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  • Faster jquery selector for finding a number of TD elements

    - by Bernard Chen
    I have a table where each row has 13 TD elements. I want to show and hide the first 10 of them when I toggle a link. These 10 TD elements all have an IDs with the prefix "foo" and a two digit number for its position (e.g., "foo01"). What's the fastest way to select them across the entire table? $("td:nth-child(-n+10)") or $("td[id^=foo]") or is it worth concatenating all of the ids? $("#foo01, #foo02, #foo03, #foo04, #foo05, #foo06, #foo07, #foo08, #foo09, #foo10") Is there another approach I should be considering as well?

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  • Public property List needs to Concat 2 types with inheritance

    - by Bernard
    I have 2 lists: one of type A and one of type Aa. type Aa is inherited from type A. So: List<A> listA = new List<A>(); List<Aa> listAa = new List<Aa>(); with class Aa : A I have: public property Lists<A> { get { List<A> newList = new List<A>(); //return concat of both lists foreach(List l in listA) { newList.Add(l); } foreach(List l in listAa) { newList.Add(l); } } Can I somehow use Concat instead of the foreach loop? i.e. get { return listA.Concat(listAa); } // this doesn't work And secondly, how do I do the set part of the property? set { //figure out the type of variable value and put into appropriate list? }

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  • Unknown error sourcing a script containing 'typeset -r' wrapped in command substitution

    - by Bernard Assaf
    I wish to source a script, print the value of a variable this script defines, and then have this value be assigned to a variable on the command line with command substitution wrapping the source/print commands. This works on ksh88 but not on ksh93 and I am wondering why. $ cat typeset_err.ksh #!/bin/ksh unset _typeset_var typeset -i -r _typeset_var=1 DIR=init # this is the variable I want to print When run on ksh88 (in this case, an AIX 6.1 box), the output is as follows: $ A=$(. ./typeset_err.ksh; print $DIR) $ echo $A init When run on ksh93 (in this case, a Linux machine), the output is as follows: $ A=$(. ./typeset_err.ksh; print $DIR) -ksh: _typeset_var: is read only $ print $A ($A is undefined) The above is just an example script. The actual thing I wish to accomplish is to source a script that sets values to many variables, so that I can print just one of its values, e.g. $DIR, and have $A equal that value. I do not know in advance the value of $DIR, but I need to copy files to $DIR during execution of a different batch script. Therefore the idea I had was to source the script in order to define its variables, print the one I wanted, then have that print's output be assigned to another variable via $(...) syntax. Admittedly a bit of a hack, but I don't want to source the entire sub-script in the batch script's environment because I only need one of its variables. The typeset -r code in the beginning is the error. The script I'm sourcing contains this in order to provide a semaphore of sorts--to prevent the script from being sourced more than once in the environment. (There is an if statement in the real script that checks for _typeset_var = 1, and exits if it is already set.) So I know I can take this out and get $DIR to print fine, but the constraints of the problem include keeping the typeset -i -r. In the example script I put an unset in first, to ensure _typeset_var isn't already defined. By the way I do know that it is not possible to unset a typeset -r variable, according to ksh93's man page for ksh. There are ways to code around this error. The favorite now is to not use typeset, but just set the semaphore without typeset (e.g. _typeset_var=1), but the error with the code as-is remains as a curiosity to me, and I want to see if anyone can explain why this is happening. By the way, another idea I abandoned was to grep the variable I need out of its containing script, then print that one variable for $A to be set to; however, the variable ($DIR in the example above) might be set to another variable's value (e.g. DIR=$dom/init), and that other variable might be defined earlier in the script; therefore, I need to source the entire script to make sure I all variables are defined so that $DIR is correctly defined when sourcing.

