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• MySQL2 Ruby gem will not Install 10.6

  - by Kish

  I know this has been asked several times, but I searched and I tried many different things and nothing worked. ERROR: Error installing mysql2: ERROR: Failed to build gem native extension. /Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/bin/ruby extconf.rb checking for rb_thread_blocking_region()... yes checking for mysql.h... yes checking for errmsg.h... yes checking for mysqld_error.h... yes creating Makefile make gcc -I. -I/Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/include/ruby-1.9.1/x86_64-darwin10.6.0 -I/Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/include/ruby-1.9.1/ruby/backward -I/Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/include/ruby-1.9.1 -I. -DHAVE_RB_THREAD_BLOCKING_REGION -DHAVE_MYSQL_H -DHAVE_ERRMSG_H -DHAVE_MYSQLD_ERROR_H -D_XOPEN_SOURCE -D_DARWIN_C_SOURCE -I/usr/local/mysql/include -Os -g -fno-common -fno-strict-aliasing -arch i386 -fno-common -O3 -ggdb -Wextra -Wno-unused-parameter -Wno-parentheses -Wpointer-arith -Wwrite-strings -Wno-missing-field-initializers -Wshorten-64-to-32 -Wno-long-long -fno-common -pipe -Wall -funroll-loops -o client.o -c client.c In file included from /Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/include/ruby-1.9.1/ruby.h:32, from ./mysql2_ext.h:4, from client.c:1: /Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/include/ruby-1.9.1/ruby/ruby.h:108: error: size of array ‘ruby_check_sizeof_long’ is negative /Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/include/ruby-1.9.1/ruby/ruby.h:112: error: size of array ‘ruby_check_sizeof_voidp’ is negative In file included from /Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/include/ruby-1.9.1/ruby/intern.h:29, from /Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/include/ruby-1.9.1/ruby/ruby.h:1327, from /Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/include/ruby-1.9.1/ruby.h:32, from ./mysql2_ext.h:4, from client.c:1: /Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/include/ruby-1.9.1/ruby/st.h:69: error: size of array ‘st_check_for_sizeof_st_index_t’ is negative make: *** [client.o] Error 1 Gem files will remain installed in /Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/lib/ruby/gems/1.9.1/gems/mysql2-0.2.6 for inspection. Results logged to /Users/kishinmanglani/.rvm/rubies/ruby-1.9.2-p180/lib/ruby/gems/1.9.1/gems/mysql2-0.2.6/ext/mysql2/gem_make.out

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• How do I zoom an MKMapView to the users current location without CLLocationManager?

  - by Danny Tuppeny

  With the MKMapView there's an option called "Show users current location" which will automatically show a users location on the map. I'd like to zoom into this location without adding a CLLocationManager (this seems silly if the map is already taking care of this). The problem is, I can't find a nice event that fires when the map has figured out the users location, so I don't know where to put the code that will zoom/scroll. I tried using the viewForAnnotation method like this: - (MKAnnotationView *) mapView:(MKMapView *)map viewForAnnotation:(id <MKAnnotation>)annotation { if ([annotation class] == [MKUserLocation class] ) { NSLog(@"UserLocation annotation"); return [map viewForAnnotation:annotation]; } return nil; } However it doesn't fire if the users location is off the screen. Is there a way to be notified when an MKMapView has got the user location so I can zoom in to it, or do I just have to add a CLLocationManager for this?

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• PHP, Codeigniter: How to Set Date/Time based on users timezone/location globally in a web app?

  - by Abs

  Hello all, I have just realised if I add a particular record to my MySQL database - it will have a date/time of the server and not the particular user and where they are located which means my search function by date is useless! As they will not be able to search by when they have added it in their timezone rather when it was added in the servers timezone. Is there a way in Codeigniter to globally set time and date specific to a users location (maybe using their IP) and every time I call date() or time() that users timezone is used. What I am actually asking for is probably how to make my application dependent on each users timezone? Maybe its better to store each users timezone in their profile and have a standard time (servers time) and then convert the time to for each user? Thanks all

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• Best way for a remote web app to authenticate users in my current web app?

