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  • Why does this regular expression for sed break inside Makefile?

    - by jcrocholl
    I'm using GNU Make 3.81, and I have the following rule in my Makefile: jslint : java org.mozilla.javascript.tools.shell.Main jslint.js mango.js \ | sed 's/Lint at line \([0-9]\+\) character \([0-9]\+\)/mango.js:\1:\2/' This works fine if I enter it directly on the command line, but the regular expression does not match if I run it with "make jslint". However, it works if I replace \+ with \{1,\} in the Makefile: jslint : java org.mozilla.javascript.tools.shell.Main jslint.js mango.js \ | sed 's/Lint at line \([0-9]\{1,\}\) character \([0-9]\{1,\}\)/mango.js:\1:\2/' Is there some special meaning to \+ in Makefiles, or is this a bug?

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  • How does windows differentiate between regular exe and a .Net exe?

    - by Yogendra
    Hi All, This might be a duplicate entry. Excuse me if it is. I was asked in an interview as to how does Windows OS differentiate between regular exe and .Net exe. My reply was, when .Net exe is build, the compiler emits some information into the header. The information is PE32 or PE32+. windows verifies the header to determine if it needs to load the MSCOREE.dll which loads the CLR and executes the exe. I am not sure if my answer is correct ? Your inputs is appreciated.

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  • Can you use regular expressions in struts-config.xml?

    - by rquinn
    I'm trying to route these two url's to different Actions. We are using Struts 1.2: /abc-def/products /abc-def I tried putting this action first: <action path="/abc*/products" type="com.business.exampleAction"> <forward name="success" path="/go"/> </action> and then this one after: <action path="/abc*" type="com.business.differentExampleAction"> <forward name="success" path="/goElsewhere"/> </action> but it always goes to the second action (differentExampleAction in this case). I've tried various iterations for the , like . or (.*), but haven't found anything that actually works yet. From what I've read, it seems like the only regular-expression-like characters allowed in struts-config are the wildcard symbols (* and **), but I hope I am wrong.

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  • How to paste text and variables into a logical expression in R?

    - by Jasper
    I want to paste variables in the logical expression that I am using to subset data, but the subset function does not see them as column names when pasted (either with ot without quotes). I have a dataframe with columns named col1, col2 etc. I want to subset for the rows in which colx < 0.05 This DOES work: subsetdata<-subset(dataframe, col1<0.05) subsetdata<-subset(dataframe, col2<0.05) This does NOT work: for (k in 1:2){ subsetdata<-subset(dataframe, paste("col",k,sep="")<0.05) } for (k in 1:2){ subsetdata<-subset(dataframe, noquote(paste("col",k,sep=""))<0.05) } I can't find the answer; any suggestions?

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  • ASP.NET in VS IDE gives "Expression cannot be evaluated at this time."

    - by S Nash
    I read all of similar questions in SO but none seem to be an answer. Situation is simple. I have a ASP.NET webpage, In a VB code behind file I have the following line: iRendProvider = ddlProvider.SelectedItem.Value where iRendProvider is integer and ddlProvider is a dropdownlist. I put a breakpoint at the above line and code stops there. Now in the immdediate window in I type: ? ddlProvider.SelectedItem.Value I get "Expression cannot be evaluated at this time." If I step over iRendProvider = ddlProvider.SelectedItem.Value To the next line and I type ? iRendProvider , I see the correct value in the immediate window. Question is why this error and how to see contents of ddlProvider.

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  • how do i grab text of multiple tags in an xml feed using one xpath expression?

    - by Incognito
    Im trying to parse an xml feed that looks something like this: <item> <title>item title</title> <link>item link</link> <description>item description</description> </item> I'm trying to find an xpath expression that will retrieve all the details of each item so that each item in the feed is contained within its own array or grouped in some way. I tried using //item/* but the tags are not grouped, although they are correctly ordered. Is there anyway of doing that? edit: [ [title1, link1, desc1], [title2, link2, desc2], [title3, link3, desc3] ]

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  • Regular Expression; Find whether a line contains any word with more than X characters.

