I'm working on a Java Program which takes a question from a user, sends it to the Wolfram Alpha API and then cleans up the result and prints it.
If the user asks the question "Who is the President of the USA?" the result is as follows
Response: <section><title>Input interpretation</title> <sectioncontents>United States | President</sectioncontents></section><section><title>Result</title><sectioncontents>Barack Obama (from 20/01/2009 to present)</sectioncontents></section><section><title>Basic information</title><sectioncontents>official position | President (44th)..........etc
I would like to Extract "Barack Obama (from 20/01/2009 to present)"
I have been able to trim up to Barack using the below code:
String clean =response.substring(response.indexOf("Result") + 31 , response.length());
System.out.println("Response: " + clean);
How would I trim the rest of the result?
Well, in case it helps, I came up with this regex:
Result.+?>([^<]+?)<
After finding "Result" it captures the first instance of > and < with at least one character between them.
UPDATE
Below is some sample code that might be helpful:
String response = "Response: <section><title>..."
Pattern pattern = Pattern.compile("Result.+?>([^<]+?)<");
Matcher match = pattern.matcher(response);
String clean = "";
if (match.find())
clean = match.group(1);
System.out.println(clean);
The response is essentially XML.
As has been discussed endlessly in many programming fora, regular expressions are not suitable for parsing XML - you should use an XML parser.
Related
I want to extract particular fields from a url of a facebookpage. Iam not able to extract since link format is not static.eg:if I gave the below examples as input it should give the o/p as what we desire
1)https://www.facebook.com/pages/Ice-cream/109301862430120?rf=102173023157556
o/p -109301862430120
What about this type of link
can anyone help me
So in short, you want to get name after last / and (if there is any) before ? mark.
You can do it with using URI and File classes like
String data = "https://www.facebook.com/pages/Anti-Christian-sentiment/149675731889496?ref=br_tf";
System.out.println(new File(new URI(data).getRawPath()).getName());
Output: 149675731889496
If you need to use regex then you can use
([^/?]+)(\\?|$)
and just read content of group 1 (the one in first pair of parenthesis).
If you don't want to use groups, and make regex match only digit part (without including ? in match) then you can use look around mechanisms like look-ahead (?=...). Regex you would have to use would look like
[^/?]+(?=\\?|$)
Code example:
String data = "https://www.facebook.com/pages/Anti-Christian-sentiment/149675731889496?ref=br_tf";
Pattern p = Pattern.compile("([^/?]+)(\\?|$)");
Matcher m = p.matcher(data);
if (m.find()){
System.out.println(m.group(1));
}
Output:
149675731889496
Given the string
Content ID [9283745997] Content ID [9283005997] There can be text in between Content ID [9283745953] Content ID [9283741197] Content ID [928374500] There can be valid text here which should not be removed.
I want to remove the text starting Content ID followed by [9283745997] any numbers can be present between square brackets. Eventually I want the result string to be
There can be text in between There can be valid text here which should not be removed.
Could anyone please provide a valid regex to capture this recurring text but the numerals within square brackets are unique?
I appreciate your help!
My soulution to this was :
Pattern p = Pattern.compile("(Content ID \\[\\d*\\] )");
Matcher m = p.matcher(str);
StringBuffer sb = new StringBuffer();
while(m.find()) {
m.appendReplacement(sb, "");
}
m.appendTail(sb);
System.out.println(sb);
So basically you are trying to remove each of Content ID [one or more digits].
To do this you can use replaceAll("regex","replacement") method of String class. As replacement you can use empty String "".
Only problem that stays is what regex should you use.
to match Content ID just write it normally as "Content ID "
to match [ or ] you will have to add \ before each of them because they are regex metacharacters and you need to escape them (in Java you will need to write \ as "\\")
to represent one digit (character from range 0-9) regex uses \d (again in Java you will need to write \ as "\\" which will result in "\\d")
to say "one or more of previously described element" just add + after definition of such element. For example if you want to match one or more letters a you can write it as a+.
Now you should be able to create correct regex. If you will have some questions feel free to ask them in comments.