  - by jklp

  So a bit of background, I'm working on an existing web application which has a set of users, who are able to log in via a traditional login screen with a user name and password, etc. Recently we've managed to score a client (who have their own Intranet site), who are wanting to be able to have their users log into their Intranet site, and then have their users click a link on their Intranet which redirects to our application and logs them into it automatically. I've had two suggestions on how to implement this so far: Create a URL which takes 2 parameters (which are "username" and "password") and have the Intranet site pass those parameters to us (our connection is via TLS so it's all encrypted). This would work fine, but it seems a little "hacky", and also means that the logins and passwords have to be the same on both systems (and having to write some kind of web service which can update the passwords for users - which also seems a bit insecure) Provide a token to the Intranet, so when the client clicks on a link on the Intranet, it sends the token to us, along with the user name (and no password) which means they're authenticated. Again, this sounds a bit hacky as isn't that essentially the same as providing everyone with the same password to log in? So to summarise, I'm after the following things: A way for the users who are already authenticated on the Intranet to log into our system without too much messing around, and without using an external system to authenticate, i.e. LDAP / Kerberos Something which isn't too specific to this client, and can easily be implemented by other Intranets to log in

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• Can the same ssh key be used to access two different users on the same server?

  - by Nick

  I have an new ubuntu (hardy 8.04) server, it has two users, User1 and User2. User1 is listed in sudoers. I appended my public ssh key to authorized_keys in /home/user1/.ssh/authorized_keys, changed the permissions on user1/.ssh/ to 700 and user1/.ssh/authorized_keys to 600 and both file and folder are owned my User1. Then added I User1 to sshd_config (AllowUsers User1). This works and I can login into User1 debug1: Offering public key: /Users/nick/.ssh/id_rsa debug1: Server accepts key: pkalg ssh-rsa blen 277 debug1: Authentication succeeded (publickey). debug1: channel 0: new [client-session] debug1: Entering interactive session. Last login: Mon Mar 15 09:51:01 2010 from 86.141.61.197 I then copied the authorized_keys file to /home/user2/.shh/ and changed the permissions and ownership and added User2 to AllowUsers in sshd_config (AllowUsers User1 User2). Now when I try to login to User2 it will not authenticate the same public key. debug1: Offering public key: /Users/nick/.ssh/id_rsa debug1: Authentications that can continue: publickey debug1: Trying private key: /Users/nick/.ssh/identity debug1: Trying private key: /Users/nick/.ssh/id_dsa debug1: No more authentication methods to try. Permission denied (publickey). Am I missing something fundamental about the way ssh works? Thanks in advance, Nick

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• In a SQL GROUP BY query, what value is used for the non-aggregate columns?

  - by Queencity13

  Say I've got the following data back from a SQL query: Lastname Firstname Age Anderson Jane 28 Anderson Lisa 22 Anderson Jack 37 If I want to know the age of the oldest person with the last name Anderson, I can select MAX(Age) and GROUP BY Lastname. But I also want to know the first name of that oldest person. How can I make sure that, when the Firstname values are collapsed into one row by the GROUP BY, I get the Firstname value from the same row where I got the max age?

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• Linq-to-SQL: How to perform a count on a sub-select

  - by Peter Bridger

  I'm still trying to get my head round how to use LINQ-to-SQL correctly, rather than just writing my own sprocs. In the code belong a userId is passed into the method, then LINQ uses this to get all rows from the GroupTable tables matching the userId. The primary key of the GroupUser table is GroupUserId, which is a foreign key in the Group table. /// <summary> /// Return summary details about the groups a user belongs to /// </summary> /// <param name="userId"></param> /// <returns></returns> public List<Group> GroupsForUser(int userId) { DataAccess.KINv2DataContext db = new DataAccess.KINv2DataContext(); List<Group> groups = new List<Group>(); groups = (from g in db.Groups join gu in db.GroupUsers on g.GroupId equals gu.GroupId where g.Active == true && gu.UserId == userId select new Group { Name = g.Name, CreatedOn = g.CreatedOn }).ToList<Group>(); return groups; } } This works fine, but I'd also like to return the total number of Users who are in a group and also the total number of Contacts that fall under ownership of the group. Pseudo code ahoy! /// <summary> /// Return summary details about the groups a user belongs to /// </summary> /// <param name="userId"></param> /// <returns></returns> public List<Group> GroupsForUser(int userId) { DataAccess.KINv2DataContext db = new DataAccess.KINv2DataContext(); List<Group> groups = new List<Group>(); groups = (from g in db.Groups join gu in db.GroupUsers on g.GroupId equals gu.GroupId where g.Active == true && gu.UserId == userId select new Group { Name = g.Name, CreatedOn = g.CreatedOn, // ### This is the SQL I would write to get the data I want ### MemberCount = ( SELECT COUNT(*) FROM GroupUser AS GU WHERE GU.GroupId = g.GroupId ), ContactCount = ( SELECT COUNT(*) FROM Contact AS C WHERE C.OwnerGroupId = g.GroupId ) // ### End of extra code ### }).ToList<Group>(); return groups; } }

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• PowerPoint save group as picture creates asymmetric edge, how to fix?