    - by Simpsoid
    Hi, I am trying to use a Validator on a ASP.NET site and need to find whether the Street Address textbox contains a valid entry. Entries with words that are longer than X characters (in this case 25, with no punctuation or spaces) will cause the HTML on a printed A4 page to not wrap properly and therefore not to confrom to certain sizes correctly pushing the margins off. For a street address I want to match that something like "201 Long Road" is valid but "235 ReallyLongAndNarrowWindingRoadBesideTheRiver Street" is invalid. Using a Microsoft .Net Regular Expression Validator I need to know what the RegEx pattern might be. I think if it does find a match the Validator will fire correctly however if there is no match the Validator won't fire and the Update button (in this case) won't fire. Since Street addresses can contain Capital Letters and numbers etc. it will need to accomodate for that and also Spaces, Commas, Semi-Colons and Colons and Hyphens are valid characters too. Any help would be greatly appreciated as I am really stuck with this problem. Thanks, David

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  • Regular Expression to recognise truncated forms of search string?

    - by Moonshield
    I'm trying to formulate a regular expression which will recognise the search term truncated by any number of characters from the right. For example, if the search term is "pickle", the regex should recognise "pi", "pick" but not "pickaxe". Initially I came up with the following: p(i(c(k(l(e)?)?)?)?)? That works perfectly, but seems a crude way of doing it. Is there a better way of doing this? I had a look around for something similar to what I want, but I'm not entirely sure what to search for.

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  • Is it possible to define XML schemata with node names specified via regular expressions?

    - by MartyIX
    Hello, I know it is probably a question against XML philosophy but still is it possible to define schemata for XML like this: <Root> <arbitrary-name-of-node> <Name></Name> <Position></Position> <!-- ... --> </arbitrary-name-of-node> <arbitrary-name-of-node> <Name></Name> <Position></Position> <!-- ... --> </arbitrary-name-of-node> </Root> where arbitrary-name-of-node matches regular expression [a-zA-Z0-9]? Thanks for an answer!

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  • TypeError: Result of expression 'printWindow' [undefined] is not an object.

    - by orangebrainer
    I'm trying to create hidden iframes in my page dynamically to load 3 other pages so that i can grab all the HTML and combine them into 1 in a new window. However i'm stuck at this. tHe frames are created fine. But whenever the javascript runs to the part of function openNewWindow() { var printWindow = window.open("",""); printWindow.document.open(); printWindow.document.write(HTMLfromFrames); printWindow.document.close(); } i get this error: TypeError: Result of expression 'printWindow' [undefined] is not an object. but if i generate a button to call this function seperately, it works fine. however i need it to run all in one click event Anybody has any idea what's wrong? Thanks!

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  • Regular Expression to get the size value of a CSS property?

    - by thiesdiggity
    I wrote a regular expression to get the size from a CSS property: /(([+-]?\d*\.?\d+(\s)*(px|em|ex|pt|in|pc|mm|cm)?)|thin|medium|thick)(\s|;|$)/i but for some reason its not working as intended. For example, when I run the following: preg_match_all('/(([+-]?\d*\.?\d+(\s)*(px|em|ex|pt|in|pc|mm|cm)?)|thin|medium|thick)(\s|;|$)/i', "border-bottom:1px solid #99999;", $matches); It outputs: 1px 99999; But I only want the 1px value returned. I understand why its returning the above but can't seem to figure out how to only return the size and not the color value. I tried using the following negative lookbehind but it's not working either: /(((?<!#\d{3}|#\d{6})[+-]?\d*\.?\d+(\s)*(px|em|ex|pt|in|pc|mm|cm)?)|thin|medium|thick)(\s|;|$)/i I want to use preg_match_all for those CSS properties that can have multiple size values (i.e. margin). Anyone have any ideas how to get this regex to return only the size values? Thanks for your help!

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  • Get part of a string of part of regular expression?

    - by Buttink
    Im trying to make a script that will go into a directory and run my own application with each file matching a regular expression, specifically "Test[0-9]*.txt". My input filenames look like this "TestXX.txt". Now, I could just use cut and chop off the Test and .txt, but how would I do this if XX wasn't predefined to be 2 numbers? What would I do if i had "Test1.txt" ... "Test10.txt"? In other words, How would I get the [0-9]* part? Just so you know, I want to be able to make a OutputXX.txt :)

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  • How do I specify the regular expression to disable the use of underscores after the '@' sign in php for email validation?