Try this one:
(Content ID \[[0-9]+\])
You can test it here: http://regexpal.com/
I would use the regex
Content ID \[\d+\] ?
Implement it like this:
str.replaceAll("Content ID \\[\\d+\\] ?", "");
You can find an explanation and demonstration here: http://regex101.com/r/qD5rJ6
[introduction][position]Lead Researcher and Research Manager[/position] in the [affiliation]Web Search and Mining Group, Microsoft Research[/affiliation]</b>.
I am a [position]lead researcher[/position] at [affiliation]Microsoft Research[/affiliation]. I am also [position]adjunct professor[/position] of [affiliation]Peking University[/affiliation], [affiliation]Xian Jiaotong University[/affiliation] and [affiliation]Nankai University[/affiliation].
I joined [affiliation]Microsoft Research[/affiliation] in June 2001. Prior to that, I worked at the Research Laboratories of NEC Corporation.
I obtained a [bsdegree]B.S.[/bsdegree] in [bsmajor]Electrical Engineering[/bsmajor] from [bsuniv]Kyoto University[/bsuniv] in [bsdate]1988[/bsdate] and a [msdegree]M.S.[/msdegree] in [msmajor]Computer Science[/msmajor] from [msuniv]Kyoto University[/msuniv] in [msdate]1990[/msdate]. I earned my [phddegree]Ph.D.[/phddegree] in [phdmajor]Computer Science[/phdmajor] from the [phduniv]University of Tokyo[/phduniv] in [phddate]1998[/phddate].
I am interested in [interests]statistical learning[/interests], [interests]natural language processing[/interests], [interests]data mining, and information retrieval[/interests].[/introduction]
I'm able to strip all tags from the paragraph above with:
String stripped = html.replaceAll("\\[.*?\\]", "");
But I'd like to keep three pairs of tags in the paragraph, which are [bsuniv][/bsuniv],[msuniv][/msuniv] and [phduniv][/phduniv]. In other words, I don't want to strip those tags containing the keyword "univ". I can't find a convenient way to rewrite the regular expression. Anyone help me?
You can use a negative-look ahead assertion here: -
str = str.replaceAll("\\[(.(?!univ))*?\\]", "");
or: -
str = str.replaceAll("\\[((?!univ).)*?\\]", "");
Both of them will give you the desired output. There is only one difference -
The first one does a negative look-ahead, against the current character, and if it is not followed by univ, it moves to the next character.
The second one does a negative look-ahead against an empty string before every character, and if it is not followed by univ, it goes ahead to match a single character.
need to find an expression for the following problem:
String given = "{ \"questionID\" :\"4\", \"question\":\"What is your favourite hobby?\",\"answer\" :\"answer 4\"},{ \"questionID\" :\"5\", \"question\" :\"What was the name of the first company you worked at?\",\"answer\" :\"answer 5\"}";
What I want to get: "{ \"questionID\" :\"4\", \"question\":\"What is your favourite hobby?\",\"answer\" :\"*******\"},{ \"questionID\" :\"5\", \"question\" :\"What was the name of the first company you worked at?\",\"answer\" :\"******\"}";
What I am trying:
String regex = "(.*answer\"\\s:\"){1}(.*)(\"[\\s}]?)";
String rep = "$1*****$3";
System.out.println(test.replaceAll(regex, rep));
What I am getting:
"{ \"questionID\" :\"4\", \"question\":\"What is your favourite hobby?\",\"answer\" :\"answer 4\"},{ \"questionID\" :\"5\", \"question\" :\"What was the name of the first company you worked at?\",\"answer\" :\"******\"}";
Because of the greedy behaviour, the first group catches both "answer" parts, whereas I want it to stop after finding enough, perform replacement, and then keep looking further.
The pattern
("answer"\s*:\s*")(.*?)(")
Seems to do what you want. Here's the escaped version for Java:
(\"answer\"\\s*:\\s*\")(.*?)(\")
The key here is to use (.*?) to match the answer and not (.*). The latter matches as many characters as possible, the former will stop as soon as possible.
