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  • Is Family Tree Maker 2011 the right upgrade for a FTM 11 user?

    - by bill weaver
    My father has used Family Tree Maker for years, but hasn't upgraded since version 11. It is difficult to tell from reviews at Amazon and other places whether upgrading to FTM 2011 is a good choice. File incompatibilities and upgrade woes sound like customer service is lacking, and i've read reports of it uploading your data to their database but then trying to sell you a download of data. Looking at the ancestry.com site makes me think it's solely about selling add-ons and upgrades. On the other hand, the feature set seems fairly rich and the software has a pretty strong following. I was able to get Gramps working on my system, but that's not going to work for my dad. Any advice on a good upgrade path?

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  • Why I get errors when I try to out a compiler defined macro using a pragma message?

    - by bogdan
    I would like to know why the Visual C++ compiler gets me an warning/error if I use the following code: #pragma message( "You have " _MSC_FULL_VER ) Here is what I get: error C2220: warning treated as error - no 'object' file generated warning C4081: expected ':'; found ')' The problem reproduces for _MSC_FULL_VER or _MSV_VER but not if I try to use others like __FILE__ or __DATE__. These macros are defined, they are documented on msdn

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  • Connection for control user as defined in your configuration failed. xampp

    - by Zann
    when i tried to uninstall xampp and reinstall xampp.I received below error message when i go phpmyadmin Need help and guide to solve it .thanks Error MySQL said: Documentation 1045 - Access denied for user 'root'@'localhost' (using password: NO) Connection for controluser as defined in your configuration failed. phpMyAdmin tried to connect to the MySQL server, and the server rejected the connection. You should check the host, username and password in your configuration and make sure that they correspond to the information given by the administrator of the MySQL server.

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  • mysql subquery strangely slow

    - by aviv
    I have a query to select from another sub-query select. While the two queries look almost the same the second query (in this sample) runs much slower: SELECT user.id ,user.first_name -- user.* FROM user WHERE user.id IN (SELECT ref_id FROM education WHERE ref_type='user' AND education.institute_id='58' AND education.institute_type='1' ); This query takes 1.2s Explain on this query results: id select_type table type possible_keys key key_len ref rows Extra 1 PRIMARY user index first_name 152 141192 Using where; Using index 2 DEPENDENT SUBQUERY education index_subquery ref_type,ref_id,institute_id,institute_type,ref_type_2 ref_id 4 func 1 Using where The second query: SELECT -- user.id -- user.first_name user.* FROM user WHERE user.id IN (SELECT ref_id FROM education WHERE ref_type='user' AND education.institute_id='58' AND education.institute_type='1' ); Takes 45sec to run, with explain: id select_type table type possible_keys key key_len ref rows Extra 1 PRIMARY user ALL 141192 Using where 2 DEPENDENT SUBQUERY education index_subquery ref_type,ref_id,institute_id,institute_type,ref_type_2 ref_id 4 func 1 Using where Why is it slower if i query only by index fields? Why both queries scans the full length of the user table? Any ideas how to improve? Thanks.

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  • SQLAlchemy & Complex Queries

    - by user356594
    I have to implement ACL for an existing application. So I added the a user, group and groupmembers table to the database. I defined a ManyToMany relationship between user and group via the association table groupmembers. In order to protect some ressources of the app (i..e item) I added a additional association table auth_items which should be used as an association table for the ManyToMany relationship between groups/users and the specific item. item has following columns: user_id -- user table group_id -- group table item_id -- item table at least on of user_id and group_id columns are set. So it's possible to define access for a group or for a user to a specific item. I have used the AssociationProxy to define the relationship between users/groups and items. I now want to display all items which the user has access to and I have a really hard time doing that. Following criteria are used: All items which are owned by the user should be shown (item.owner_id = user.id) All public items should be shown (item.access = public) All items which the user has access to should be shown (auth_item.user_id = user.id) All items which the group of the user has access to should be shown. The first two criteria are quite straightforward, but I have a hard time doing the 3rd one. Here is my approach: clause = and_(item.access == 'public') if user is not None: clause = or_(clause,item.owner == user,item.users.contains(user),item.groups.contains(group for group in user.groups)) The third criteria produces an error. item.groups.contains(group for group in user.groups) I am actually not sure if this is a good approach at all. What is the best approach when filtering manytomany relationships? How I can filter a manytomany relationship based on another list/relationship? Btw I am using the latest sqlalchemy (6.0) and elixir version Thanks for any insights.

