<=sentence) is a positive lookbehind assertion. This matches at a certain position in the string, namely at a position right after the text sentence without making that text itself part of the match. Consequently, (?<=sentence).* will match any text after sentence.
This is quite a nice feature of regex. However, in Java this will only work for finite-length subexpressions, i. e. (?<=sentence|word|(foo){1,4}) is legal, but (?<=sentence\s*) isn't.
Your regex "sentence(.*)" is right. To retrieve the contents of the group in parenthesis, you would call:
Pattern p = Pattern.compile( "sentence(.*)" );
Matcher m = p.matcher( "some lame sentence that is awesome" );
if ( m.find() ) {
String s = m.group(1); // " that is awesome"
}
Note the use of m.find() in this case (attempts to find anywhere on the string) and not m.matches() (would fail because of the prefix "some lame"; in this case the regex would need to be ".*sentence(.*)")
if Matcher is initialized with str, after the match, you can get the part after the match with
str.substring(matcher.end())
Sample Code:
final String str = "Some lame sentence that is awesome";
final Matcher matcher = Pattern.compile("sentence").matcher(str);
if(matcher.find()){
System.out.println(str.substring(matcher.end()).trim());
}
Output:
that is awesome
You need to use the group(int) of your matcher - group(0) is the entire match, and group(1) is the first group you marked. In the example you specify, group(1) is what comes after "sentence".
You just need to put "group(1)" instead of "group()" in the following line and the return will be the one you expected:
System.out.println("I found the text: " + matcher.group(**1**).toString());
I tried looking for an answer to this question but just couldn't finding anything and I hope that there's an easy solution for this. I have and using the following code in C#,
String pattern = ("(hello|hello world)");
Regex regex = new Regex(pattern, RegexOptions.IgnoreCase);
var matches = regex.Matches("hello world");
Question is, is there a way for the matches method to return the longest pattern first? In this case, I want to get "hello world" as my match as opposed to just "hello". This is just an example but my pattern list consist of decent amount of words in it.
If you already know the lengths of the words beforehand, then put the longest first. For example:
String pattern = ("(hello world|hello)");
The longest will be matched first. If you don't know the lengths beforehand, this isn't possible.
An alternative approach would be to store all the matches in an array/hash/list and pick the longest one manually, using the language's built-in functions.
Regular expressions (will try) to match patterns from left to right. If you want to make sure you get the longest possible match first, you'll need to change the order of your patterns. The leftmost pattern is tried first. If a match is found against that pattern, the regular expression engine will attempt to match the rest of the pattern against the rest of the string; the next pattern will be tried only if no match can be found.
String pattern = ("(hello world|hello wor|hello)");
Make two different regex matches. The first will match your longer option, and if that does not work, the second will match your shorter option.
string input = "hello world";
string patternFull = "hello world";
Regex regexFull = new Regex(patternFull, RegexOptions.IgnoreCase);
var matches = regexFull.Matches(input);
if (matches.Count == 0)
{
string patternShort = "hello";
Regex regexShort = new Regex(patternShort, RegexOptions.IgnoreCase);
matches = regexShort.Matches(input);
}
At the end, matches will be be the output of "full" or "short", but "full" will be checked first and will short-circuit if it is true.
You can wrap the logic in a function if you plan on calling it many times. This is something I came up with (but there are plenty of other ways you can do this).
public bool HasRegexMatchInOrder(string input, params string[] patterns)
{
foreach (var pattern in patterns)
{
Regex regex = new Regex(pattern, RegexOptions.IgnoreCase);
if (regex.IsMatch(input))
{
return true;
}
}
return false;
}
string input = "hello world";
bool hasAMatch = HasRegexMatchInOrder(input, "hello world", "hello", ...);
What is the regular expression (in JavaScript if it matters) to only match if the text is an exact match? That is, there should be no extra characters at other end of the string.
For example, if I'm trying to match for abc, then 1abc1, 1abc, and abc1 would not match.
Use the start and end delimiters: ^abc$
It depends. You could
string.match(/^abc$/)
But that would not match the following string: 'the first 3 letters of the alphabet are abc. not abc123'
I think you would want to use \b (word boundaries):
var str = 'the first 3 letters of the alphabet are abc. not abc123';
var pat = /\b(abc)\b/g;
console.log(str.match(pat));
Live example: http://jsfiddle.net/uu5VJ/
If the former solution works for you, I would advise against using it.
