Search Results

Search found 5968 results on 239 pages for 'generator expression'.

Page 39/239 | < Previous Page | 35 36 37 38 39 40 41 42 43 44 45 46  | Next Page >

  • php replace regular expression

    - by Enkay
    I need to use php to add a space between a period and the next word/letter when there's none. For example, "This is a sentence.This is the next one." needs to become "This is a sentence. This is the next one." Notice the added space after the first period. My problem is that even if I'm able to make a regular expression that finds every dot followed by a letter, how do I then replace that dot with a "dot + space" and keep the letter? Also it needs to keep the case of the letter, lower or upper. Thanks for your input.

    Read the article

  • Javascript Regular expression to enforce 2 digits after decimal points

    - by Manas Saha
    I need to validate a text box where users are supposed to enter numeric values with 2 decimal places. I am validating the text in client side javascript. The validation will pass only if the number has precisely 2 decimal places, and there is at least 1 digit before the decimal point. (could be zero) the number before the decimal point can be 0, but can not be multiple zeroes. like 00 or 000 The number before the decimal point can not begin with more than 1 zero. Example of Passed validation: 0.01 0.12 111.23 1234.56 012345.67 123.00 0.00 Example of failed validation .12 1.1 0.0 00.00 1234. 1234.567 1234 00123.45 abcd.12 12a4.56 1234.5A I have tried with [0-9][.][0-9][0-9]$ But it is allowing alphabets before decimal point. like 12a4.56 Can anyone please fix the expression? thanks a lot in advance.

    Read the article

  • "Data type mismatch in criteria expression"

    - by simon
    Hey guys ! I have a problem when i want to insert values from textboxes to my access database ! When i want to save i get that error ("Data type mismatch in criteria expression") The code: string conString = "Provider=Microsoft.Jet.OLEDB.4.0;" + "Data Source=C:\Users\Simon\Desktop\test5\test5\test5\save.mdb"; OleDbConnection empConnection = new OleDbConnection(conString); string insertStatement = "INSERT INTO aktivnosti_save " + "([ID_uporabnika],[ID_aktivnosti],[kalorij]) " + "VALUES (@ID_uporabnika,@ID_aktivnosti,@kalorij)"; OleDbCommand insertCommand = new OleDbCommand(insertStatement, empConnection); insertCommand.Parameters.Add("@ID_uporabnika", OleDbType.Char).Value = textBox3.Text; insertCommand.Parameters.Add("@ID_zivila", OleDbType.Char).Value = iDTextBox.Text; insertCommand.Parameters.Add("@kalorij", OleDbType.Char).Value = textBox2.Text; empConnection.Open(); try { int count = insertCommand.ExecuteNonQuery(); } catch (OleDbException ex) { MessageBox.Show(ex.Message); } finally { empConnection.Close(); textBox1.Clear(); textBox2.Clear(); } }

    Read the article

  • Passing expression as a parameter in Call by reference

    - by darkie15
    All, When we are passing an expression as a parameter, how does the evaluation occur? Here is a small example. This is just a pseudocode kind of example: f (x,y) { y = y+1; x = x+y; } main() { a = 2; b = 2; f(a+b, a) print a; } When accessing variable x in f, does it access the address of the temp variable which contains the result of a+b or will it access the individual addresses of a and b and then evaluate the value of a+b Please help. Regards, darkie15

    Read the article

  • Binding expression failing.

    - by Rohit Kandhal
    My output window is flooded with binding expression failures like : "Cannot find source for binding with reference 'RelativeSource FindAncestor, AncestorType='System.Windows.Controls.UserControl', AncestorLevel='1''. BindingExpression:Path=DataContext.CurrentActivationObjectType; DataItem=null; target element is 'StackPanel' (Name=''); target property is 'NoTarget' (type 'Object')" and similar binding issues. Also i found that these failure occurs only when the screen loads (i.e. first time) No failures there after. I've checked the binding it is fine . Binding :

    Read the article

  • Why does this regular expression fail?

    - by Stephen
    I have a password validation script in PHP that checks a few different regular expressions, and throws a unique error message depending on which one fails. Here is an array of the regular expressions and the error messages that are thrown if the match fails: array( 'rule1' => array( '/^.*[\d].*$/i', 'Password must contain at least one number.' ), 'rule2' => array( '/^.*[a-z].*$/i', 'Password must contain at least one lowercase letter' ), 'rule3' => array( '/^.*[A-Z].*$/i', 'Password must contain at least one uppercase letter' ), 'rule4' => array( '/^.*[~!@#$%^&*()_+=].*$/i', 'Password must contain at least one special character [~!@#$%^&*()_+=]' ) ); For some reason, no matter what I pass through the validation, the "Special Characters" rule fails. I'm guessing it's a problem with the expression. If there's a better (or correct) way to write these expressions, I'm all ears!

