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  • How do I locate a particular word in a text file using .NET

    - by cmrhema
    I am sending mails (in asp.net ,c#), having a template in text file (.txt) like below User Name :<User Name> Address : <Address>. I used to replace the words within the angle brackets in the text file using the below code StreamReader sr; sr = File.OpenText(HttpContext.Current.Server.MapPath(txt)); copy = sr.ReadToEnd(); sr.Close(); //close the reader copy = copy.Replace(word.ToUpper(),"#" + word.ToUpper()); //remove the word specified UC //save new copy into existing text file FileInfo newText = new FileInfo(HttpContext.Current.Server.MapPath(txt)); StreamWriter newCopy = newText.CreateText(); newCopy.WriteLine(copy); newCopy.Write(newCopy.NewLine); newCopy.Close(); Now I have a new problem, the user will be adding new words within an angle, say for eg, they will be adding <Salary>. In that case i have to read out and find the word <Salary>. In other words, I have to find all the words, that are located with the angle brackets (<). How do I do that?

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  • Not-quite-JSON string deserialization in Python

    - by cpharmston
    I get the following text as a string from an XML-based REST API 'd':4 'ca':5 'sen':1 'diann':2,6,8 'feinstein':3,7,9 that I'm looking to deserialize into a pretty little Python dictionary: { 'd': [4], 'ca': [5], 'sen': [1], 'diann': [2, 6, 8], 'feinstein': [3, 7, 9] } I'm hoping to avoid using regular expressions or heavy string manipulation, as this format isn't documented and may change. The best I've been able to come up with: members = {} for m in elem.text.split(' '): m = m.split(':') members[m[0].replace("'", '')] = map(int, m[1].split(',')) return members Obviously a terrible approach, but it works, and that's better than anything else I've got right now. Any suggestions on better approaches?

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  • multiline perl search and replace (one-liner)

    - by yaya3
    I want to perform the following vim substitution as a one-liner in the terminal with perl. I would prefer to allow for any occurences of white space and/or new lines, rather than explicitly catering for them as I am below. %s/blockDontForget">\n*\s*<p><span><a\(.*\)<\/span>/blockDontForget"><p><a\1/g I've tried this: perl -pi -e 's/blockDontForget"><p><span><a(.*)<\/span>/blockDontForget"><p><a$1/msg' I presume I am misinterpreting the flags. Where am I going wrong? Thanks. EDIT: The above example is to strip the spans out of the following html: <div class="block blockDontForget"> <p><span><a href="../../../foo/bar/x/x.html">Lorem Ipsum</a></span></p> </div>

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  • Java RegExp ViewState

    - by CDSO1
    I am porting some functionality from a C++ application to java. This involves reading non-modifiable data files that contain regular expressions. A lot of the data files contain regular expressions that look similar to the following: (?<=id="VIEWSTATE".*?value=").*?(?=") These regular expressions produce the following error: "Look-behind group does not have an obvious maximum length near index XX" In C++ the engine being used supported these expressions. Is there another form of regexp that can produce the same result that can be generated using expressions like my example as input?

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  • UPDATE REGEX MYSQL

    - by Simon
    I have a table of contacts and a table of postcode data. I need to match the first part of the postcode and the join that with the postcode table... and then perform an update... I want to do something like this... UPDATE `contacts` LEFT JOIN `postcodes` ON PREG_GREP("/^[A-Z]{1,2}[0-9][0-9A-Z]{0,1}/", `contacts`.`postcode`) = `postcodes`.`postcode` SET `contacts`.`lat` = `postcode`.`lat`, `contacts`.`lng` = `postcode`.`lng` Is it possible?? Or do I need to use an external script? Many thanks.

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  • parse search string

    - by Benjamin Ortuzar
    I have search strings, similar to the one bellow: energy food "olympics 2010" Terrorism OR "government" OR cups NOT transport and I need to parse it with PHP5 to detect if the content belongs to any of the following clusters: AllWords array AnyWords array NotWords array These are the rules i have set: If it has OR before or after the word or quoted words if belongs to AnyWord. If it has a NOT before word or quoted words it belongs to NotWords If it has 0 or more more spaces before the word or quoted phrase it belongs to AllWords. So the end result should be something similar to: AllWords: (energy, food, "olympics 2010") AnyWords: (terrorism, "government", cups) NotWords: (Transport) What would be a good way to do this?

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  • which regular expression will capture this sequence?

