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• Extract multiple values from one column in MySql

  - by Neil

  I've noticed that MySql has an extensive search capacity, allowing both wildcards and regular expressions. However, I'm in somewhat in a bind since I'm trying to extract multiple values from a single string in my select query. For example, if I had the text "<span>Test</span> this <span>query</span>", perhaps using regular expressions I could find and extract values "Test" or "query", but in my case, I have potentially n such strings to extract. And since I can't define n columns in my select statement, that means I'm stuck. Is there anyway I could have a list of values (ideally separated by commas) of any text contained with span tags? In other words, if I ran this query, I would get "Test,query" as the value of spanlist: select <insert logic here> as spanlist from HtmlPages ...

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• Java Matcher groups: Understanding The difference between "(?:X|Y)" and "(?:X)|(?:Y)"

  - by user358795

  Can anyone explain: Why the two patterns used below give different results? (answered below) Why the 2nd example gives a group count of 1 but says the start and end of group 1 is -1? public void testGroups() throws Exception { String TEST_STRING = "After Yes is group 1 End"; { Pattern p; Matcher m; String pattern="(?:Yes|No)(.*)End"; p=Pattern.compile(pattern); m=p.matcher(TEST_STRING); boolean f=m.find(); int count=m.groupCount(); int start=m.start(1); int end=m.end(1); System.out.println("Pattern=" + pattern + "\t Found=" + f + " Group count=" + count + " Start of group 1=" + start + " End of group 1=" + end ); } { Pattern p; Matcher m; String pattern="(?:Yes)|(?:No)(.*)End"; p=Pattern.compile(pattern); m=p.matcher(TEST_STRING); boolean f=m.find(); int count=m.groupCount(); int start=m.start(1); int end=m.end(1); System.out.println("Pattern=" + pattern + "\t Found=" + f + " Group count=" + count + " Start of group 1=" + start + " End of group 1=" + end ); } } Which gives the following output: Pattern=(?:Yes|No)(.*)End Found=true Group count=1 Start of group 1=9 End of group 1=21 Pattern=(?:Yes)|(?:No)(.*)End Found=true Group count=1 Start of group 1=-1 End of group 1=-1

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• How to find if dataTable contains column which name starts with abc

  - by VilemRousi

  In my program I have a dataTable and I´d like to know if is there a column which name starts with abc. For example I have a DataTable and its name is abcdef. I like to find this column using something like this: DataTable.Columns.Constains(ColumnName.StartWith(abc)) Because I know only part of the column name, I cannot use a Contains method. Is there any simple way how to do that? Thanks a lot.

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• Regular expression to match any table tag

  - by keeg

  I'm trying to write a regular expression to see if a string contains any of the typical table tags: <table></table> <td></td> <th></th> <tr></tr> <thead></thead> <tfoot></tfoot> <tbody></tbody> Along with tags that may contain other attributes e.g: <table border="1"> I've come up with this so far, however, it matches <br /> tag and I'm not sure why: /<\/?[table|td|th|tr|tfoot|thead|tbody]{1,}>?/ http://www.rexfiddle.net/20Xtqka

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• Can a repeated piece of regular expression create multiple groups? Such as this example...

  - by Yousui

  Hi guys, I'm using RUBY 's regular expression to deal with text such as ${1:aaa|bbbb} ${233:aaa | bbbb | ccc ccccc } ${34: aaa | bbbb | cccccccc |d} ${343: aaa | bbbb | cccccccc |dddddd ddddddddd} ${3443:a aa|bbbb|cccccccc|d} ${353:aa a| b b b b | c c c c c c c c | dddddd} I want to get the trimed text between each pipe line. For example, for the first line of my upper example, I want to get the result aaa and bbbb, for the second line, I want aaa, bbbb and ccc ccccc. Now I have wrote a piece of regular expression and a piece of ruby code to test it: array = "${33:aaa|bbbb|cccccccc}".scan(/\$\{\s*(\d+)\s*:(\s*[^\|]+\s*)(?:\|(\s*[^\|]+\s*))+\}/) puts array Now my problem is the (?:\|(\s*[^\|]+\s*))+ part can't create multiple groups. I don't know how to solve this problem, because the number of text I need in each line is variable. Anyone can help? Great thanks.