That means you may have something like the following:
var strs = ['abc', 'abc1', 'abc2']
for (var i = 0; i < strs.length; i++) {
if (strs[i] == 'abc') {
//do something
}
else {
//do something else
}
}
While you could use
if (str[i].match(/^abc$/g)) {
//do something
}
It would be considerably more resource-intensive. For me, a general rule of thumb is for a simple string comparison use a conditional expression, for a more dynamic pattern use a regular expression.
More on JavaScript regexes: https://developer.mozilla.org/en/JavaScript/Guide/Regular_Expressions
"^" For the begining of the line "$" for the end of it. Eg.:
var re = /^abc$/;
Would match "abc" but not "1abc" or "abc1". You can learn more at https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions
I need to determine whether a string begins with a number - I've tried the following to no avail:
if (matches("^[0-9].*)", upper(text))) str = "Title"""
I'm new to DXL and Regex - what am I doing wrong?
You need the caret character to indicate a match only at the start of a string. I added the plus character to match all the numbers, although you might not need it for your situation. If you're only looking for numbers at the start, and don't care if there is anything following, you don't need anymore.
string str1 = "123abc"
string str2 = "abc123"
string strgx = "^[0-9]+"
Regexp rgx = regexp2(strgx)
if(rgx(str1)) { print str1[match 0] "\n" } else { print "no match\n" }
if(rgx(str2)) { print str2[match 0] "\n" } else { print "no match\n" }
The code block above will print:
123
no match
#mrhobo is correct, you want something like this:
Regexp numReg = "^[0-9]"
if(numReg text) str = "Title"
You don't need upper since you are just looking for numbers. Also matches is more for finding the part of the string that matches the expression. If you just want to check that the string as a whole matches the expression then the code above would be more efficient.
Good luck!
At least from example I found this example should work:
Regexp plural = regexp "^([0-9].*)$"
if plural "15systems" then print "yes"
Resource:
http://www.scenarioplus.org.uk/papers/dxl_regexp/dxl_regexp.htm
I have a string like '[1]-[2]-[3],[4]-[5],[6,7,8],[9]' or '[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]', I'd like the Pattern to get the list result, but don't know how to figure out the pattern. Basically the comma is the split, but [6,7,8] itself contains the comma as well.
the string: [1]-[2]-[3],[4]-[5],[6,7,8],[9]
the result:
[1]-[2]-[3]
[4]-[5]
[6,7,8]
[9]
or
the string: [Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]
the result:
[Computers]-[Apple]-[Laptop]
[Cables]-[Cables,Connectors]
[Adapters]
,(?=\[)
This pattern splits on any comma that is followed by a bracket, but keeps the bracket within the result text.
The (?=*stuff*) is known as a "lookahead assertion". It acts as a condition for the match but is not itself part of the match.
In C# code:
String inputstring = "[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]";
foreach(String s in Regex.Split(inputstring, #",(?=\[)"))
System.Console.Out.WriteLine(s);
In Java code:
String inputstring = "[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]";
Pattern p = Pattern.compile(",(?=\\[)"));
for(String s : p.split(inputstring))
System.out.println(s);
Either produces:
[Computers]-[Apple]-[Laptop]
[Cables]-[Cables,Connectors]
[Adapters]
Although I believe the best approach here is to use split (as presented by #j__m's answer), here's an approach that uses matching rather than splitting.
Regex:
(\[.*?\](?!-))
Example usage:
String input = "[Computers]-[Apple]-[Laptop],[Cables]-[Cables,Connectors],[Adapters]";
Pattern p = Pattern.compile("(\\[.*?\\](?!-))");
Matcher m = p.matcher(input);
while (m.find()) {
System.out.println(m.group(1));
}
Resulting output:
[Computers]-[Apple]-[Laptop]
[Cables]-[Cables,Connectors]
[Adapters]
An answer that doesn't use regular expressions (if that's worth something in ease of understanding what's going on) is:
substitute "]#[" for "],["
split on "#"