    - by John Smith
    The text follows this pattern <tr class="text" (any sequence of characters here, except ABC)ABC(any sequence of characters here, except ABC) <tr class="text" (any sequence of characters here, except ABC)ABC(any sequence of characters here, except ABC) <tr class="text" (any sequence of characters here, except ABC)ABC(any sequence of characters here, except ABC) <tr class="text" (any sequence of characters here, except ABC)ABC(any sequence of characters here, except ABC) so basically the above line might repeat itself multiple times, and the idea is to retrieve the first 3 characters immediately after ABC. I have tried regular expressions along the lines of \<tr class="text" [.]+ABC(?<capture>[.]{3}) but they all fail. Can someone give me a hint?

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  • Problem with Javascript RegExp-mask

    - by OrjanL
    I have a string that looks something like this: {theField} > YEAR (today, -3) || {theField} < YEAR (today, +3) I want it to be replaced into: {theField} > " + YEAR (today, -3) + " || {theField} < " + YEAR (today, +3) + " I have tried this: String.replace(/(.*)(YEAR|MONTH|WEEK|DAY+)(.*[)]+)/g, "$1 \" + $2 $3 + \"") But that gives me: {theField} > YEAR (today, +3) || {theField} > " + YEAR (today, +3) + " Does anyone have any ideas?

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  • preg_replace only part of match

    - by Tony Vipros
    Hi, I'm using preg_replace to create urls for modrewrite based paging links. I use: $nextURL = preg_replace('%/([\d]+)/%','/'.($pageNumber+1).'/',$currentURL); which works fine, however I was wondering if there is a better way without having to include the '/' in the replacement parameter. I need to match the number as being between two / as the URLs can sometimes contain numbers other than the page part. These numbers are never only numbers however, so have /[\d]+/ stops them from getting replaced.

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  • In Python BeautifulSoup How to move tags

    - by JJ
    I have a partially converted XML document in soup coming from HTML. After some replacement and editing in the soup, the body is essentially - <Text...></Text> # This replaces <a href..> tags but automatically creates the </Text> <p class=norm ...</p> <p class=norm ...</p> <Text...></Text> <p class=norm ...</p> and so forth. I need to "move" the <p> tags to be children to <Text> or know how to suppress the </Text>. I want - <Text...> <p class=norm ...</p> <p class=norm ...</p> </Text> <Text...> <p class=norm ...</p> </Text> I've tried using item.insert and item.append but I'm thinking there must be a more elegant solution. for item in soup.findAll(['p','span']): if item.name == 'span' and item.has_key('class') and item['class'] == 'section': xBCV = short_2_long(item._getAttrMap().get('value','')) if currentnode: pass currentnode = Tag(soup,'Text', attrs=[('TypeOf', 'Section'),... ]) item.replaceWith(currentnode) # works but creates end tag elif item.name == 'p' and item.has_key('class') and item['class'] == 'norm': childcdatanode = None for ahref in item.findAll('a'): if childcdatanode: pass newlink = filter_hrefs(str(ahref)) childcdatanode = Tag(soup, newlink) ahref.replaceWith(childcdatanode) Thanks

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  • strange behavior in vim with negative look-behind

    - by João Portela
    So, I am doing this search in vim: /\(\(unum\)\|\(player\)=\)\@<!\"1\" and as expected it does not match lines that have: player="1" but matches lines that have: unum="1" what am i doing wrong? isn't the atom to be negated all of this: \(\(unum\)\|\(player\)=\) naturally just doing: /\(\(unum\)\|\(player\)=\) matches unum= or player=.

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  • Regexp match in Java

    - by tinti
    Regexp in Java I want to make a regexp who do this verify if a word is like [0-9A-Za-z][._-'][0-9A-Za-z] example for valid words A21a_c32 daA.da2 das'2 dsada ASDA 12SA89 non valid words dsa#da2 34$ Thanks

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  • Replaceing <a href="mailto: with just email aadress

    - by Lauri
    I want to replace all "mailto:" links in html with plain emails. In: text .... <a href="mailto:[email protected]">not needed</a> text Out: text .... [email protected] text I did this: $str = preg_replace("/\<a.+href=\"mailto:(.*)\".+\<\/a\>/", "$1", $str); But it fails if there are multiple emails in string or html inside "a" tag In: <a href="mailto:[email protected]">not needed</a><a href="mailto:[email protected]"><font size="3">[email protected]</font></a> Out: [email protected]">

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  • Invert regexp in vim

    - by Chris J
    There's a few "how do I invert a regexp" questions here on stackoverflow, but I can't find one for vim (if it does exist, by goggle-fu is lacking today). In essence I want to match all non-printable characters and delete them. I could write a short script, or drop to a shell and use tr or something similar to delete, but a vim solution would be dandy :-) Vim has the atom \p to match printable characters, however trying to do this :s/[^\p]//g to match the inverse failed and just left me with every 'p' in the file. I've seen the (?!xxx) sequence in other questions, and vim seems to not recognise this sequence. I've not found seen an atom for non-printable chars. In the interim, I'm going to drop to external tools, but if anyone's got any trick up their sleeve to do this, it'd be welcome :-) Ta!