    - by Qlidnaque
    Currently the following email in the script validates even though there are no underscores indicated in the second part of the regular expression validation after the '@' sign. How do I make underscores invalid in the second part of the email? <?php $email = 'firstname.lastname@a_aa.bbb.co_m'; $regexp = "/^[^0-9][A-z0-9_]+([.][A-z0-9_]+)*[@][A-z0-9]+([.][A-z0-9]+)*[.][A-z]{2,4}$/"; if (preg_match($regexp, $email)) { echo "Email address is valid."; } else { echo "Email address is <u>not</u> valid."; } ?>

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  • Is it possible to use a back reference to specify the number of replications in a regular expression

    - by user307894
    Is it possible to use a back reference to specify the number of replications in a regular expression? foo= 'ADCKAL+2AG.+2AG.+2AG.+2AGGG+.+G+3AGGa4.' The substrings that start with '+[0-9]' followed by '[A-z]{n}.' need to be replaced with simply '+' where the variable n is the digit from earlier in the substring. Can that n be back referenced? For example (doesn't work) '+([0-9])[A-z]{/1}.' is the pattern I want replaced with "+" (that last dot can be any character and represents a quality score) so that foo should come out to ADCKAL++++G.G+. foo = 'ADCKAL+2AG.+2AG.+2AG.+2AGGG^+.+G+3AGGa4.' indelpatt = re.compile('\+([0-9])') while indelpatt.search(foo): indelsize=int(indelpatt.search(foo).group(1)) new_regex = '\+%s[ACGTNacgtn]{%s}.' % (indelsize,indelsize) newpatt=re.compile(new_regex) foo = newpatt.sub("+", foo) I'm probably missing an easier way to parse the string.

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  • How to check that a regular expression fall into infty loop or not?

    - by Sorush Rabiee
    I'm trying to learn and hopefully use regular expression with Qt. I found that some patterns cause handler to fall into loop. for example searching \b\b in a text, never will terminate. and number of these expressions is infinite (\b\b\b,\b\b\b\b and so on). I'd like to determine and control all invalid expressions. Is there an algorithm to distinguish valid and invalid expressions? If not, how can I prevent my program from falling into an infinite loop?

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  • Is using a DataSet's column Expression works in background same as manual calculation?

    - by Harikrishna
    I have one datatable which is not bindided and records are coming from the file by parsing it in the datatable dynamically every time. Now there is three columns in the datatable Marks1,Marks2 and FinalMarks. And their types is decimal. Now for making addition of columns Marks1 and Marks2 's records and store it into FinalMarks column,For that what I do is : datatableResult.Columns["FinalMarks"].Expression="Marks1+Marks2"; It's works properly. It can be done in other way also is foreach (DataRow r in datatableResult.Rows) { r["FinalMarks"]=Convert.ToDecimal(r["Marks1"])+Convert.ToDecimal(r["Marks2"]); } Is first approach same as second approach in background means is both approach same or what? EDIT: I want to know that first approach works in background as second approach.

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  • ypeError: Result of expression 'printWindow' [undefined] is not an object.

    - by orangebrainer
    I'm trying to create hidden iframes in my page dynamically to load 3 other pages so that i can grab all the HTML and combine them into 1 in a new window. However i'm stuck at this. tHe frames are created fine. But whenever the javascript runs to the part of openNewWindow(); { var printWindow = window.open("",""); printWindow.document.open(); printWindow.document.write(HTMLfromFrames); printWindow.document.close(); } i get this error: TypeError: Result of expression 'printWindow' [undefined] is not an object. but if i generate a button to call this function seperately, it works fine. however i need it to run all in one click event Anybody as any idea what's wrong? Thanks!

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  • Is it possible to define XML scheme with node names specified via regular expressions?

    - by MartyIX
    Hello, I know it is probably a question against XML philosophy but still is it possible to define scheme for XML like this: <Root> <arbitrary-name-of-node> <Name></Name> <Position></Position> <!-- ... --> </arbitrary-name-of-node> <arbitrary-name-of-node> <Name></Name> <Position></Position> <!-- ... --> </arbitrary-name-of-node> </Root> where arbitrary-name-of-node matches regular expression [a-zA-Z0-9]? Thanks for an answer!

