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Java Forum / General / January 2006find a string in line
you have a sentence and you need to find wether it contains a word; for example "he is great " .containts("is") returns true, but "his great".contains("is") also returns true, but I want it to return false I tried regexpe but it didnt work String patterns="[^A-Za-z]+str+[^A-Za-z]" ? but it didnt work. > you have a sentence and you need to find wether it contains a word; for > example "he is great " .containts("is") returns true, but "his [quoted text clipped - 4 lines] > > but it didnt work. Use the whitespace character class: ".*\\sis\\s.*" will match zero or more characters, a whitespace, the string "is", a whitespace, and zero or more characters. Note the double backslashes.  SignatureBeware the False Authority Syndrome > > you have a sentence and you need to find wether it contains a word; for > > example "he is great " .containts("is") returns true, but "his [quoted text clipped - 16 lines] > -- > Beware the False Authority Syndrome but I also want it to match things such as ,is, is. >> > you have a sentence and you need to find wether it contains a word; >> > for example "he is great " .containts("is") returns true, but "his [quoted text clipped - 11 lines] > > but I also want it to match things such as ,is, is. Then maybe the \W construct can help. This will match any non-word character. ".*\\W+" + searchString + "\\W+.*" non-word characters are defined as anything other than an alfanumeric character or an underscore. So this would return true for "what is, once was." but not for "his word" There may be a problem if the search string is at the end or beginning (or both) of the line you're searching, but you can check for that with String:startsWith and String:endsWith SignatureBeware the False Authority Syndrome > Then maybe the \W construct can help. This will match any non-word > character. > > ".*\\W+" + searchString + "\\W+.*" Just to be paranoid, make that: ".*\\W+" + Pattern.quote(searchString) + "\\W+.*" Note that Pattern.quote is only available in Java 1.5. Prior to Java 1.5, it's exceedingly difficult to search for arbitrary substrings using regular expressions, and you'd be better of using String.indexOf(String) and checking the surrounding characters on your own. > There may be a problem if the search string is at the end or beginning > (or both) of the line you're searching, but you can check for that with > String:startsWith and String:endsWith Or, since you've got a Java regular expression anyway: "(^|.*\\W+)" + Pattern.quote(searchString) + "($|\\W+.*)" Signaturewww.designacourse.com The Easiest Way To Train Anyone... Anywhere. Chris Smith - Lead Software Developer/Technical Trainer MindIQ Corporation > > Then maybe the \W construct can help. This will match any non-word > > character. [quoted text clipped - 15 lines] > > Or, since you've got a Java regular expression anyway that is slow with JFC... any old fashion ways to accomplish this task? Thanks > "(^|.*\\W+)" + Pattern.quote(searchString) + "($|\\W+.*)" > [quoted text clipped - 4 lines] > Chris Smith - Lead Software Developer/Technical Trainer > MindIQ Corporation > > Then maybe the \W construct can help. This will match any non-word > > character. [quoted text clipped - 24 lines] > Chris Smith - Lead Software Developer/Technical Trainer > MindIQ Corporation regular expressiona are quite slow with jfc.... can anyone suggest a quick indexof variant or lagacy variant? > regular expressiona are quite slow with jfc.... can anyone suggest a > quick indexof variant or lagacy variant? Huh? Is this some "jfc" that I'm not familiar with? Regular expressions are no slower or faster than normal with the Java Foundation Classes (that is, Swing and some related APIs). In fact, the two have little to do with each other. In any case, Noodles Jefferson already gave you a solution without using regular expressions. You seemed no happier with that, because it didn't work precisely the way you want. If you're that unable to write your own code, perhaps its time to think about why you're involved in programming. An if statement probably won't kill you. Signaturewww.designacourse.com The Easiest Way To Train Anyone... Anywhere. Chris Smith - Lead Software Developer/Technical Trainer MindIQ Corporation > Then maybe the \W construct can help. This will match any non-word > character. [quoted text clipped - 8 lines] > (or both) of the line you're searching, but you can check for that with > String:startsWith and String:endsWith You can use boundary matches to overcome this problem. The \b construct matches a word boundary, so modifying your expression to ".*\\b" + searchString + "\\b.*" matches searchString between word boundaries, including at the beginning or the end. SignatureJaakko Kangasharju, Helsinki Institute for Information Technology Will you be my friend, please? > You can use boundary matches to overcome this problem. The \b > construct matches a word boundary, so modifying your expression to > ".*\\b" + searchString + "\\b.*" matches searchString between word > boundaries, including at the beginning or the end. So if that works, then the correct version can be written as: ".*\\b" + Pattern.quote(searchString) + "\\b.*" The Pattern.quote could technically be omitted if searchString were guaranteed to contain only word characters... but it would need to be accompanied with copious documentation to explain that fact. Signaturewww.designacourse.com The Easiest Way To Train Anyone... Anywhere. Chris Smith - Lead Software Developer/Technical Trainer MindIQ Corporation >> You can use boundary matches to overcome this problem. The \b >> construct matches a word boundary, so modifying your expression to [quoted text clipped - 8 lines] > guaranteed to contain only word characters... but it would need to be > accompanied with copious documentation to explain that fact. It seems patterns are indeed a complex subject, with lots of near-identical alternatives. Btw, anyone know how this works with strings with international content? The Pattern JavaDoc states that a word character is [a-ZA-Z_0-9], so accented characters won't work here - and I'm not even talking about non-latin script. SignatureBeware the False Authority Syndrome > > you have a sentence and you need to find wether it contains a word; for > > example "he is great " .containts("is") returns true, but "his [quoted text clipped - 16 lines] > -- > Beware the False Authority Syndrome pattern=".*[^a-z]" +"is"+"[^a-z].*"; line.matches(patern); returns few results Note: both line and is are lower case. > Use the whitespace character class: > > ".*\\sis\\s.*" doesn't find cases such as ",is" or ",is." In article <1136340802.254990.89700@g43g2000cwa.googlegroups.com>, puzzlecracker took the hamburger, threw it on the grill, and I said "Oh wow"... > you have a sentence and you need to find wether it contains a word; for > example "he is great " .containts("is") returns true, but "his [quoted text clipped - 4 lines] > > but it didnt work. Not Tested: public boolean findAWord(String aString, String aWord) { boolean hasWord = false; int i = 0; String[] sa = aString.split(); while (i < sa.length) { if (sa[i].equals(aWord)) { hasWord = true; break; } i++; } return hasWord; } SignatureNoodles Jefferson mhm31x9 Smeeter#29 WSD#30 sTaRShInE_mOOnBeAm aT HoTmAil dOt CoM NP: "The Road to Chicago" -- Thomas Newman (Road to Perdition Soundtrack) "Our earth is degenerate in these latter days, bribery and corruption are common, children no longer obey their parents and the end of the world is evidently approaching." --Assyrian clay tablet 2800 B.C. > public boolean findAWord(String aString, String aWord) { > [quoted text clipped - 15 lines] > > } Not going to work, it will only find word delemited by space such as ".. word .." but not ",word" > return hasWord; > [quoted text clipped - 12 lines] > world is evidently approaching." > --Assyrian clay tablet 2800 B.C. > > public boolean findAWord(String aString, String aWord) { > > [quoted text clipped - 35 lines] > > world is evidently approaching." > > --Assyrian clay tablet 2800 B.C. didn't work either pattern=".*\\W?"+s+"\\W.*?"; Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. Simply choose a title, complete the application form and submit it. Read more ...
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