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  • How to include the login form on the Home index page in MVC

    - by Bernard Larouche
    Hi guys I really need your help for this. I am relatively new to programming and I need help to something that could be easy for a experienced programmer. I would like to get the login form that we get for free in an MVC application on the left sidebar of my Home index page instead of the usual Account/Login page. I am facing some problems. First I need a product object to be displayed on my Home Index page as well. What I did is that I added a product object to the LogOnModel that they provide in the AccountModels class and I created a UserControl (partial view) copying the content of the LogOn.aspx view. Now my Home index.aspx as well as my partial view inherits the LogOnModel class. I can see the Login form on my Home Index page as well as my product object BUT the login Form is never empty. The last username and password always appear there. I know I must have forgotten something or have done something wrong or the way did it is completely wrong !! Please could you give me some advice Thks <%@ Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<CoderForTradersSite.Models.LogOnModel>" %> <h4>Login Form</h4> <p> Please enter your username and password. <%= Html.ActionLink("Register", "Register") %> if you don't have an account. </p> <% using (Html.BeginForm()) { %> <%= Html.ValidationSummary(true, "Login was unsuccessful. Please correct the errors and try again.") %> <div> <fieldset> <legend>Account Information</legend> <div class="editor-label"> <%= Html.LabelFor(m => m.UserName) %> </div> <div class="editor-field"> <%= Html.TextBoxFor(m => m.UserName) %> <%= Html.ValidationMessageFor(m => m.UserName) %> </div> <div class="editor-label"> <%= Html.LabelFor(m => m.Password) %> </div> <div class="editor-field"> <%= Html.PasswordFor(m => m.Password) %> <%= Html.ValidationMessageFor(m => m.Password) %> </div> <div class="editor-label"> <%= Html.CheckBoxFor(m => m.RememberMe) %> <%= Html.LabelFor(m => m.RememberMe) %> </div> <p> <input type="submit" value="Log On" /> </p> </fieldset> </div> <% } %>

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  • Problem with NHibernate

    - by Bernard Larouche
    I am trying to get a list of Products that share the Category. NHibernate returns no product which is wrong. Here is my Criteria API method : public IList<Product> GetProductForCategory(string name) { return _session.CreateCriteria(typeof(Product)) .CreateCriteria("Categories") .Add(Restrictions.Eq("Name", name)) .List<Product>(); } Here is my HQL method : public IList<Product> GetProductForCategory(string name) { return _session.CreateQuery("select from Product p, p.Categories.elements c where c.Name = :name").SetString("name",name).List<Product>(); } Both methods return no product when they should return 2 products. Here is the Mapping for the Product class : <?xml version="1.0" encoding="utf-8" ?> <hibernate-mapping xmlns="urn:nhibernate-mapping-2.2" assembly="CBL.CoderForTraders.DomainModel" namespace="CBL.CoderForTraders.DomainModel"> <class name="Product" table="Products" > <id name="_persistenceId" column="ProductId" type="Guid" access="field" unsaved-value="00000000-0000-0000-0000-000000000000"> <generator class="assigned" /> </id> <version name="_persistenceVersion" column="RowVersion" access="field" type="int" unsaved-value="0" /> <property name="Name" column="ProductName" type="String" not-null="true"/> <property name="Price" column="BasePrice" type="Decimal" not-null="true" /> <property name="IsTaxable" column="IsTaxable" type="Boolean" not-null="true" /> <property name="DefaultImage" column="DefaultImageFile" type="String"/> <bag name="Descriptors" table="ProductDescriptors"> <key column="ProductId" foreign-key="FK_Product_Descriptors"/> <one-to-many class="Descriptor"/> </bag> <bag name="Categories" table="Categories_Products" > <key column="ProductId" foreign-key="FK_Products_Categories"/> <many-to-many class="Category" column="CategoryId"></many-to-many> </bag> <bag name="Orders" generic="true" table="OrderProduct" > <key column="ProductId" foreign-key="FK_Products_Orders"/> <many-to-many column="OrderId" class="Order" /> </bag> </class> </hibernate-mapping> And finally the mapping for the Category class : <?xml version="1.0" encoding="utf-8" ?> <hibernate-mapping xmlns="urn:nhibernate-mapping-2.2" assembly="CBL.CoderForTraders.DomainModel" namespace="CBL.CoderForTraders.DomainModel" default-access="field.camelcase-underscore" default-lazy="true"> <class name="Category" table="Categories" > <id name="_persistenceId" column="CategoryId" type="Guid" access="field" unsaved-value="00000000-0000-0000-0000-000000000000"> <generator class="assigned" /> </id> <version name="_persistenceVersion" column="RowVersion" access="field" type="int" unsaved-value="0" /> <property name="Name" column="Name" type="String" not-null="true"/> <property name="IsDefault" column="IsDefault" type="Boolean" not-null="true" /> <property name="Description" column="Description" type="String" not-null="true" /> <many-to-one name="Parent" column="ParentID"></many-to-one> <bag name="SubCategories" inverse="true"> <key column="ParentID" foreign-key="FK_Category_ParentCategory" /> <one-to-many class="Category"/> </bag> <bag name="Products" table="Categories_Products"> <key column="CategoryId" foreign-key="FK_Categories_Products" /> <many-to-many column="ProductId" class="Product"></many-to-many> </bag> </class> </hibernate-mapping> Can you see what could be the problem ?

