What i want is to check if there is a number followed by spaces and another number, without any "," in between, anywhere in the String
Currently i am doing this:
Pattern pattern = Pattern.compile("[0-9][\" \"]+[0-9]");
Matcher matcher = pattern.matcher(input);
if(matcher.find()) return false;
and it works just fine. But i was wondering if there is any other simpler way of achieving this?
Since it's an assignment, I won't write out the code, but an alternative solution is:
Split the string on the , token (using String.split())
For each member of the resulting split array:
Trim the leading and trailing spaces from the member
If the trimmed member is an integer (I'll let you figure out how to determine that):
It doesn't meet the criteria you specified
Else:
It's possible that the token could meet your criteria (of containing multiple integers and spaces but no commas. There are several ways you could determine this: do a split on " "; use a while loop, or maybe something else. I'll let you figure that out.
Your solution is good enough, you could try by the positive way like
Pattern pattern = Pattern.compile("^[1-9](,[1-9])*$");
Matcher matcher = pattern.matcher(input);
if(matcher.matches()) return true;
Related
So I am working on regex comparing phone numbers and this is the result:
(?:(?:0{2}|\+)?([1-9][0-9]))? ?([1-9][0-9])? ?([1-9][0-9]{5})
As you can see there are spaces between the numbers. I want them to appear only when there is some other number before the space so:
"0022 45 432345" - should match
"45 345678" or "560032" - should match
" 324400" - shouldn't match because of the space in the beginning
I've been reading different tutorials about regexes and found out about look-behinds, but simple construction like that(just for test):
Pattern p2 = Pattern.compile("(?<=abc)aa");
Matcher m2 = p2.matcher("abcaa");
doesn't work.
Can you tell me what's wrong?
Another problem is - I want a character only happen when it is THE FIRST character in a string, otherwise it shouldn't occur. So the code:
0043 022 234567 should not work, but 022 123450 should match.
I'm stuck right now and would appreciate any help a lot.
This should work just fine. The spaces are moved into the optional groups and are themselves optional. This way, they only match if the group before them is present, but even then they are still optional. No look-behind required.
(?:(?:(?:00|\+)?([1-9][0-9]) ?)?([1-9][0-9]) ?)?([1-9][0-9]{5})
Lookbehind is a zero length match.
The javadoc for the Matcher.matches method determines if the whole String is a match.
What you're looking for is something the Matcher.find and Matcher.group methods. Something like:
final Pattern pattern = Pattern.compile("(?<=abc)aa");
final Matcher matcher = pattern.matcher("abaca");
final String subMatch;
if (matcher.find()) {
subMatch = matcher.group();
} else {
subMatch = "";
}
System.out.println(subMatch);
Example.
I'm using Java and I would like to implement a code whose output is PRP I when the input is (NP (PRP I)).
My current implementation is like the following:
Pattern pattern = Pattern.compile("\\((.?)\\)");
Matcher matcher = pattern.matcher(noun_phrase);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
and its output is NP (PRP I.
I know that one possibility would be to count the parentheses, but I'm wondering if there is any way to get just the string inside the nested parentheses using regex.
This should work
Pattern pattern = Pattern.compile("\\(.*?\\((.*?)\\)\\)");
Matcher matcher = pattern.matcher("(NP (PRP I))");
while (matcher.find()) {
System.out.println(matcher.group(1));
}
You can use following sites to experiment with Regular expressions.
https://regex101.com/r/cE0dM7/1
http://leaverou.github.io/regexplained/
https://www.debuggex.com/r/gfVglXkY1Cw5D6Mb
You need to add another braces around the group. Also, you need to make sure that between the fixed parentheses you don't match the parentheses:
String noun_phrase = "(NP (PRP I))";
Pattern pattern = Pattern.compile("\\([^(]*\\(([^)]*)\\)[^)]*\\)");
Matcher matcher = pattern.matcher(noun_phrase);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
The negated character classes [^(] and [^)] make sure you don't match parentheses too eagerly.
Well, as I don't know how deep you can go with your parantheses, I will suggest two possible solutions.
Solution 1: Assuming the depth's exactly as in your question.
This regex will work: Pattern pattern = Pattern.compile("\\(([^()]*)\\)").
Solution 2: Assuming the depths arbitrary (but at least the most inner string is surrounded by parantheses).
In this case, you will have to make some more changes. First, your pattern will look like this: Pattern pattern = Pattern.compile("(\\(.*)*\\(([^)]*)\\)"). See the difference? You now have two groups, the first matching on all but the innermost part surrounded by parantheses, the second group is exactly the one you want. That means, in your loop, you have to change matcher.group(1) to matcher.group(2). Furthermore, [^)] makes sure, you don't have any closing parantheses in your group.