    Read the article

  • 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.

    Read the article

  • 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?

    Read the article

  • MySQL cross table regular expression match

    - by Josef Sábl
    I have a web application and I am working on engine that analyzes referals. Now I have table with pageviews along with referes that looks something like this: pv_id referer ------------------------------------------------------------ 5531854534 http://www.google.com/search?ie=UTF-8... 8161876343 http://google.cn/search?search=human+rights 8468434831 http://search.yahoo.com/search;_... The second table contains sources definitions like: source regex ------------------------------------------------------------ Google ^https?:\/\/[^\/]*google\.([a-z]{2,4})(\/.*|)$ Yahoo ^https?:\/\/[^\/]*yahoo\.com(\/.*|)$ What I want is third table created by joinin these two: pv_id source ------------------------------------------------------------ 5531854534 Google 8161876343 Google 8468434831 Yahoo How to join these tables with regular expression?

    Read the article

  • Regular expression problem (PHP)

    - by Marcos
    Hello all. I have a little problem with my regular expression, that I use in PHP. My code identify all tags of my content and add a link in this image. My code is working when I use dinamycally, without any defined image. When I try with a imapge path, the code does not work. How can I solve this problem? Working code: $content = preg_replace('/(<img .*?src="(.+?)".*?>)/','<a class="nyromodal foto" href="'.$imagem_wordpress.'">\1</a>', $content); Problem code: $content = preg_replace('/(<img .*?src="ttp://mysite.com/files/2010/04/bac-gallery-site-matters-saline-project1.jpg".*?>)/','<a class="nyromodal foto" href="'.$imagem_wordpress.'">\1</a>', $content);

    Read the article

  • xquery expression to return a link text only if it contains within it a specific string

    - by Arvind
    I want to extract some links from a XML document (links are in same format as on html pages). Now for eg a link is "http://xyz.com/start/tyu/a.html" and another is "http://ert.com/tyu/b.html" while a third link is "http://asdf.com/ghjk/c.html" From the above 3 links (which I have with me using a for clause in a FLWOR expression)...I want only the links which contain within them a string "tyu" to be selected-- I thought of using substring for this, but substring requires start and end positions to be specified- whereas in my scenario, I dont know at which position the desired string will be. How do I do substring matching in such a scenario, i.e. where exact position for occurrence of substring, is not known? I can use XQuery 1.0 for this purpose. Finally, I want to extract the link URL, as well as link text...

    Read the article

  • 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

    Read the article

  • Detecting regular expression in content during parse

    - by sonofdelphi
    I am writing a parser for C. I was just running it with some other language files (for fun, to see the extent C-likeness). It breaks down if the code being parsed contains regular expressions... Case 1: For example, while parsing the JavaScript code snippet, var phone="(304)434-5454" phone=phone.replace(/[\(\)-]/g, "") //Returns "3044345454" (removes "(", ")", and "-") The '(', '[' etc get matched as starters of new scopes, which may never be closed. Case 2: And, for the Perl code snippet, # Replace backslashes with two forward slashes # Any character can be used to delimit the regex $FILE_PATH =~ s@\\@//@g; The // gets matched as a comment... How can I detect a regular expression within the content text of a "C-like" program-file?

    Read the article

  • Regular expression in Ruby

    - by Sainath Mallidi
    Hi, Could anybody help me make a proper regular expression from a bunch of text in Ruby. I tried a lot but I don't know how to handle variable length titles. The string will be of format <sometext>title:"<actual_title>"<sometext>. I want to extract actual_title from this string. I tried /title:"."/ but it doesnt find any matches as it expects a closing quotation after one variable from opening quotation. I couldn't figure how to make it check for variable length of string. Any help is appreciated. Thanks.