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  • VHDL - Problem with std_logic_vector

    - by wretrOvian
    Hi, i'm coding a 4-bit binary adder with accumulator: library ieee; use ieee.std_logic_1164.all; entity binadder is port(n,clk,sh:in bit; x,y:inout std_logic_vector(3 downto 0); co:inout bit; done:out bit); end binadder; architecture binadder of binadder is signal state: integer range 0 to 3; signal sum,cin:bit; begin sum<= (x(0) xor y(0)) xor cin; co<= (x(0) and y(0)) or (y(0) and cin) or (x(0) and cin); process begin wait until clk='0'; case state is when 0=> if(n='1') then state<=1; end if; when 1|2|3=> if(sh='1') then x<= sum & x(3 downto 1); y<= y(0) & y(3 downto 1); cin<=co; end if; if(state=3) then state<=0; end if; end case; end process; done<='1' when state=3 else '0'; end binadder; The output : -- Compiling architecture binadder of binadder ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(15): No feasible entries for infix operator "xor". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(15): Type error resolving infix expression "xor" as type std.standard.bit. ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): No feasible entries for infix operator "and". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): Bad expression in right operand of infix expression "or". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): No feasible entries for infix operator "and". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): Bad expression in left operand of infix expression "or". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): Bad expression in right operand of infix expression "or". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(16): Type error resolving infix expression "or" as type std.standard.bit. ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(28): No feasible entries for infix operator "&". ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(28): Type error resolving infix expression "&" as type ieee.std_logic_1164.std_logic_vector. ** Error: C:/Modeltech_pe_edu_6.5a/examples/binadder.vhdl(39): VHDL Compiler exiting I believe i'm not handling std_logic_vector's correctly. Please tell me how? :(

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  • SQL Query - Count column values separately

    - by user575535
    I have a hard time getting a Query to work right. This is the DDL for my Tables CREATE TABLE Agency ( id SERIAL not null, city VARCHAR(200) not null, PRIMARY KEY(id) ); CREATE TABLE Customer ( id SERIAL not null, fullname VARCHAR(200) not null, status VARCHAR(15) not null CHECK(status IN ('new','regular','gold')), agencyID INTEGER not null REFERENCES Agency(id), PRIMARY KEY(id) ); Sample Data from the Tables AGENCY id|'city' 1 |'London' 2 |'Moscow' 3 |'Beijing' CUSTOMER id|'fullname' |'status' |agencyid 1 |'Michael Smith' |'new' |1 2 |'John Doe' |'regular'|1 3 |'Vlad Atanasov' |'new' |2 4 |'Vasili Karasev'|'regular'|2 5 |'Elena Miskova' |'gold' |2 6 |'Kim Yin Lu' |'new' |3 7 |'Hu Jintao' |'regular'|3 8 |'Wen Jiabao' |'regular'|3 I want to produce the following output, but i need to count separately for ('new','regular','gold') 'city' |new_customers|regular_customers|gold_customers 'Moscow' |1 |1 |1 'Beijing'|1 |2 |0 'London' |1 |1 |0

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  • Converting F# Quotations into LINQ Expressions

    - by forki23
    Hi, I can convert a quotation of type Expr<'a -> 'b> to a Linq expression via the following snippet: /// Converts a F# Expression to a LINQ Lambda let toLambda (exp:Expr) = let linq = exp.ToLinqExpression() :?> MethodCallExpression linq.Arguments.[0] :?> LambdaExpression /// Converts a Lambda quotation into a Linq Lamba Expression with 1 parameter let ToLinq (exp : Expr<'a -> 'b>) = let lambda = toLambda exp Expression.Lambda<Func<'a, 'b>>(lambda.Body, lambda.Parameters) Now I want to convert a quotation of type Expr<'a * 'b -> 'c> or maybe even Expr<'a -> 'b -> 'c> to a Linq Lambda Expression of type Expression<Func<'a,'b'c>>. How can I do this? Regards, forki

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  • Undefined Behavior and Sequence Points Reloaded