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  • How can I use Perl regular expressions to parse XML data?

    - by Luke
    I have a pretty long piece of XML that I want to parse. I want to remove everything except for the subclass-code and city. So that I am left with something like the example below. EXAMPLE TEST SUBCLASS|MIAMI CODE <?xml version="1.0" standalone="no"?> <web-export> <run-date>06/01/2010 <pub-code>TEST <ad-type>TEST <cat-code>Real Estate</cat-code> <class-code>TEST</class-code> <subclass-code>TEST SUBCLASS</subclass-code> <placement-description></placement-description> <position-description>Town House</position-description> <subclass3-code></subclass3-code> <subclass4-code></subclass4-code> <ad-number>0000284708-01</ad-number> <start-date>05/28/2010</start-date> <end-date>06/09/2010</end-date> <line-count>6</line-count> <run-count>13</run-count> <customer-type>Private Party</customer-type> <account-number>100099237</account-number> <account-name>DOE, JOHN</account-name> <addr-1>207 CLARENCE STREET</addr-1> <addr-2> </addr-2> <city>MIAMI</city> <state>FL</state> <postal-code>02910</postal-code> <country>USA</country> <phone-number>4014612880</phone-number> <fax-number></fax-number> <url-addr> </url-addr> <email-addr>[email protected]</email-addr> <pay-flag>N</pay-flag> <ad-description>DEANESTATES2BEDS2BATHSAPPLIANCED</ad-description> <order-source>Import</order-source> <order-status>Live</order-status> <payor-acct>100099237</payor-acct> <agency-flag>N</agency-flag> <rate-note></rate-note> <ad-content> MIAMI&#47;Dean Estates&#58; 2 beds&#44; 2 baths&#46; Applianced&#46; Central air&#46; Carpets&#46; Laundry&#46; 2 decks&#46; Pool&#46; Parking&#46; Close to everything&#46;No smoking&#46; No utilities&#46; &#36;1275 mo&#46; 401&#45;578&#45;1501&#46; </ad-content> </ad-type> </pub-code> </run-date> </web-export> PERL So what I want to do is open an existing file read the contents then use regular expressions to eliminate the unnecessary XML tags. open(READFILE, "FILENAME"); while(<READFILE>) { $_ =~ s/<\?xml version="(.*)" standalone="(.*)"\?>\n.*//g; $_ =~ s/<subclass-code>//g; $_ =~ s/<\/subclass-code>\n.*/|/g; $_ =~ s/(.*)PJ RER Houses /PJ RER Houses/g; $_ =~ s/\G //g; $_ =~ s/<city>//g; $_ =~ s/<\/city>\n.*//g; $_ =~ s/<(\/?)web-export>(.*)\n.*//g; $_ =~ s/<(\/?)run-date>(.*)\n.*//g; $_ =~ s/<(\/?)pub-code>(.*)\n.*//g; $_ =~ s/<(\/?)ad-type>(.*)\n.*//g; $_ =~ s/<(\/?)cat-code>(.*)<(\/?)cat-code>\n.*//g; $_ =~ s/<(\/?)class-code>(.*)<(\/?)class-code>\n.*//g; $_ =~ s/<(\/?)placement-description>(.*)<(\/?)placement-description>\n.*//g; $_ =~ s/<(\/?)position-description>(.*)<(\/?)position-description>\n.*//g; $_ =~ s/<(\/?)subclass3-code>(.*)<(\/?)subclass3-code>\n.*//g; $_ =~ s/<(\/?)subclass4-code>(.*)<(\/?)subclass4-code>\n.*//g; $_ =~ s/<(\/?)ad-number>(.*)<(\/?)ad-number>\n.*//g; $_ =~ s/<(\/?)start-date>(.*)<(\/?)start-date>\n.*//g; $_ =~ s/<(\/?)end-date>(.*)<(\/?)end-date>\n.*//g; $_ =~ s/<(\/?)line-count>(.*)<(\/?)line-count>\n.*//g; $_ =~ s/<(\/?)run-count>(.*)<(\/?)run-count>\n.*//g; $_ =~ s/<(\/?)customer-type>(.*)<(\/?)customer-type>\n.*//g; $_ =~ s/<(\/?)account-number>(.*)<(\/?)account-number>\n.*//g; $_ =~ s/<(\/?)account-name>(.*)<(\/?)account-name>\n.*//g; $_ =~ s/<(\/?)addr-1>(.*)<(\/?)addr-1>\n.*//g; $_ =~ s/<(\/?)addr-2>(.*)<(\/?)addr-2>\n.*//g; $_ =~ s/<(\/?)state>(.*)<(\/?)state>\n.*//g; $_ =~ s/<(\/?)postal-code>(.*)<(\/?)postal-code>\n.*//g; $_ =~ s/<(\/?)country>(.*)<(\/?)country>\n.*//g; $_ =~ s/<(\/?)phone-number>(.*)<(\/?)phone-number>\n.*//g; $_ =~ s/<(\/?)fax-number>(.*)<(\/?)fax-number>\n.*//g; $_ =~ s/<(\/?)url-addr>(.*)<(\/?)url-addr>\n.*//g; $_ =~ s/<(\/?)email-addr>(.*)<(\/?)email-addr>\n.*//g; $_ =~ s/<(\/?)pay-flag>(.*)<(\/?)pay-flag>\n.*//g; $_ =~ s/<(\/?)ad-description>(.*)<(\/?)ad-description>\n.*//g; $_ =~ s/<(\/?)order-source>(.*)<(\/?)order-source>\n.*//g; $_ =~ s/<(\/?)order-status>(.*)<(\/?)order-status>\n.*//g; $_ =~ s/<(\/?)payor-acct>(.*)<(\/?)payor-acct>\n.*//g; $_ =~ s/<(\/?)agency-flag>(.*)<(\/?)agency-flag>\n.*//g; $_ =~ s/<(\/?)rate-note>(.*)<(\/?)rate-note>\n.*//g; $_ =~ s/<ad-content>(.*)\n.*//g; $_ =~ s/\t(.*)\n.*//g; $_ =~ s/<\/ad-content>(.*)\n.*//g; } close( READFILE1 ); Is there an easier way of doing this? I don't want to use any modules. I know that it might make this easier but the file I am reading has a lot of data in it.