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  • Extend the bootstrap-typeahead in order to take an object instead of a string

    - by Lorraine Bernard
    _.extend($.fn.typeahead.Constructor.prototype, { render: function (items) { var that = this; items = $(items).map(function (i, item) { i = $(that.options.item) .attr('data-value', item[that.options.display]) .attr('data-id', item.id); i.find('a').html(that.highlighter(item)); return i[0]; }); items.first().addClass('active'); this.$menu.html(items); return this; }, select: function () { var val = this.$menu.find('.active').attr('data-value'), id = this.$menu.find('.active').attr('data-id'); this.$element .val(this.updater(val, id)) .change(); return this.hide() } }); return function (element, options) { var getSource = function () { var users = app.userCollection.filter(function (model) { if (options && options.excludeCurrentUser) { return model.id !== app.currentUser.id; } }); return _.map(users, function (user) { return { id: user.get('id'), full_name: user.get('first_name') + ' ' + user.get('last_name') }; }); }; element.typeahead({ minLength: 3, source: getSource, display: 'full_name', sorter: function (items) { var beginswith = [], caseSensitive = [], caseInsensitive = [], item, itemDisplayed; while (item = items.shift()) { itemDisplayed = item[this.options.display]; if (!itemDisplayed.toLowerCase().indexOf(this.query.toLowerCase())) { beginswith.push(item); } else if (~itemDisplayed.indexOf(this.query)) { caseSensitive.push(item); } else { caseInsensitive.push(item); } } return beginswith.concat(caseSensitive, caseInsensitive); }, highlighter: function (item) { var query = this.query.replace(/[\-\[\]{}()*+?.,\\\^$|#\s]/g, '\\$&'); return item[this.options.display].replace(new RegExp('(' + query + ')', 'ig'), function ($1, match) { return '<strong>' + match + '</strong>'; }); }, matcher: function (item) { var value = item[this.options.display]; return { value: ~value.toLowerCase().indexOf(this.query.toLowerCase()), id: item.id }; }, updater: function (item, userId) { options.hiddenInputElement.val(userId); return item; } }); };

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  • How do I flag a folder as being a package?

    - by Pierre Bernard
    I used to think that folders needed to have an extension so that they are recognized as packages by the Finder. That extension would be declared in the owning application's Info.plist. Obviously there is another, more elegant way, but I can't figure out how it is done. E.g. the iPhoto Library is being treated as a package by the Finder. Yet it has no extension. mdls reveals that it indeed has "com.apple.package" in the content type tree. The actual content type is dynamically assigned. How did iPhoto go about to create such a directory?