    - by Nawaz
    Consider this topic a sequel of the following topic: Previous Installment Undefined Behavior and Sequence Points Let's revisit this funny and convoluted expression (the italicized phrases are taken from the above topic *smile* ): i += ++i; We say this invokes undefined-behavior. I presume that when say this, we implicitly assume that type of i is one of built-in types. So my question is: what if the type of i is a user-defined type? Say it's type is Index which is defined later in this post (see below). Would it still invoke undefined-behavior? If yes, why? Is it not equivalent to writing i.operator+=(i.operator ++()); or even syntactically simpler i.add(i.inc());? Or, do they too invoke undefined-behavior? If no, why not? After all, the object i gets modified twice between consecutive sequence points. Please recall the rule of thumb : an expression can modify an object's value only once between consecutive "sequence points. And if i += ++i is an expression, then it must invoke undefined-behavior. If so, then it's equivalents i.operator+=(i.operator ++()); and i.add(i.inc()); must also invoke undefined-behavior which seems to be untrue! (as far as I understand) Or, i += ++i is not an expression to begin with? If so, then what is it and what is the definition of expression? If it's an expression, and at the same time, it's behavior is also well-defined, then it implies that number of sequence points associated with an expression somehow depends on the type of operands involved in the expression. Am I correct (even partly)? By the way, how about this expression? a[++i] = i; //taken from the previous topic. but here type of `i` is Index. class Index { int state; public: Index(int s) : state(s) {} Index& operator++() { state++; return *this; } Index& operator+=(const Index & index) { state+= index.state; return *this; } operator int() { return state; } Index & add(const Index & index) { state += index.state; return *this; } Index & inc() { state++; return *this; } };

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  • I need help translating this portion of the ECMAScript grammar?

    - by ChaosPandion
    I've been working on my own implementation of ECMAScript for quite some time now. I have basically done everything by hand to help gain a deep understanding of the process. Repeated attempts to analyze and understand this portion of the grammar have failed so I've been working on the run time instead. Now I am at a point were I will be working on object literals so I really need to polish my syntactic analyzer. Can anyone put this in terms a language parser novice could understand? My biggest source of confusion is the following: new MemberExpression Arguments This is supposed to be a member expression, but this seemingly conflicts with the following: NewExpression : MemberExpression new NewExpression Is a new expression a member expression or a left hand side expression? To be honest I am having trouble laying out the proper C# classes for the concrete grammar. MemberExpression : PrimaryExpression FunctionExpression MemberExpression [ Expression ] MemberExpression . IdentifierName new MemberExpression Arguments NewExpression : MemberExpression new NewExpression CallExpression : MemberExpression Arguments CallExpression Arguments CallExpression [ Expression ] CallExpression . IdentifierName LeftHandSideExpression : NewExpression CallExpression

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  • What is the RFC complicant and working regular expression to check if a string is a valid URL