    Read the article

  • ms-access: missing operator in query expression

    - by every_answer_gets_a_point
    i have this sql statement in access: SELECT * FROM (SELECT [Occurrence Number], [1 0 Preanalytical (Before Testing)], NULL, NULL,NULL FROM [Lab Occurrence Form] WHERE NOT ([1 0 Preanalytical (Before Testing)] IS NULL) UNION SELECT [Occurrence Number], NULL, [2 0 Analytical (Testing Phase)], NULL,NULL FROM [Lab Occurrence Form] WHERE NOT ([2 0 Analytical (Testing Phase)] IS NULL) UNION SELECT [Occurrence Number], NULL, NULL, [3 0 Postanalytical ( After Testing)],NULL FROM [Lab Occurrence Form] WHERE NOT ([3 0 Postanalytical ( After Testing)] IS NULL) UNION SELECT [Occurrence Number], NULL, NULL,NULL [4 0 Other] FROM [Lab Occurrence Form] WHERE NOT ([4 0 Other] IS NULL) ) AS mySubQuery ORDER BY mySubQuery.[Occurrence Number]; everything was fine until i added the last line: SELECT [Occurrence Number], NULL, NULL,NULL [4 0 Other] FROM [Lab Occurrence Form] WHERE NOT ([4 0 Other] IS NULL) i get this error: syntax error (missing operator) in query expression 'NULL [4 0 Other]' anyone have any clues why i am getting this error?

    Read the article

  • Regular Expression to return the contents of a HTML tag received as a string of text

    - by Nathan Hernandez
    I have a string in my code that I receive that contains some html tags. It is not part of the HTML page being displayed so I cannot grab the html tag contents using the DOM (i.e. document.getElementById('tag id').firstChild.data); So, for example within the string of text would appear a tag like this: 12 My question is how would I use a regular expression to access the '12' numeric digit in this example? This quantity could be any number of digits (i.e. it is not always a double digit). I have tried some regular expressions, but always end up getting the full span tag returned along with the contents. I only want the '12' in the example above, not the surrounding tag. The id of the tags will always be 'myQty' in the string of text I receive. Thanks in advance for any help!

    Read the article

  • ASP.NET 4.0 Route expression builder inside Listview control

    - by Carlos Lone
    One of the features of ASP.NET 4.0 is Route Expression builder which allows you to set up hyperlinks like this: <asp:HyperLink runat="server" NavigateUrl="<%$ RouteUrl:RouteName=productos,categoria=Cereales,id=2 %>" >Productos</asp:HyperLink> Now I'm wondering if I can use this sort of syntax inside a ListView Control, I know is possible, but the tricky thing is that I want to genereate de route key value dynamically. So instead to write id=2 I would like to write id=<%# Eval("CategoryID") % . Can I do that?, if so, how should I write it. Thanks for your help!

    Read the article

  • python: multiline regular expression

    - by facha
    Hi, everyone I have a piece of text and I've got to parse usernames and hashes out of it. Right now I'm doing it with two regular expressions. Could I do it with just one multiline regular expression? #!/usr/bin/env python import re test_str = """ Hello, UserName. Please read this looooooooooooooooong text. hash Now, write down this hash: fdaf9399jef9qw0j. Then keep reading this loooooooooong text. Hello, UserName2. Please read this looooooooooooooooong text. hash Now, write down this hash: gtwnhton340gjr2g. Then keep reading this loooooooooong text. """ logins = re.findall('Hello, (?P<login>.+).',test_str) hashes = re.findall('hash: (?P<hash>.+).',test_str)

    Read the article

  • Result of expression 'xxxx' is not a constructor in JS

    - by Pselus
    Trying to create an object in Javascript (for Appcelerator/Titanium). The "object" is defined like this: function server () { this.cacheimages = 0; this.login = ""; this.name = ""; this.root = ""; this.signup = ""; this.useimages = 0; this.userexists = ""; this.isdefault = 0; return this; } In the same file, in another function when I run this line: var server = new server(); I get the error Result of expression 'server' is not a constructor. I have tried it with and without the "return" line, neither work. What am I doing wrong?

    Read the article

  • Use 'let' in 'if' expression

    - by demas
    I need a function that works like this: foo :: Integer -> Integer -> [Integer] foo a b = do let result = [] let Coord x y = boo a b if x > 0 let result = result ++ [3] if y > 0 let result = result ++ [5] if x < a let result = result ++ [7] if y < b let result = result ++ [9] result I can not use the guards because the result can have more then one element. But as I see I can not use 'let' in the 'if' expression: all_possible_combinations.hs:41:14: parse error on input `let' How can I check multiple expressions and add new elements in the list? I search not only imperative solution, but the functional one.