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  • Leveraging hobby experience to get a job

    - by Bernard
    Like many other's I began programming at an early age. I started when I was 11 and I learned C when I was 14 (now 26). While most of what I did were games just to entertain myself I did everything from low level 2D graphics, and binary I/O, to interfacing with free API's, custom file systems, audio, 3D animations, OpenGL, web sites, etc. I worked on a wide variety of things trying to make various games. Because of this experience I have tested out of every college level C/C++ programming course I have ever been offered. In the classes I took, my classmates would need a week to do what I finished in class with an hour or two of work. I now have my degree now and I have 2 years of experience working full time as a web developer however I would like to get back into C++ and hopefully do simulation programming. Unfortunately I have yet to do C++ as a job, I have only done it for testing out of classes and doing my senior project in college. So most of what I have in C++ is still hobby experience and I don't know how to best convey that so that I don't end up stuck doing something too low level for me. Right now I see a job offer that requires 2 years of C++ experience, but I have at least 9 (I didn't do C++ everyday for the last 14 years). How do I convey my experience? How much is it truly worth? and How do I get it's full value? The best thing that I can think of is a demo and a portfolio, however that only comes into play after an interview has been secured. I used a portfolio to land my current job. All answers and advice are appreciated.

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  • subversion 1.6.x losing changes on check-in

    - by Bernard
    I'm trying to figure out if this is a known issue with SVN 1.6.x Developer A modifies a file and commits it. Developer B modifies the same file. Tries to commit it and gets told local copy out of date so does an update and then a commit. However the changes from Developer A are lost so the resulting file only contains the version that Developer B checked in. We can see this in the logs. It seems to happen when the same file is modified but in different places. Anyone else experienced this? We've had it happen 4 or 5 times in the past few weeks and we've lost a half day or so each time trying to figure out what's been lost, etc. We're starting to lose confidence in SVN. Should we be thinking of moving to GIT or Mecurial? Would that sort out this problem?

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  • Is there a difference between `==` and `is` in python?

    - by Bernard
    My Google-fu has failed me. In Python, are these: n = 5 # Test one. if n == 5: print 'Yay!' # Test two. if n is 5: print 'Yay!' two tests for equality equivalent (ha!)? Does this hold true for objects where you would be comparing instances (a list say)? Okay, so this kind of answers my question: l = list() l.append(1) if l == [1]: print 'Yay!' # Holds true, but... if l is [1]: print 'Yay!' # Doesn't. So == tests value where is tests to see if they are the same object?

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  • Getting access to a custom Master page from a user control

    - by Bernard
    Hi We have created a Master page that inherits off the asp.net Master class. We have also got ui controls that inherit off the standard asp.net ui control class. Our Master page has a public member variable. We need to be able to access that member variable from the ui controls that we use. However we can't seem to get at it? Is it our architecture that is wrong? Or the idea itself - user control getting acces to Master page variables?

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  • select xml data column into flat reporting table

    - by Bernard
    We have a xml column in SQL Server 2008. We need to do reporting off the data in the xml so we're going to select the xml into a flat table. The flat table has columns that correspond to various nodes in the xml. What is the best way to do this using SSIS? Is this a good approach? Or should we just try and write the reports directly off the xml column?

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  • What is the command-line input to produce the javadoc?

    - by Bernard
    After writing all the comments inside the code about the javadoc such as /** * This method compares the student's answer to the standard answer * @param ans The student's answer * @return True for correct answer; False for incorrect answer */ boolean compareAnswer(int ans); I guess it starts with : javadoc [optionss] [packages|files] I'm not sure what is the regular or default [option] and how can I say to produce it in my current home directory?

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  • nginx reload failing: `object version does not match bootstrap parameter`

    - by Jean Jordaan
    I added a server stanza to my virtual.conf, and now nginx seems to have a problem reloading the config. At this point I don't know what exactly is going wrong or how to debug better. Any help would be most appreciated. The config test succeeds: root@server:~# service nginx configtest nginx: the configuration file /etc/nginx/nginx.conf syntax is ok nginx: configuration file /etc/nginx/nginx.conf test is successful I'm tailing the logfile. Upon reload, the following error is logged. As far as I can see, the new config is not used. root@server:~# service nginx reload Reloading nginx: [ OK ] root@server:~# ==> /var/log/nginx/error.log <== nginx object version 0.8.54 does not match bootstrap parameter 1.0.15 at /usr/lib64/perl5/XSLoader.pm line 94. Compilation failed in require. BEGIN failed--compilation aborted. 2012/10/18 12:31:07 [alert] 9620#0: perl_parse() failed: 2 This is the version of nginx I'm running: root@server:~# yum info nginx Loaded plugins: fastestmirror, presto Loading mirror speeds from cached hostfile * base: ftp.udc.es * epel: mirror.nl.leaseweb.net * extras: ftp.udc.es * updates: ftp.cica.es Installed Packages Name : nginx Arch : x86_64 Version : 1.0.15 Release : 2.el6 [...] Server OS: CentOS release 6.3 (Final)