    - by bestis
    There is question by the almost the same name already: What is the best regular expression to check if a string is a valid URL I don't understand this stackoverflow. It seems like I need reputation to comment an answer. As I don't have it, I don't know how to tell/ask that the proposed solution doesn't seem to work. So I'm forced to make a new question and ask for the solution this way? But that regexp seems to fail in input which has IPv6 address in it: For example facebook's IPv6 address: http://2620:0:1cfe:face:b00c::3/ Also link to localhost fails: http://::1/ Or is PHP to blame? /** * Validate URL - RFC 3987 (IRI) * * http://stackoverflow.com/questions/161738/what-is-the-best-regular-expression-to-check-if-a-string-is-a-valid-url * * @param string $str_url * @return boolean */ function is_url($str_url) { // RFC 3987 For absolute IRIs (internationalized): // @todo FIXME - Has bugs in IPv6 (http://2620:0:1cfe:face:b00c::3/) fails return (bool) preg_match('/^[a-z](?:[-a-z0-9\+\.])*:(?:\/\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:])*@)?(?:\[(?:(?:(?:[0-9a-f]{1,4}:){6}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|::(?:[0-9a-f]{1,4}:){5}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:){4}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:[0-9a-f]{1,4}:[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:){3}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,2}[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:){2}(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,3}[0-9a-f]{1,4})?::[0-9a-f]{1,4}:(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,4}[0-9a-f]{1,4})?::(?:[0-9a-f]{1,4}:[0-9a-f]{1,4}|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3})|(?:(?:[0-9a-f]{1,4}:){0,5}[0-9a-f]{1,4})?::[0-9a-f]{1,4}|(?:(?:[0-9a-f]{1,4}:){0,6}[0-9a-f]{1,4})?::)|v[0-9a-f]+[-a-z0-9\._~!\$&\'\(\)\*\+,;=:]+)\]|(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:\.(?:[0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3}|(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=@])*)(?::[0-9]*)?(?:\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))*)*|\/(?:(?:(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))+)(?:\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))*)*)?|(?:(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))+)(?:\/(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@]))*)*|(?!(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@])))(?:\?(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@])|[\x{E000}-\x{F8FF}\x{F0000}-\x{FFFFD}|\x{100000}-\x{10FFFD}\/\?])*)?(?:\#(?:(?:%[0-9a-f][0-9a-f]|[-a-z0-9\._~\x{A0}-\x{D7FF}\x{F900}-\x{FDCF}\x{FDF0}-\x{FFEF}\x{10000}-\x{1FFFD}\x{20000}-\x{2FFFD}\x{30000}-\x{3FFFD}\x{40000}-\x{4FFFD}\x{50000}-\x{5FFFD}\x{60000}-\x{6FFFD}\x{70000}-\x{7FFFD}\x{80000}-\x{8FFFD}\x{90000}-\x{9FFFD}\x{A0000}-\x{AFFFD}\x{B0000}-\x{BFFFD}\x{C0000}-\x{CFFFD}\x{D0000}-\x{DFFFD}\x{E1000}-\x{EFFFD}!\$&\'\(\)\*\+,;=:@])|[\/\?])*)?$/iu',$str_url); } Here is the test for it: $urls=array('http://www.example.org/','http://www.example.org:80/','example.org','ftp://user:[email protected]/','http://example.org/?cat=5&test=joo','http://www.fi/?cat=5&amp;test=joo','http://::1/','http://2620:0:1cfe:face:b00c::3/','http://2620:0:1cfe:face:b00c::3:80/'); foreach ($urls as $a) { echo $a."\n"; $a=is_url($a); var_dump($a); } And that outputs: > `http://www.example.org/` bool(true) > `http://www.example.org:80/` bool(true) > example.org bool(false) > `ftp://user:[email protected]/` > bool(true) > `http://example.org/?cat=5&test=joo` > bool(true) > `http://www.fi/?cat=5&amp;test=joo` > bool(true) `http://::1/` bool(false) > `http://2620:0:1cfe:face:b00c::3/` > bool(false) > `http://2620:0:1cfe:face:b00c::3:80/` > bool(false) And it also seems that stackoverflow's code is miss behaving on those :) So what is the RFC compilicant and working regexp? ps. If you close this, please then tell me how this situation should be handled? I don't think that the answer is, just earn your reputation. Who wants to do that if they cannot even tell that some proposed solution isn't working correctly. pps. "we're sorry, but as a spam prevention mechanism, new users can only post a maximum of one hyperlink. Earn more than 10 reputation to post more hyperlinks.". Oh C'mon, I'm fine with plain text :D

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  • regular expression to remove original message from reply mail using in java ?

    - by ravi ravi
    In my forum, users can reply through email. I am handling mails from their reply. When they are replying the original message getting appended. I want to get only the reply message not the original message. I have to write regular expression for gmail & hotmail. I written regex for gmail as follows : \n.wrote:(?s).--End of Post-- It is removing the original message except date. I want to remove the date also. before removing the original message : " hi 33 On Tue, May 11, 2010 at 4:18 PM, Mmmmm, Rrrrr wrote: The following update has been posted to this discussion: test as user 222 [$MESSAGE_SIGNATURE_HEADER$] --End of Post-- " When I use the above regex it is filtering as follows : " hi 33 On Tue, May 11, 2010 at 4:18 PM, Mmmmm, Rrrrr " Here i want only the actual message 'hi 33' not that date. How can I filter the date using above regex? Also I need regex for Hotmail also. I appreciate for any reply. Thanks in advance.

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