Lets say that i'm getting a text and I need to have some regex on it which goes as follows:
String aContent = " title='111' title='222' ";
Pattern pattern = Pattern.compile("\\s{1,}(title=){1}+(.){1,}'{1}");
Matcher matcher = pattern.matcher(aTagContent);
And the data is being found/matched by using find()
How can I know how many groups I assume to get from this regex?
I know that there is matcher.groupCount() so this is not the answer i'm looking for.
What i'm actully asking is:
How this text will be splitted? how can I know that without using matcher.groupCount() ?
Matcher.groupCount() returns the number of groups in your Pattern, not in the result.
Matcher.matches() tries to match the entire input string against your pattern, Matcher.find() will sequentially try to match only part of your input string. The latter typically being used in a while-loop, so there's no prior knowledge about the amount of matches.
You can remove the trivial {1} quantifier, it makes your pattern overly verbose. Also, {1,} can be replaced by +. The first quote is missing from your pattern so it won't match your input string. Maybe something like this works for you:
Pattern pattern = Pattern.compile("\\s+(title)='([^']+)'");
Matcher matcher = pattern.matcher(" title='111' title='222' ");
while (matcher.find()) {
System.out.println("attribute: " + matcher.group(1) + ", value: " + matcher.group(2));
}
Can you consider using String.split("\\s") first and iterate over the returned String array? At least you'll know the number of attribute-value pairs in advance.
I currently have this string:
"display_name":"test","game":"test123"
and I want to split the string so I can get the value test. I have looked all over the internet and tried some things, but I couldn't get it to work.
I found that splitting using quotation marks could be done using this regex: \"([^\"]*)\". So I tried this regex: display_name:\":\"([^\"]*)\"game\", but this returned null. I hope that someone could explain me why my regex didn't work and how it should be done.
You forget to include the ",comma before "game" and also you need to remove the extra colon after display_name
display_name\":\"([^\"]*)\",\"game\"
or
\"display_name\":\"([^\"]*)\",\"game\"
Now, print the group index 1.
DEMO
Matcher m = Pattern.compile("\"display_name\":\"([^\"]*)\",\"game\"").matcher(str);
while(m.find())
{
System.out.println(m.group(1))
}
I think you could do it easier, like this:
/(\w)+/g
This little regex will take all your strings.
Your java code should be something like:
Pattern pattern = Pattern.compile("(\w)+");
Matcher matcher = pattern.matcher(yourText);
while (matcher.find()) {
System.out.println("Result: " + matcher.group(2));
}
I also want to note as #AbishekManoharan noted that it looks like JSON
I have the following string:
http://xxx/Content/SiteFiles/30/32531a5d-b0b1-4a8b-9029-b48f0eb40a34/05%20%20LEISURE.mp3?&mydownloads=true
How can I extract the part after 30/? In this case, it's 32531a5d-b0b1-4a8b-9029-b48f0eb40a34.I have another strings having same part upto 30/ and after that every string having different id upto next / which I want.
You can do like this:
String s = "http://xxx/Content/SiteFiles/30/32531a5d-b0b1-4a8b-9029-b48f0eb40a34/05%20%20LEISURE.mp3?&mydownloads=true";
System.out.println(s.substring(s.indexOf("30/")+3, s.length()));
split function of String class won't help you in this case, because it discards the delimiter and that's not what we want here. you need to make a pattern that looks behind. The look behind synatax is:
(?<=X)Y
Which identifies any Y that is preceded by a X.
So in you case you need this pattern:
(?<=30/).*
compile the pattern, match it with your input, find the match, and catch it:
String input = "http://xxx/Content/SiteFiles/30/32531a5d-b0b1-4a8b-9029-b48f0eb40a34/05%20%20LEISURE.mp3?&mydownloads=true";
Matcher matcher = Pattern.compile("(?<=30/).*").matcher(input);
matcher.find();
System.out.println(matcher.group());
Just for this one, or do you want a generic way to do it ?
String[] out = mystring.split("/")
return out[out.length - 2]
I think the / is definitely the delimiter you are searching for.
I can't see the problem you are talking about Alex
EDIT : Ok, Python got me with indexes.
Regular expression is the answer I think. However, how the expression is written depends on the data (url) format you want to process. Like this one:
Pattern pat = Pattern.compile("/Content/SiteFiles/30/([a-z0-9\\-]+)/.*");
Matcher m = pat.matcher("http://xxx/Content/SiteFiles/30/32531a5d-b0b1-4a8b-9029-b48f0eb40a34/05%20%20LEISURE.mp3?&mydownloads=true");
if (m.find()) {
System.out.println(m.group(1));
}