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• Mod rewrite with multiple query strings

  - by Boris

  Hi, I'm a complete n00b when it comes to regular expressions. I need these redirects: (1) www.mysite.com/products.php?id=001&product=Product-Name&source=Source-Name should become -> www.mysite.com/Source-Name/001-Product-Name (2) www.mysite.com/stores.php?id=002&name=Store-Name should become -> www.mysite.com/002-Store-Name Any help much appreciated :)

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• How to change source order of <div> in less steps/automatically?

  - by metal-gear-solid

  How can i do this task automate. i need to change source order of div, which has same id in above 100 pages. i created example This is default condition <div class="identification"> <div class="number">Number 1</div> </div> <div class="identification"> <div class="number">Number 2</div> </div> <div class="identification"> <div class="number">Number 3</div> </div> <div class="identification"> <div class="number">Number 4</div> </div> <div class="identification"> <div class="number">Number 5</div> </div> <div class="identification"> <div class="number">Number 6</div> </div> I need lik this <div class="identification"> <div class="number">Number 1</div> </div> <div class="identification"> <div class="number">Number 3</div> </div> <div class="identification"> <div class="number">Number 2</div> </div> <div class="identification"> <div class="number">Number 6</div> </div> <div class="identification"> <div class="number">Number 4</div> </div> <div class="identification"> <div class="number">Number 5</div> </div> Is the manual editing only option? I use dreamweaver.

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• js regexp problem

  - by Alexander

  I have a searching system that splits the keyword into chunks and searches for it in a string like this: var regexp_school = new RegExp("(?=.*" + split_keywords[0] + ")(?=.*" + split_keywords[1] + ")(?=.*" + split_keywords[2] + ").*", "i"); I would like to modify this so that so that I would only search for it in the beginning of the words. For example if the string is: "Bbe be eb ebb beb" And the keyword is: "be eb" Then I want only these to hit "be ebb eb" In other words I want to combine the above regexp with this one: var regexp_school = new RegExp("^" + split_keywords[0], "i"); But I'm not sure how the syntax would look like. I'm also using the split fuction to split the keywords, but I dont want to set a length since I dont know how many words there are in the keyword string. split_keywords = school_keyword.split(" ", 3); If I leave the 3 out, will it have dynamic lenght or just lenght of 1? I tried doing a alert(split_keywords.lenght); But didnt get a desired response

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• Flex 3 Regular Expression Problem

  - by Tommy

  I've written a url validator for a project I am working on. For my requirements it works great, except when the last part for the url goes longer than 22 characters it breaks. My expression: /((https?):\/\/)([^\s.]+.)+([^\s.]+)(:\d+\/\S+)/i It expects input that looks like "http(s)://hostname:port/location". When I give it the input: https://demo10:443/111112222233333444445 it works, but if I pass the input https://demo10:443/1111122222333334444455 it breaks. You can test it out easily at http://ryanswanson.com/regexp/#start. Oddly, I can't reproduce the problem with just the relevant (I would think) part /(:\d+\/\S+)/i. I can have as many characters after the required / and it works great. Any ideas or known bugs?

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• How Do I Remove The First 4 Characters From A String If It Matches A Pattern In Ruby

  - by James

  I have the following string: "h3. My Title Goes Here" I basically want to remove the first 4 characters from the string so that I just get back: "My Title Goes Here". The thing is I am iterating over an array of strings and not all have the h3. part in front so I can't just ditch the first 4 characters blindly. I have checked the docs and the closest think I could find was chomp, but that only works for the end of a string. Right now I am doing this: "h3. My Title Goes Here".reverse.chomp(" .3h").reverse This gives me my desired output, but there has to be a better way right? I mean I don't want to reverse a string twice for no reason. I am new to programming so I might have missed something obvious, but I didn't see the opposite of chomp anywhere in the docs. Is there another method that will work? Thanks!

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• Need a regular expression for an Irish phone number

  - by Eoghan O'Brien

  I need to validate an Irish phone number but I don't want to make it too user unfriendly, many people are used to writing there phone number with brackets wrapping their area code followed by 5 to 7 digits for their number, some add spaces between the area code or mobile operator. The format of Irish landline numbers is an area code of between 1 and 4 digits and a number of between 5 to 8 digits. e.g. (021) 9876543 (01)9876543 01 9876543 (0402)39385 I'm looking for a regular expression for Javascript/PHP. Thanks.