    Read the article

  • convert string to dict using list comprehension in python

    - by Pavel
    I have came across this problem a few times and can't seem to figure out a simple solution. Say I have a string string = "a=0 b=1 c=3" I want to convert that into a dictionary with a, b and c being the key and 0, 1, and 3 being their respective values (converted to int). Obviously I can do this: list = string.split() dic = {} for entry in list: key, val = entry.split('=') dic[key] = int(val) But I don't really like that for loop, It seems so simple that you should be able to convert it to some sort of list comprehension expression. And that works for slightly simpler cases where the val can be a string. dic = dict([entry.split('=') for entry in list]) However, I need to convert val to an int on the fly and doing something like this is syntactically incorrect. dic = dict([[entry[0], int(entry[1])] for entry.split('=') in list]) So my question is: is there a way to eliminate the for loop using list comprehension? If not, is there some built in python method that will do that for me?

    Read the article

  • Jointure in linq with a regular expression

    - by Graveen
    I'm actually using a join in linqtosql (via dblinq). I'm trying to include a regular expression in the join part of the linq query. from i in collectiona join j in collectionb on Regex.IsMatch(i.name, j.jokered_name) equals true (...) I agree i can push the RegExp check in the where part of the linq query, but i was wondering if it is possible in the join part ? The above code wants an "i equals j" code structure. One thing i think to perform is overriding Equals() which 'll contains the RegEx.IsMatch() stuff and put a simple i equals j in the join part. Any suggestions about my problem ?

    Read the article

  • One regular expression to match parts in various positions

    - by richsage
    Hi all, I'm trying to parse a DSN (from a Symfony application) using regular expressions, in order to link with a secondary application, but using the same database. The DSN I currently have is: mysql:dbname=my_db_name;host=localhost with a regex of: /^(\w+):(dbname=(\w+))?;?(host=(\w+))?/ (using preg_match()). This matches OK, but fails in my test environment because the DSN elements are switched around, thus: mysql:host=localhost;dbname=my_testdb_name I could just switch them round, yes :-) but I'm sure that extraction of the host and dbname parts from both DSNs is possible with a single regular expression, and I'd like to be able to enhance my knowledge at the same time ;-) Is there a way I can do this?

    Read the article

  • Regular expression match, extracting only wanted segments of string

    - by Ben Carey
    I am trying to extract three segments from a string. As I am not particularly good with regular expressions, I think what I have done could probably be done better... I would like to extract the bold parts of the following string: SOMETEXT: ANYTHING_HERE (Old=ANYTHING_HERE, New=ANYTHING_HERE) Some examples could be: ABC: Some_Field (Old=,New=123) ABC: Some_Field (Old=ABCde,New=1234) ABC: Some_Field (Old=Hello World,New=Bye Bye World) So the above would return the following matches: $matches[0] = 'Some_Field'; $matches[1] = ''; $matches[2] = '123'; So far I have the following code: preg_match_all('/^([a-z]*\:(\s?)+)(.+)(\s?)+\(old=(.+)\,(\s?)+new=(.+)\)/i',$string,$matches); The issue with the above is that it returns a match for each separate segment of the string. I do not know how to ensure the string is the correct format using a regular expression without catching and storing the match if that makes sense? So, my question, if not already clear, how I can retrieve just the segments that I want from the above string?

    Read the article

  • error: polymorphic expression with default arguments

    - by 0__
    This following bugs me: trait Foo[ A ] class Bar[ A ]( set: Set[ Foo[ A ]] = Set.empty ) This yields <console>:8: error: polymorphic expression cannot be instantiated to expected type; found : [A]scala.collection.immutable.Set[A] required: Set[Foo[?]] class Bar[ A ]( set: Set[ Foo[ A ]] = Set.empty ) ^ It is quite annoying that I have to repeat the type parameter in Set.empty. Why does the type inference fail with this default argument? The following works: class Bar[ A ]( set: Set[ Foo[ A ]] = { Set.empty: Set[ Foo[ A ]]}) Please note that this has nothing to do with Set in particular: case class Hallo[ A ]() class Bar[ A ]( hallo: Hallo[ A ] = Hallo.apply ) // nope Strangely not only this works: class Bar[ A ]( hallo: Hallo[ A ] = Hallo.apply[ A ]) ...but also this: class Bar[ A ]( hallo: Hallo[ A ] = Hallo() ) // ???

    Read the article

< Previous Page | 35 36 37 38 39 40 41 42 43 44 45 46  | Next Page >