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  • how to setup kismet.conf on Ubuntu

    - by Registered User
    I installed Kismet on my Ubuntu 10.04 machine as apt-get install kismet every thing seems to work fine. but when I launch it I see following error kismet Launching kismet_server: //usr/bin/kismet_server Suid priv-dropping disabled. This may not be secure. No specific sources given to be enabled, all will be enabled. Non-RFMon VAPs will be destroyed on multi-vap interfaces (ie, madwifi-ng) Enabling channel hopping. Enabling channel splitting. NOTICE: Disabling channel hopping, no enabled sources are able to change channel. Source 0 (addme): Opening none source interface none... FATAL: Please configure at least one packet source. Kismet will not function if no packet sources are defined in kismet.conf or on the command line. Please read the README for more information about configuring Kismet. Kismet exiting. Done. I followed this guide http://www.ubuntugeek.com/kismet-an-802-11-wireless-network-detector-sniffer-and-intrusion-detection-system.html#more-1776 how ever in kismet.conf I am not clear with following line source=none,none,addme as to what should I change this to. lspci -vnn shows 0c:00.0 Network controller [0280]: Broadcom Corporation BCM4312 802.11b/g [14e4:4315] (rev 01) Subsystem: Dell Device [1028:000c] Flags: bus master, fast devsel, latency 0, IRQ 17 Memory at f69fc000 (64-bit, non-prefetchable) [size=16K] Capabilities: [40] Power Management version 3 Capabilities: [58] Vendor Specific Information <?> Capabilities: [e8] Message Signalled Interrupts: Mask- 64bit+ Queue=0/0 Enable- Capabilities: [d0] Express Endpoint, MSI 00 Capabilities: [100] Advanced Error Reporting <?> Capabilities: [13c] Virtual Channel <?> Capabilities: [160] Device Serial Number Capabilities: [16c] Power Budgeting <?> Kernel driver in use: wl Kernel modules: wl, ssb and iwconfig shows lo no wireless extensions. eth0 no wireless extensions. eth1 IEEE 802.11bg ESSID:"WIKUCD" Mode:Managed Frequency:2.462 GHz Access Point: <00:43:92:21:H5:09> Bit Rate=11 Mb/s Tx-Power:24 dBm Retry min limit:7 RTS thr:off Fragment thr:off Encryption key:off Power Managementmode:All packets received Link Quality=1/5 Signal level=-81 dBm Noise level=-90 dBm Rx invalid nwid:0 Rx invalid crypt:0 Rx invalid frag:0 Tx excessive retries:169 Invalid misc:0 Missed beacon:0 So what should I be putting in place of source=none,none,addme with output I mentioned above ?

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  • PHP 5.3: Late static binding doesn't work for properties when defined in parent class while missing in child class

    - by DavidPesta
    Take a look at this example, and notice the outputs indicated. <?php class Mommy { protected static $_data = "Mommy Data"; public static function init( $data ) { static::$_data = $data; } public static function showData() { echo static::$_data . "<br>"; } } class Brother extends Mommy { } class Sister extends Mommy { } Brother::init( "Brother Data" ); Sister::init( "Sister Data" ); Brother::showData(); // Outputs: Sister Data Sister::showData(); // Outputs: Sister Data ?> My understanding was that using the static keyword would refer to the child class, but apparently it magically applies to the parent class whenever it is missing from the child class. (This is kind of a dangerous behavior for PHP, more on that explained below.) I have the following two things in mind for why I want to do this: I don't want the redundancy of defining all of the properties in all of the child classes. I want properties to be defined as defaults in the parent class and I want the child class definition to be able to override these properties wherever needed. The child class needs to exclude properties whenever the defaults are intended, which is why I don't define the properties in the child classes in the above example. However, if we are wanting to override a property at runtime (via the init method), it will override it for the parent class! From that point forward, child classes initialized earlier (as in the case of Brother) unexpectedly change on you. Apparently this is a result of child classes not having their own copy of the static property whenever it isn't explicitly defined inside of the child class--but instead of throwing an error it switches behavior of static to access the parent. Therefore, is there some way that the parent class could dynamically create a property that belongs to the child class without it appearing inside of the child class definition? That way the child class could have its own copy of the static property and the static keyword can refer to it properly, and it can be written to take into account parent property defaults. Or is there some other solution, good, bad, or ugly?