  - by Se Norm

  I created tons of figures for my thesis in PowerPoint and now I realized that when I try to save the grouped items (= one figure) as a picture (EMF), it somehow asymmetrically adds a border on the left and the bottom. First one is original group, second is the same pasted as a picture. Original group: Pasted as a picture: Does anyone have an idea how to fix that for a huge number of figures? I think it only started happening when I used a page size of 1m x 1m in PowerPoint to be able to zoom in more for some figures. However, I cannot not simply change the page size now as it messes up font and object sizes. Also, copying it into a smaller page and then saving as EMF doesn't do the trick. Maybe it is not related to the page size after all. Cropping every figure individually would be a lot of work, so I hope there is a different solution. I found the origin of the problem: the text label in the left bottom corner of each image (0s, 8s, 16s). I still do not understand why it is happening though, since the text label does not expand over the edge of the image (it was aligned using the align left function). It would still be great if there was an easy way to fix this, especially as I want to keep the text where it is.

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• Delphi: How to set the default project in a project group?

  - by Ian Boyd

  i have two projects in a project group: ProjectA ProjectB Whenever i open the ProjectGroup.bpg in Delphi, it always starts with the 2nd project as the active one: ProjectA ProjectB And every time i have to flip it to the the "real" project: ProjectA ProjectB How can i make ProjectA the default project that opens with the project group? ProjectGroup.bpg #------------------------------------------------------------------------------ VERSION = BWS.01 #------------------------------------------------------------------------------ !ifndef ROOT ROOT = $(MAKEDIR)\.. !endif #------------------------------------------------------------------------------ MAKE = $(ROOT)\bin\make.exe -$(MAKEFLAGS) -f$** DCC = $(ROOT)\bin\dcc32.exe $** BRCC = $(ROOT)\bin\brcc32.exe $** #------------------------------------------------------------------------------ PROJECTS = ProjectA.exe ProjectB.exe #------------------------------------------------------------------------------ default: $(PROJECTS) #------------------------------------------------------------------------------ ProjectA.exe: ProjectA.dpr $(DCC) ProjectB.exe: childfolder\ProjectB.dpr $(DCC)

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• SSH Public Key Authentication only works if active session exists before