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  • bootmgr is missing

    - by jean
    I have a Toshiba and it is windows 7 and as soon as i turn my computer on it says bootmgr is missing. the only thing i can get into is the setup menu. does anyone know what might be wrong because my step brother thinks that it might be that everything was erased off my hard drive. the last thing he did when he used it was the toshiba updates and restarted the computer. so if any one knows what might be wrong or how i can get my computer up and running please let me know.

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  • Migrating from Apache2 to Lighttpd creating errors in PHP/mySQL?

    - by Jean-Philippe Murray
    Ok, I've been using basics ubuntu LAMP setups for years now, and I wanted to give lighttpd a try. My LAMP setup run in a virtual machine with scripts running just fine. So I created a new virtual machine, starting with a fresh install of ubuntu and made my setups. On this new VM, lighttpd + php works just fine. (Or at least it seems...) Problem occurs when I take the scripts from my LAMP setup and upload them to the new VM. I'm getting : Warning: mysql_real_escape_string(): Access denied for user 'www-data'@'localhost' (using password: NO) My lighttpd setup is configured as php-cgi but not my apache2 setup. Could this be the source of the problem? I think that scripts would be independent of the server configuration, so I doubt it. Also, I know that my DB connexion informations are good (as I can log in via phpmyadmin perfectly). I'm in the dark here, any pointers ? Thanks,

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  • I am unable to get the subdomain from the URL in NGINX

    - by Jean-Nicolas Boulay Desjardins
    I am unable to get the subdomain from the URL in NGINX. Here is my config: server { listen 80; server_name ~^(?<appname>)\.example\.com$; rewrite ^ https://$appname.example.com$request_uri? permanent; } When I do: http://bob.example.com/ I am sent to: https://.example.com/ I don't know what I am doing wrong. I am using NGiNX 1.2.7. I have another config for the: http://example.com/ So I have one server block for the domain without the subdomain and the second with the subdomain... This is about the subdomain.

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  • Can't upgrade my Ubuntu server, it gets stuck on openjdk-6-jre-headless

    - by Jean-Nicolas Boulay Desjardins
    I am using Ubuntu Server. When I do: apt-get upgrade it gets stuck on: Setting up openjdk-6-jre-headless (6b20-1.9.7-0ubuntu1) ... Why? And what can I do to stop it? I tried removing it with apt-get... I get this error: E: dpkg was interrupted, you must manually run 'sudo dpkg --configure -a' to correct the problem. So then I tried this: dpkg --purge openjdk-6-jre-headless I got this: dpkg: dependency problems prevent removal of openjdk-6-jre-headless: openjdk-6-jre-lib depends on openjdk-6-jre-headless (>= 6b17). ca-certificates-java depends on openjdk-6-jre-headless (>= 6b16-1.6.1-2) | java6-runtime-headless; however: Package openjdk-6-jre-headless is to be removed. Package java6-runtime-headless is not installed. Package openjdk-6-jre-headless which provides java6-runtime-headless is to be removed. ca-certificates-java depends on openjdk-6-jre-headless (>= 6b16-1.6.1-2) | java6-runtime-headless; however: Package openjdk-6-jre-headless is to be removed. Package java6-runtime-headless is not installed. Package openjdk-6-jre-headless which provides java6-runtime-headless is to be removed. dpkg: error processing openjdk-6-jre-headless (--purge): dependency problems - not removing Errors were encountered while processing: openjdk-6-jre-headless The thing is I think my DB is using it... Not sure... I am using Cassandra with Thrift... Yes, it's getting a bit more complex... # dpkg --configure -a I get: dpkg: dependency problems prevent configuration of openjdk-6-jre: openjdk-6-jre depends on openjdk-6-jre-headless (>= 6b20-1.9.7-0ubuntu1); however: Package openjdk-6-jre-headless is not configured yet. dpkg: error processing openjdk-6-jre (--configure): dependency problems - leaving unconfigured Processing triggers for libc-bin ... ldconfig deferred processing now taking place dpkg: dependency problems prevent configuration of libaccess-bridge-java: libaccess-bridge-java depends on default-jre | openjdk-6-jre | sun-java6-jre; however: Package default-jre is not installed. Package openjdk-6-jre is not configured yet. Package sun-java6-jre is not installed. dpkg: error processing libaccess-bridge-java (--configure): dependency problems - leaving unconfigured dpkg: dependency problems prevent configuration of icedtea-6-jre-cacao: icedtea-6-jre-cacao depends on openjdk-6-jre-headless (= 6b20-1.9.7-0ubuntu1); however: Package openjdk-6-jre-headless is not configured yet. dpkg: error processing icedtea-6-jre-cacao (--configure): dependency problems - leaving unconfigured dpkg: dependency problems prevent configuration of libaccess-bridge-java-jni: libaccess-bridge-java-jni depends on libaccess-bridge-java (>= 1.26.2-5); however: Package libaccess-bridge-java is not configured yet. dpkg: error processing libaccess-bridge-java-jni (--configure): dependency problems - leaving unconfigured Errors were encountered while processing: openjdk-6-jre libaccess-bridge-java icedtea-6-jre-cacao libaccess-bridge-java-jni Thanks again for any help.