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  • Use System Restore to rescue lost user profile in Win XP?

    - by im_chc
    Hi! My win XP account profile has recently been "reset". Many app settings are lost. For example, the "recent project" list in VS 2005 is empty. There should be lots of other stuffs that are painfully lost without me knowing! What can I do? Can I retrieve the app settings from System Restore? I don't have much confidence on this util, even tho I think restoring to a point when the profile still works, and back up away the C:\Documents and Settings (is it where all the app setting files are located?), that should work... Is it reliable to restore to a previous restore pt and then goes back to the latest RP? I've googled on System Restore, looks like what the util does is just back up some physical files, and restore them when doing System Restore. That sounds quite safe, but I am still uncomfortable to this. Thx for u guys' help in advance!

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  • Hibernate - strange order of native SQL parameters

    - by Xorty
    Hello, I am trying to use native MySQL's MD5 crypto func, so I defined custom insert in my mapping file. <hibernate-mapping package="tutorial"> <class name="com.xorty.mailclient.client.domain.User" table="user"> <id name="login" type="string" column="login"></id> <property name="password"> <column name="password" /> </property> <sql-insert>INSERT INTO user (login,password) VALUES ( ?, MD5(?) )</sql-insert> </class> </hibernate-mapping> Then I create User (pretty simple POJO with just 2 Strings - login and password) and try to persist it. session.beginTransaction(); // we have no such user in here yet User junitUser = (User) session.load(User.class, "junit_user"); assert (null == junitUser); // insert new user junitUser = new User(); junitUser.setLogin("junit_user"); junitUser.setPassword("junitpass"); session.save(junitUser); session.getTransaction().commit(); What actually happens? User is created, but with reversed parameters order. He has login "junitpass" and "junit_user" is MD5 encrypted and stored as password. What did I wrong? Thanks EDIT: ADDING POJO class package com.xorty.mailclient.client.domain; import java.io.Serializable; /** * POJO class representing user. * @author MisoV * @version 0.1 */ public class User implements Serializable { /** * Generated UID */ private static final long serialVersionUID = -969127095912324468L; private String login; private String password; /** * @return login */ public String getLogin() { return login; } /** * @return password */ public String getPassword() { return password; } /** * @param login the login to set */ public void setLogin(String login) { this.login = login; } /** * @param password the password to set */ public void setPassword(String password) { this.password = password; } /** * @see java.lang.Object#toString() * @return login */ @Override public String toString() { return login; } /** * Creates new User. * @param login User's login. * @param password User's password. */ public User(String login, String password) { setLogin(login); setPassword(password); } /** * Default constructor */ public User() { } /** * @return hashCode * @see java.lang.Object#hashCode() */ @Override public int hashCode() { final int prime = 31; int result = 1; result = prime * result + ((null == login) ? 0 : login.hashCode()); result = prime * result + ((null == password) ? 0 : password.hashCode()); return result; } /** * @param obj Compared object * @return True, if objects are same. Else false. * @see java.lang.Object#equals(java.lang.Object) */ @Override public boolean equals(Object obj) { if (this == obj) { return true; } if (obj == null) { return false; } if (!(obj instanceof User)) { return false; } User other = (User) obj; if (login == null) { if (other.login != null) { return false; } } else if (!login.equals(other.login)) { return false; } if (password == null) { if (other.password != null) { return false; } } else if (!password.equals(other.password)) { return false; } return true; } }

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  • User not in the sudoers file. This incident will be reported

    - by Sergiy Byelozyorov
    I need to install a package. For that I need root access. However the system says that I am not in sudoers file. When trying to edit one, it complains alike! How I am supposed to add myself to the sudoers file if I don't have the right to edit one? I have installed this system and only administrator. What can I do? Edit: I have tried visudo already. It requires me to be in sudoers in the first place. amarzaya@linux-debian-gnu:/$ sudo /usr/sbin/visudo We trust you have received the usual lecture from the local System Administrator. It usually boils down to these three things: #1) Respect the privacy of others. #2) Think before you type. #3) With great power comes great responsibility. [sudo] password for amarzaya: amarzaya is not in the sudoers file. This incident will be reported. amarzaya@linux-debian-gnu:/$