  - by Webx10

  I have a rather strange problem with my SSH configuration. I set up my server with the help of a Remote Access Card and configured everything with a KVM viewer. So while being logged into the server via the KVM Viewer I configured SSH with only pubkey and tried to login from my local laptop. It worked fine. If I quit the KVM Session (or logout with the user in the KVM session) I cannot login via ssh anymore (pubkey denied). SSH login only works as long as the user is somewhere still logged in. Any hints what the problem might be? Console output for a failed login (all personal data exchanged): OpenSSH_6.2p2, OSSLShim 0.9.8r 8 Dec 2011 debug1: Reading configuration data /Users/mylocaluser/.ssh/config debug1: Reading configuration data /etc/ssh_config debug1: /etc/ssh_config line 20: Applying options for * debug1: /etc/ssh_config line 103: Applying options for * debug1: Connecting to 100.100.100.100 [100.100.100.100] port 12345. debug1: Connection established. debug1: identity file /Users/mylocaluser/.ssh/id_rsa type 1 debug1: identity file /Users/mylocaluser/.ssh/id_rsa-cert type -1 debug1: identity file /Users/mylocaluser/.ssh/id_dsa type -1 debug1: identity file /Users/mylocaluser/.ssh/id_dsa-cert type -1 debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_6.2 debug1: Remote protocol version 2.0, remote software version OpenSSH_6.6.1p1 Ubuntu-2ubuntu2 debug1: match: OpenSSH_6.6.1p1 Ubuntu-2ubuntu2 pat OpenSSH* debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-ctr [email protected] none debug1: kex: client->server aes128-ctr [email protected] none debug1: SSH2_MSG_KEX_DH_GEX_REQUEST(1024<1024<8192) sent debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP debug1: SSH2_MSG_KEX_DH_GEX_INIT sent debug1: expecting SSH2_MSG_KEX_DH_GEX_REPLY debug1: Server host key: RSA ab:12:23:34:45:56:67:78:89:90:12:23:34:45:56:67 debug1: Host '[100.100.100.100]:12345' is known and matches the RSA host key. debug1: Found key in /Users/mylocaluser/.ssh/known_hosts:36 debug1: ssh_rsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: Roaming not allowed by server debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey debug1: Next authentication method: publickey debug1: Offering RSA public key: /Users/mylocaluser/.ssh/id_rsa debug1: Authentications that can continue: publickey debug1: Offering RSA public key: /Users/mylocaluser/.ssh/id_rsa2 debug1: Authentications that can continue: publickey debug1: Trying private key: /Users/mylocaluser/.ssh/id_dsa debug1: No more authentication methods to try. Permission denied (publickey). Console output for a successfull login (only possible while "active session" exists): OpenSSH_6.2p2, OSSLShim 0.9.8r 8 Dec 2011 debug1: Reading configuration data /Users/mylocaluser/.ssh/config debug1: Reading configuration data /etc/ssh_config debug1: /etc/ssh_config line 20: Applying options for * debug1: /etc/ssh_config line 103: Applying options for * debug1: Connecting to 100.100.100.100 [100.100.100.100] port 12345. debug1: Connection established. debug1: identity file /Users/mylocaluser/.ssh/id_rsa type 1 debug1: identity file /Users/mylocaluser/.ssh/id_rsa-cert type -1 debug1: identity file /Users/mylocaluser/.ssh/id_dsa type -1 debug1: identity file /Users/mylocaluser/.ssh/id_dsa-cert type -1 debug1: Enabling compatibility mode for protocol 2.0 debug1: Local version string SSH-2.0-OpenSSH_6.2 debug1: Remote protocol version 2.0, remote software version OpenSSH_6.6.1p1 Ubuntu-2ubuntu2 debug1: match: OpenSSH_6.6.1p1 Ubuntu-2ubuntu2 pat OpenSSH* debug1: SSH2_MSG_KEXINIT sent debug1: SSH2_MSG_KEXINIT received debug1: kex: server->client aes128-ctr [email protected] none debug1: kex: client->server aes128-ctr [email protected] none debug1: SSH2_MSG_KEX_DH_GEX_REQUEST(1024<1024<8192) sent debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP debug1: SSH2_MSG_KEX_DH_GEX_INIT sent debug1: expecting SSH2_MSG_KEX_DH_GEX_REPLY debug1: Server host key: RSA ab:12:23:34:45:56:67:78:89:90:12:23:34:45:56:67 debug1: Host '[100.100.100.100]:12345' is known and matches the RSA host key. debug1: Found key in /Users/mylocaluser/.ssh/known_hosts:36 debug1: ssh_rsa_verify: signature correct debug1: SSH2_MSG_NEWKEYS sent debug1: expecting SSH2_MSG_NEWKEYS debug1: SSH2_MSG_NEWKEYS received debug1: Roaming not allowed by server debug1: SSH2_MSG_SERVICE_REQUEST sent debug1: SSH2_MSG_SERVICE_ACCEPT received debug1: Authentications that can continue: publickey debug1: Next authentication method: publickey debug1: Offering RSA public key: /Users/mylocaluser/.ssh/id_rsa debug1: Server accepts key: pkalg ssh-rsa blen 279 debug1: Authentication succeeded (publickey). Authenticated to 100.100.100.100 ([100.100.100.100]:12345). debug1: channel 0: new [client-session] debug1: Requesting [email protected] debug1: Entering interactive session. debug1: Sending environment. debug1: Sending env LANG = de_DE.UTF-8 Welcome to Ubuntu 14.04.1 LTS

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• How to update a table using a select group by in a second one as the data source in MySQL?

  - by Jader Dias

  I can't do this in MySQL UPDATE tableA, tableB SET tableA.column1 = SUM(tableB.column2) WHERE tableA.column3 = tableB.column4 GROUP BY tableB.column4 ; Neither can I UPDATE tableA, ( SELECT SUM(tableB.column2) sumB, tableB.column4 FROM tableB GROUP BY tableB.column4 ) t1 SET tableA.column1 = sumB WHERE tableA.column3 = column4 ; Besides it being illegal code, I think you can understand what I tried to do with the queries above. Both of them had the same intent. How can I do that in MySQL?