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  • Intersection of two regular expressions

    - by Henry
    Hi, Im looking for function (PHP will be the best), which returns true whether exists string matches both regexpA and regexpB. Example 1: $regexpA = '[0-9]+'; $regexpB = '[0-9]{2,3}'; hasRegularsIntersection($regexpA,$regexpB) returns TRUE because '12' matches both regexps Example 2: $regexpA = '[0-9]+'; $regexpB = '[a-z]+'; hasRegularsIntersection($regexpA,$regexpB) returns FALSE because numbers never matches literals. Thanks for any suggestions how to solve this. Henry

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  • Bash: Extract Range with Regular Expressioin (maybe sed?)

    - by sixtyfootersdude
    I have a file that is similar to this: <many lines of stuff> SUMMARY: <some lines of stuff> END OF SUMMARY I want to extract just the stuff between SUMMARY and END OF SUMMARY. I suspect I can do this with sed but I am not sure how. I know I can modify the stuff in between with this: sed "/SUMMARY/,/END OF SUMMARY/ s/replace/with/" fileName (But not sure how to just extract that stuff). I am Bash on Solaris.

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  • Find ASCII "arrows" in text

    - by ulver
    I'm trying to find all the occurrences of "Arrows" in text, so in "<----=====><==->>" the arrows are: "<----", "=====>", "<==", "->", ">" This works: String[] patterns = {"<=*", "<-*", "=*>", "-*>"}; for (String p : patterns) { Matcher A = Pattern.compile(p).matcher(s); while (A.find()) { System.out.println(A.group()); } } but this doesn't: String p = "<=*|<-*|=*>|-*>"; Matcher A = Pattern.compile(p).matcher(s); while (A.find()) { System.out.println(A.group()); } No idea why. It often reports "<" instead of "<====" or similar. What is wrong?

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  • regular expression of 0's and 1's

    - by Lopa
    Hello all I got this question which asks me to figure out why is it foolish to write a regular expression for the language that consists of strings of 0's and 1's that are palindromes( they read the same backwards and forwards). part 2 of the question says using any formal mechanism of your choice, show how it is possible to express the language that consists of strings of 0's and 1's that are palindromes?

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  • Cleaning strings in R: add punctuation w/o overwriting last character

    - by spearmint
    I'm new to R and unable to find other threads with a similar issue. I'm cleaning data that requires punctuation at the end of each line. I am unable to add, say, a period without overwriting the final character of the line preceding the carriage return + line feed. Sample code: Data1 <- "%trn: dads sheep\r\n*MOT: hunn.\r\n%trn: yes.\r\n*MOT: ana mu\r\n%trn: where is it?" Data2 <- gsub("[^[:punct:]]\r\n\\*", ".\r\n\\*", Data1) The contents of Data2: [1] "%trn: dads shee.\r\n*MOT: hunn.\r\n%trn: yes.\r\n*MOT: ana mu\r\n%trn: where is it?" Notice the "p" of sheep was overwritten with the period. Any thoughts on how I could avoid this?

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