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  • Nested Resource testing RSpec

    - by Joseph DelCioppio
    I have two models: class Solution < ActiveRecord::Base belongs_to :owner, :class_name => "User", :foreign_key => :user_id end class User < ActiveRecord::Base has_many :solutions end with the following routing: map.resources :users, :has_many => :solutions and here is the SolutionsController: class SolutionsController < ApplicationController before_filter :load_user def index @solutions = @user.solutions end private def load_user @user = User.find(params[:user_id]) unless params[:user_id].nil? end end Can anybody help me with writing a test for the index action? So far I have tried the following but it doesn't work: describe SolutionsController do before(:each) do @user = Factory.create(:user) @solutions = 7.times{Factory.build(:solution, :owner => @user)} @user.stub!(:solutions).and_return(@solutions) end it "should find all of the solutions owned by a user" do @user.should_receive(:solutions) get :index, :user_id => @user.id end end And I get the following error: Spec::Mocks::MockExpectationError in 'SolutionsController GET index, when the user owns the software he is viewing should find all of the solutions owned by a user' #<User:0x000000041c53e0> expected :solutions with (any args) once, but received it 0 times Thanks in advance for all the help. Joe

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  • How is it possible to list all folders that a particular user/group has permissions on?

    - by Lord Torgamus
    Is it possible to list all folders/files that a given group has explicit permissions on, for a machine running Windows Server 2003? If so, how? It would be nice to see inherited permissions as well, but I could do with just explicit permissions. A little background: I'm trying to update groups/permissions on a test server. One of the groups, Devs, wasn't implemented correctly when it was created, and my goal is to remove it from the system. It has been replaced by LeadDevelopers, which has permissions on many — but naturally not all — of the same folders. I want to make sure that I don't accidentally orphan any folders or cause any other issues when I remove Devs. It did have some admin-level permissions.

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  • How to cd into smb://[email protected] from terminal?

    - by John
    I am using ubuntu and gnome on my computer. When I open up File Browser, on the left hand rail, I see conveniently a folder called "Work Server". When I mouse over it, the following caption appears "smb://[email protected]". If I click on that folder, then I can see the contents of that folder. Everything is great. So now when I open up a terminal/shell, I type in cd smb://[email protected] I get an error saying the directory doesn't exist. How do I enter this directory via shell/terminal?

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  • How to copy symbolic links?

    - by Basilevs
    I have directory that contains some symbolic links: user@host:include$ find .. -type l -ls 4737414 0 lrwxrwxrwx 1 user group 13 Dec 9 13:47 ../k0607-lsi6/camac -> ../../include 4737415 0 lrwxrwxrwx 1 user group 14 Dec 9 13:49 ../k0607-lsi6/linux -> ../../../linux 4737417 0 lrwxrwxrwx 1 user group 12 Dec 9 13:57 ../k0607-lsi6/dfc -> ../../../dfc 4737419 0 lrwxrwxrwx 1 user group 17 Dec 9 13:57 ../k0607-lsi6/dfcommon -> ../../../dfcommon 4737420 0 lrwxrwxrwx 1 user group 19 Dec 9 13:57 ../k0607-lsi6/dfcommonxx -> ../../../dfcommonxx 4737421 0 lrwxrwxrwx 1 user group 17 Dec 9 13:57 ../k0607-lsi6/dfcompat -> ../../../dfcompat I need to copy them to the current directory. The resulting links should be independent from their prototypes and lead directly to their target objects. cp -s creates links to links that is not appropriate behavior. cp -s -L refuses to copy links to directories cp -s -L -r refuses to copy relative links to non-working directory What should I do?

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  • In FitNesse, can variables be defined in terms of other variables?

    - by Dan Haywood
    In FitNesse, can variables be defined in terms of other variables? I want to do the equivalent of: int a=3; int b=a; To make this concrete, I have a variable defining the date: !define clock.date {2/2/2009} I then want to define some other variable ${other.date} based on it, something like: !define other.date {=${clock.date}=} However, this doesn't work. Is there any way to do this?

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