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• How do I programmatically create Excel row groups with the +/- box at the top of the group, not bott

  - by Alex

  I am creating an Excel sheet programmatically. I like to show group totals above each group, and for this to look best the +/- sign that opens and closes the rows should be at the top, where the total row is. Is there a way to accomplish this? TIA, Alex

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• PHPUnit+ZendFramework ::How I can test multu users session in the same time?

  - by Yosef

  Hi, How I can create test for testing multi users sessions in the same time? (PHPUnit+ZendFramework) basic tests examples that I thinking to do (I am not QA - I soory if i wrong): users logins in the same time users buy process - only 1 user can write and the rest only read. How much session the server can handle in the same time. etc.. Thanks

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• How to group two CABasicAnimation animations and kick them off at the exact same time?

  - by mystify

  I know there is some kind of animation grouping mechanism in core animation. So lets say I have two CABasicAnimation firstAnimation and secondAnimation. How would I group these and how would I kick off the group to start animating?

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• How to access items by "group" in an Outlook VB macro?

  - by Noah Yetter

  By "group" I mean the collapsible classifications that you get when you enable View-Arrange By-Show in Groups. This divides e.g. messages in a folder into Today, Yesterday, Last Week, Two Weeks Ago, and so on. What I'd like to be able to do is iterate over the messages that are currently classified within a given group. Is this possible?

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• How to construct this query? (Ordering by COUNT() and joining with users table)

  - by Andrew

  users table: id-user-other columns scores table: id-user_id-score-other columns They're are more than one rows for each user, but there's only two scores you can have. (0 or 1, == win or loss). So I want to output all the users ordered by the number of wins, and all the users ordered by the numbers of losses. I know how to do this by looping through each user, but I was wondering how to do it with one query. Any help is appreciated!

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• Help With LINQ: Mixed Joins and Specifying Default Values

  - by Corey O.

  I am trying to figure out how to do a mixed-join in LINQ with specific access to 2 LINQ objects. Here is an example of how the actual TSQL query might look: SELECT * FROM [User] AS [a] INNER JOIN [GroupUser] AS [b] ON [a].[UserID] = [b].[UserID] INNER JOIN [Group] AS [c] ON [b].[GroupID] = [c].[GroupID] LEFT JOIN [GroupEntries] AS [d] ON [a].[GroupID] = [d].[GroupID] WHERE [a].[UserID] = @UserID At the end, basically what I would like is an enumerable object full of GroupEntry objects. What am interested is the last two tables/objects in this query. I will be displaying Groups as a group header, and all of the Entries underneath their group heading. If there are no entries for a group, I still want to see that group as a header without any entries. Here's what I have so far: So from that I'd like to make a function: public void DisplayEntriesByUser(int user_id) { MyDataContext db = new MyDataContext(); IEnumberable<GroupEntries> entries = ( from user in db.Users where user.UserID == user_id join group_user in db.GroupUsers on user.UserID = group_user.UserID into a from join1 in a join group in db.Groups on join1.GroupID equals group.GroupID into b from join2 in b join entry in db.Entries.DefaultIfEmpty() on join2.GroupID equals entry.GroupID select entry ); Group last_group_id = 0; foreach(GroupEntry entry in entries) { if (last_group_id == 0 || entry.GroupID != last_group_id) { last_group_id = entry.GroupID; System.Console.WriteLine("---{0}---", entry.Group.GroupName.ToString().ToUpper()); } if (entry.EntryID) { System.Console.WriteLine(" {0}: {1}", entry.Title, entry.Text); } } } The example above does not work quite as expected. There are 2 problems that I have not been able to solve: I still seem to be getting an INNER JOIN instead of a LEFT JOIN on the last join. I am not getting any empty results, so groups without entries do not appear. I need to figure out a way so that I can fill in the default values for blank sets of entries. That is, if there is a group without an entry, I would like to have a mostly blank entry returned, except that I'd want the EntryID to be null or 0, the GroupID to be that of of the empty group that it represents, and I'd need a handle on the entry.Group object (i.e. it's parent, empty Group object). Any help on this would be greatly appreciated. Note: Table names and real-world representation were derived purely for this example, but their relations simplify what I'm trying to do.