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• Nullability (Regular Expressions)

  - by danportin

  In Brzozowski's "Derivatives of Regular Expressions" and elsewhere, the function d(R) returning ? if a R is nullable, and Ø otherwise, includes clauses such as the following: d(R1 + R2) = d(R1) + d(R2) d(R1 · R2) = d(R1) ? d(R2) Clearly, if both R1 and R2 are nullable then (R1 · R2) is nullable, and if either R1 or R2 is nullable then (R1 + R2) is nullable. It is unclear to me what the above clauses are supposed to mean, however. My first thought, mapping (+), (·), or the Boolean operations to regular sets is nonsensical, since in the base case, d(a) = Ø (for all a ? S) d(?) = ? d(Ø) = Ø and ? is not a set (nor is the return type of d, which is a regular expression). Furthermore, this mapping isn't indicated, and there is a separate notation for it. I understand nullability, but I'm lost on the definition of the sum, product, and Boolean operations in the definition of d: how are ? or Ø returned from d(R1) ? d(R2), for instance, in the definition off d(R1 · R2)?

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• PHP regular expression for positive number with 0 or 2 decimal places

  - by Peter

  Hi I am trying to use the following regular expression to check whether a string is a positive number with either zero decimal places, or 2: ^\d+(\.(\d{2}))?$ When I try to match this using preg_match, I get the error: Warning: preg_match(): No ending delimiter '^' found in /Library/WebServer/Documents/lib/forms.php on line 862 What am I doing wrong?

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• Dreamweaver regular expression substitution followed by number

  - by mark

  Hi. I'm using Dreamweaver to update copyright dates across my site. I want to preserve the existing spacing (or lack thereof) between years. Examples: © 2002-2008 should update to © 2002-2009 © 2003 - 2008 should update to © 2003 - 2009 This is the regular expression I'm using to accomplish this in Dreamweaver's find & replace function Find: ©\s*(\d{4}\s*-\s*)\d{3}[^9] Replace: © $1 2009 Here's the PROBLEM: This expression works, but has that that extra space between the hyphen and 2009. If I write the replace expression without the space, as © $12009 then dreamweaver looks for the 12,009th substitution in the find expression, and, not finding one, prints $12009. Any ideas?

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• php preg_replace, regexp

  - by Michael

  I'm trying to extract the postal codes from yell.com using php and preg_replace. I successfully extracted the postal code but only along with the address. Here is an example $URL = "http://www.yell.com/ucs/UcsSearchAction.do?scrambleSeed=17824062&keywords=shop&layout=&companyName=&location=London&searchType=advance&broaderLocation=&clarifyIndex=0&clarifyOptions=CLOTHES+SHOPS|CLOTHES+SHOPS+-+LADIES|&ooa=&M=&ssm=1&lCOption32=RES|CLOTHES+SHOPS+-+LADIES&bandedclarifyResults=1"; //get yell.com page in a string $htmlContent = $baseClass-getContent($URL); //get postal code along with the address $result2 = preg_match_all("/(.*)/", $htmlContent, $matches); print_r($matches); The above code ouputs something like Array ( [0] = Array ( [0] = 7, Royal Parade, Chislehurst, Kent BR7 6NR [1] = 55, Monmouth St, London, WC2H 9DG .... the problem that I have is that I don't know how to extract the the postal code because it doesn't have an exact number of digits (sometimes it has 6 digits and sometimes has only 5 times). Basically I should extract the lasted 2 words from each array . Thank you in advance for any help !

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• Regular Expression to remove Div tags

  - by BABA

  Hi, I have a div tag, nested with many span and div tags inside it. now i want a regular expression in javascript which will strip the div and the content inside it. thanks in advance..

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• Python RegExp exception

  - by Jasie

  How do I split on all nonalphanumeric characters, EXCEPT the apostrophe? re.split('\W+',text) works, but will also split on apostrophes. How do I add an exception to this rule? Thanks!