The above pattern won't work if there are double quotes in the answer. Here's a more complex version that will allow them:
("answer"\s*:\s*")((.*?)[^\\])?(")
You'll have to use $4 instead of $3 in the replacement pattern.
The following regex works for me :
regex = "(?<=answer\"\\s:\")(answer.*?)(?=\"})";
rep = "*****";
replaceALL(regex,rep);
The \ and " might be incorrectly escaped since I tested without java.
http://regexr.com?303mm
The following code blocks on my system. Why?
System.out.println( Pattern.compile(
"^((?:[^'\"][^'\"]*|\"[^\"]*\"|'[^']*')*)/\\*.*?\\*/(.*)$",
Pattern.MULTILINE | Pattern.DOTALL ).matcher(
"\n\n\n\n\n\nUPDATE \"$SCHEMA\" SET \"VERSION\" = 12 WHERE NAME = 'SOMENAMEVALUE';"
).matches() );
The pattern (designed to detect comments of the form /*...*/ but not within ' or ") should be fast, as it is deterministic...
Why does it take soooo long?
You're running into catastrophic backtracking.
Looking at your regex, it's easy to see how .*? and (.*) can match the same content since both also can match the intervening \*/ part (dot matches all, remember). Plus (and even more problematic), they can also match the same stuff that ((?:[^'"][^'"]*|"[^"]*"|'[^']*')*) matches.
The regex engine gets bogged down in trying all the permutations, especially if the string you're testing against is long.
I've just checked your regex against your string in RegexBuddy. It aborts the match attempt after 1.000.000 steps of the regex engine. Java will keep churning on until it gets through all permutations or until a Stack Overflow occurs...
You can greatly improve the performance of your regex by prohibiting backtracking into stuff that has already been matched. You can use atomic groups for this, changing your regex into
^((?>[^'"]+|"[^"]*"|'[^']*')*)(?>/\*.*?\*/)(.*)$
or, as a Java string:
"^((?>[^'\"]+|\"[^\"]*\"|'[^']*')*)(?>/\\*.*?\\*/)(.*)$"
This reduces the number of steps the regex engine has to go through from > 1 million to 58.
Be advised though that this will only find the first occurrence of a comment, so you'll have to apply the regex repeatedly until it fails.
Edit: I just added two slashes that were important for the expressions to work. Yet I had to change more than 6 characters.... :(
I recommend that you read Regular Expression Matching Can Be Simple And Fast (but is slow in Java, Perl, PHP, Python, Ruby, ...).
I think it's because of this bit:
(?:[^'\"][^'\"]*|\"[^\"]*\"|'[^']*')*
Removing the second and third alternatives gives you:
(?:[^'\"][^'\"]*)*
or:
(?:[^'\"]+)*
Repeated repeats can take a long time.
For comment /* and */ detection I would suggest having a code like this:
String str = "\n\n\n\n\n\nUPDATE \"$SCHEMA\" /*a comment\n\n*/ SET \"VERSION\" = 12 WHERE NAME = 'SOMENAMEVALUE';";
Pattern pt = Pattern.compile("\"[^\"]*\"|'[^']*'|(/\\*.*?\\*/)",
Pattern.MULTILINE | Pattern.DOTALL);
Matcher matcher = pt.matcher(str);
boolean found = false;
while (matcher.find()) {
if (matcher.group(1) != null) {
found = true;
break;
}
}
if (found)
System.out.println("Found Comment: [" + matcher.group(1) + ']');
else
System.out.println("Didn't find Comment");
For above string it prints:
Found Comment: [/*a comment
*/]
But if I change input string to:
String str = "\n\n\n\n\n\nUPDATE \"$SCHEMA\" '/*a comment\n\n*/' SET \"VERSION\" = 12 WHERE NAME = 'SOMENAMEVALUE';";
OR
String str = "\n\n\n\n\n\nUPDATE \"$SCHEMA\" \"/*a comment\n\n*/\" SET \"VERSION\" = 12 WHERE NAME = 'SOMENAMEVALUE';";
Output is:
Didn't find Comment