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• How do you encourage users to fill out their profile?

  - by mattdell

  Hello, I wanted to open up the topic to discuss ways to encourage or incentivize users to fill in information in a user profile on a website, such as skills, location, organization, etc. More information in a user profile can give a website an improved capability for its users to search, network, and collaborate. Without bugging users to fill in their profiles (ie - via annoying e-mail reminders), what other ways have you guys come up with to encourage user input? Best, -Matt

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• How to you find out what group the current user belongs to via c++?

  - by ace

  Using my c++ program how can I find out what group the current user running my program belongs to? So my program need to figure out a couple of things : The current username of the user The group the user belongs to How can do the above 2 using c++ on a RedHat / Linux machine?

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• How to use Linq to select and group complex child object from a parents list.

  - by Daoming Yang

  How to use Linq to select and group complex child object from a parents list. I have an OrderList each of order object has a OrderProductVariantList(OrderLineList), and each of OrderProductVariant object has ProductVariant, and then the ProductVariant object will have a Product object which contains product information. My goal is to select and group the most popular products from the order list. Can anyone help me with this? Many thanks.

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• UI - How I can make users effectively read what my program says?

  - by Magnetic_dud

  I have a simple form that searches through the 2000+ issues of a 3rd party webcomic. (Easy, it's like xkcd: http://url/number That form is as easy as possible, is like this: What number do you want? User writes a number, clicks ok, and goes on the 3rd party website on a new tab Then, my form asks a question: "Did you find that issue memorable? Enter the name here, and we will add it to the "best issues" in home page" When the user will write the name of the issue, it is added to the database (pending moderation by me) So, I supposed this design is the easiest and convenient that users can find. Unfortunately, NONE of the users (maybe a 2% behaved correctly) will actually read what I asked. Some of the issues are offline, and gives a 404. On that issues users will write in the textbox a completely wrong title, and correctly capitalized! It's like if i would name http://xkcd.com/627/ as "The Great Adventures of Jack Smith" Users are from around all over the country, with different browsers, and have a different cookie. I cannot believe that my users will not read what I ask, it is a WHITE PAGE with a button that disappears when clicked and a textbox.... easier than that??? Maybe i should put a checkbox with "I acknowledge that this form is for submitting memorable issues, not for fun"? Oh, who will read that? Or maybe i could enable the textbox only if the user has effectively clicked the link?

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• Firefox url / link to a group of saved bookmarks?

  - by This_Is_Fun

  In Firefox you can easily save a group of tabs together. When (re-)accessing this group, the 'cascading' bookmark menu shows each individual bookmark (and under a line) it says "open all in tabs" I'm looking for a way to launch those tabs without going up through the bookmark menu. Possible options: A) Record a simple macro w/ any number of "superuser" utilities* ('A' is not the preferred option, since many 'little-macros' are hard to keep track of) b) Use Autohotkey (similar to option 'A' and more flexible once you learn the basics) c) How does Firefox load all those tabs? The info must be stored somewhere (as a type of URL??) Quick Summary: The moment I click on "open all in tabs", I am clicking on something very similar to a hyper-link. How do I find the content (exact code) of that 'hyper-link', and / or "How do I easily launch the tabs?" .. . New EDIT #1: I'm looking for a way to launch those tabs without going up through the bookmark menu, or cluttering the bookmarks toolbar which I hide anyway :o) .. . New EDIT #2: I tried to keep the question simple and not mentioning Autohotkey programming. The objective is to launch all tabs using a button on an AHK gui. When grawity said, "It's just an ordinary folder containing ordinary bookmarks," he (she) reminds me I can easily find the folder / Now how to launch to urls inside that folder? .. FYI: (Basic-level) AHK works like this: ; Open one folder ButtonWinMerge_Files: Run, C:\Program Files\WinMerge\ Return .. ; Use default web browser for one link ButtonGoogle: Run, http://google.com Return .. . Question still open: The moment I click on "open all in tabs", I am clicking on something very similar to a hyper-link. "How to 'replicate' the way Firefox launches the tabs with one click?"

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• How do I link ASP.NET membership/role users to tables in db?