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• Delete all characters in a multline string up to a given pattern

  - by biffabacon

  Using Python I need to delete all charaters in a multiline string up to the first occurrence of a given pattern. In Perl this can be done using regular expressions with something like: #remove all chars up to first occurrence of cat or dog or rat $pattern = 'cat|dog|rat' $pagetext =~ s/(.*?)($pattern)/$2/xms; What's the best way to do it in Python?

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• List files with two dots in their names using java regular expressions

  - by Nivas

  I was trying to match files in a directory that had two dots in their name, something like theme.default.properties I thought the pattern .\\..\\.. should be the required pattern [. matches any character and \. matches a dot] but it matches both oneTwo.txt and theme.default.properties I tried the following: [resources/themes has two files oneTwo.txt and theme.default.properties] 1. public static void loadThemes() { File themeDirectory = new File("resources/themes"); if(themeDirectory.exists()) { File[] themeFiles = themeDirectory.listFiles(); for(File themeFile : themeFiles) { if(themeFile.getName().matches(".\\..\\..")); { System.out.println(themeFile.getName()); } } } } This prints nothing and the following File[] themeFiles = themeDirectory.listFiles(new FilenameFilter() { public boolean accept(File dir, String name) { return name.matches(".\\..\\.."); } }); for (File file : themeFiles) { System.out.println(file.getName()); } prints both oneTwo.txt theme.default.properties I am unable to find why these two give different results and which pattern I should be using to match two dots... Can someone help?

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• How does one escape backslashes and forward slashes in VIM find/search?

  - by stormist

  For instance, if I wanted to a find and replace with strings containing backward or forward slashes, how would this be accomplished in vim? Thank you! Examples Find & Replace is : :%s/foo/bar/g what if I wanted to find all occurences of <dog/> and replace it with <cat\>

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• Split string on non-alphanumerics in PHP? Is it possible with php's native function?

  - by Jehanzeb.Malik

  I was trying to split a string on non-alphanumeric characters or simple put I want to split words. The approach that immediately came to my mind is to use regular expressions. Example: $string = 'php_php-php php'; $splitArr = preg_split('/[^a-z0-9]/i', $string); But there are two problems that I see with this approach. It is not a native php function, and is totally dependent on the PCRE Library running on server. An equally important problem is that what if I have punctuation in a word Example: $string = 'U.S.A-men's-vote'; $splitArr = preg_split('/[^a-z0-9]/i', $string); Now this will spilt the string as [{U}{S}{A}{men}{s}{vote}] But I want it as [{U.S.A}{men's}{vote}] So my question is that: How can we split them according to words? Is there a possibility to do it with php native function or in some other way where we are not dependent? Regards

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• How to match exact string in a webpage in PHP?

  - by Karthik

  We can easily check for a match in a string if (preg_match("/happy/i", "happy is he who has ")) { echo "match found."; } else { echo "match not found."; } ?> But how to check for the occurrence of match in a webpage or given a url?

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• Django - urls.py - Filenames with a hash/pound (#) sign?

  - by miya

  I'm using django and realized that when the filename that the user wants to access (let's say a photo) has the pound sign, the entry in the url.py does not match. Any ideas? url(r'^static/(?P<path>.*)$', 'django.views.static.serve', {'document_root': MEDIA_ROOT}, it just says: "/home/user/project/static/upload/images/hello" does not exist when actually the name of the file is: hello#world.jpg Thanks, Nico

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• Jakarta Regexp 1.5 Backreferences?

  - by Matt Smith

  Why does this match: String str = "099.9 102.2" + (char) 0x0D; RE re = new RE("^([0-9]{3}.[0-9]) ([0-9]{3}.[0-9])\r$"); System.out.println(re.match(str)); But this does not: String str = "099.9 102.2" + (char) 0x0D; RE re = new RE("^([0-9]{3}.[0-9]) \1\r$"); System.out.println(re.match(str)); The back references don't seem to be working... What am I missing?

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• Extract a sentence out of sentences separated by delimitors

  - by Laura

  Below is a sample line I have extracted from a website: below a satisfactory level; &quot;an off year for tennis&quot;; &quot;his performance was off&quot; The output displays as: below a satisfactory level; "an off year for tennis"; "his performance was off" I want to get only the first sentence "below a satisfactory level"; Here is the code I have tried after exploring many stackoverflow posts: $data=explode('; ',$str); echo $data[0]; But somehow it is not working. Thanks in advance.

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