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  • Existing connexion on Apache and mod_proxy_balancer don't fail over second JBoss node

    - by Jean-Rémy Revy
    I have a Jboss farm, load balanced by Apache HTTP + mod_proxy_balancer and mod_proxy_ajp, with the following configuration : <VirtualHost *:80> ServerName web-gui-acceptance.myorg.com ServerAlias web-gui-acceptance ProxyRequests Off ProxyPass /web-gui balancer://jbosscluster/web-gui stickysession=JSESSIONID nofailover=On ProxyPassReverse /web-gui http://srvlnx01.myorg.com:8080/web-gui ProxyPassReverse /web-gui http://srvlnx02.myorg.com:8080/web-gui <Proxy *> AuthType Kerberos [...] </Proxy> <Proxy balancer://jbosscluster> BalancerMember ajp://srvlnx01.myorg.com:8009 route=SRVLNX01_node1 BalancerMember ajp://srvlnx01.myorg.com:8009 route=SRVLNX02_node1 ProxySet lbmethod=byrequests </Proxy> </VirtualHost> When the first JBoss node fail (the hosting VM is down), my existing connexions don't fail over the second node ... the fist route is keeped (in table / .shm ?) and that provide me 503 errors. Can someone tell me what I missed ?

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  • Errors related to python version added to error log when I start apache2

    - by Jean-Nicolas Boulay Desjardins
    When I start apache I am getting those errors: [Tue Jun 14 02:28:58 2011] [error] python_init: Python version mismatch, expected '2.6.5', found '2.6.6'. [Tue Jun 14 02:28:58 2011] [error] python_init: Python executable found '/usr/bin/python'. [Tue Jun 14 02:28:58 2011] [error] python_init: Python path being used '/usr/lib/python2.6/:/usr/lib/python2.6/plat-linux2:/usr/lib/python2.6/lib-tk:/usr/lib/python2.6/lib-old:/usr/lib/python2.6/lib-dynload'. [Tue Jun 14 02:28:58 2011] [notice] mod_python: Creating 8 session mutexes based on 150 max processes and 0 max threads. [Tue Jun 14 02:28:58 2011] [notice] mod_python: using mutex_directory /tmp [Tue Jun 14 02:28:58 2011] [notice] Apache/2.2.16 (Ubuntu) PHP/5.3.3-1ubuntu9.5 with Suhosin-Patch mod_python/3.3.1 Python/2.6.6 configured -- resuming normal operations I am using Ubuntu Server... Thanks in advance for any help.

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