  - by SnapConfig.com

  I am going to use forms authentication but I want to be able to link the asp.net users to some tables in the db for example If I have a class and students (as roles) I'll have a class students table. I'm planning to put in a Users table containing a simple int userid and ASP.NET username in there and put userid wherever I want to link the users. Does that sound good? any other options of modeling this? it does sound a bit convoluted?

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• MediaWiki: how to hide users from the user list?

  - by Dave Everitt

  I've set up Mediawiki 1.15.1 for a client who has added two users by mistake. They now want to hide these users from the user list. It seems this is done via the $wgGroupPermissions array with $wgGroupPermissions['suppress']['hideuser'] = true;, but it isn't at all clear what entry this needs for the hiding to work, or whether a new group ('hidden' or whatever) has to be created first with $wgAddGroups['bureaucrat'] = true;. For now, I've added the two users to be hidden to the 'Oversight' group which explains 'Block a username, hiding it from the public (hideuser)', but they still appear on the Special:ListUsers page. At a loss as to how the MediWiki arrays alter options displayed in the interface, so far I've added this to LocalSettings.php: $wgGroupPermissions['suppress']['hideuser'] = true; $wgAddGroups['supress'] = true; Or - since they haven't actually added anything to the wiki - could they simply be removed from the MySQL users table - although MediaWiki warns against this? Has anyone else done this successfully? Update - this is a hole in MediaWiki admin (although there are workarounds). See this thread on MediaWIki Users and the note to the reply below.

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• Cakephp beforeFind() - How do I add a JOIN condition AFTER the belongsTo association is added?

  - by michael

  I'm in Model-beforeFind($queryData), trying to add a JOIN condition to the queryData on a model which has belongsTo associations. Unfortunately, the new JOIN references a table in the belongsTo association, so it must appear AFTER the belongsTo in the query. Here is my Tagged-belongsTo association: app\plugins\tags\models\tagged.php (line 192) Array ( [Tag] => Array ( [className] => Tag [foreignKey] => tag_id [conditions] => [fields] => [order] => [counterCache] => ) [Group] => Array ( [className] => Group [foreignKey] => foreign_key [conditions] => Array ( [Tagged.model] => Group ) [fields] => [order] => [counterCache] => ) ) Here is the JOIN added in Tagged-beforeFind(), notice that the belongsTo joins have not yet been added: app\plugins\tags\models\tagged.php (line 194) Array ( [conditions] => Array ( [Tag.keyname] => europe ) [fields] => Array ( [0] => DISTINCT Group.* [1] => GroupPermission.* ) [joins] => Array ( [0] => Array ( [table] => permissions [alias] => GroupPermission [foreignKey] => [type] => INNER [conditions] => Array ( [GroupPermission.model] => Group [0] => GroupPermission.foreignId = Group.id [or] => Array ( ... ) ) ) ) [limit] => [offset] => [order] => Array ( [0] => ) [page] => 1 [group] => [callbacks] => 1 [by] => europe [model] => Group ) When I check the output, it fails with "1054: Unknown column 'Group.id' in 'on clause'" because the Permissions join appeared BEFORE the Groups join. SELECT DISTINCT `Group`.*, `GroupPermission`.* FROM `tagged` AS `Tagged` INNER JOIN permissions AS `GroupPermission` ON (`GroupPermission`.`model` = 'Group' AND `GroupPermission`.`foreignId` = `Group`.`id` AND (...)) LEFT JOIN `tags` AS `Tag` ON (`Tagged`.`tag_id` = `Tag`.`id`) LEFT JOIN `groups` AS `Group` ON (`Tagged`.`foreign_key` = `Group`.`id` AND `Tagged`.`model` = 'Group') WHERE `Tag`.`keyname` = 'europe' But this SQL (with Permissions joined moved to the end) works fine: SELECT DISTINCT `Group`.*, `GroupPermission`.* FROM `tagged` AS `Tagged` LEFT JOIN `tags` AS `Tag` ON (`Tagged`.`tag_id` = `Tag`.`id`) LEFT JOIN `groups` AS `Group` ON (`Tagged`.`foreign_key` = `Group`.`id` AND `Tagged`.`model` = 'Group') INNER JOIN permissions AS `GroupPermission` ON (`GroupPermission`.`model` = 'Group' AND `GroupPermission`.`foreignId` = `Group`.`id` AND (...)) WHERE `Tag`.`keyname` = 'europe' How do I add my join in beforeFind() after the belongsTo join?

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