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  • Mailto links are not working in Chrome

    - by sfoatl
    Mailto links are not working at all in Chrome (8.0.552.224). These links are working in IE 7&8, Safari, and Firefox. In Chrome, we are clicking on mailto links (we have tested this about 25 times now), and things just go into the ether. In all other browsers, we click on the mailto links, they open up email clients (gmail, outlook, and others), and we can send the email. But in Chrome, we click the mailto links and nothing happens...

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  • Change stack order in mobile view at 1140 grid?

    - by iHaveacomputer
    I want to implement the 1140grid at my site. The layout is pretty simple: 100% header 25% sidebar 75% page 100% footer see also http://jsfiddle.net/KB5Nq/ the problem is that i would like to change the stack order when the site is in mobile view: 100% header 100% page 100% sidebar 100% footer however, by default it arranges the blocks in the same order as they appear in the source code: header, sidebar, page, footer. is there an easy css-only fix for that?

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  • Looking for work with startups in exchange for Equity [closed]

    - by SteveG
    Hi guys, I'm looking for a startup to get involved with for an equity stake. I'm a software enginner/architect with 25 years experience. I've got extensive experience with Java, C# and LAMP and have worked with serveral startups over the last couple of years. If you're interested but would like to know more about me I can send you my resume. Steve

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  • why can't I call methods on a for-yield expression?

    - by 1984isnotamanual
    Say I have some scala code like this: // outputs 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 println( squares ) def squares = { val s = for ( count <- 1 to 10 ) yield { count * count } s.mkString(", "); } Why do I have to use the temporary val s? I tried this: def squares = for ( count <- 1 to 10 ) yield { count * count }.mkString(", ") That fails to compile with this error message: error: value mkString is not a member of Int def squares = for ( count <- 1 to 10 ) yield { count * count }.mkString(", ") Shouldn't mkString be called on the collection returned by the for loop?

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  • How do I simulate a progress counter in a command line application in Python?

    - by CRP
    My Python program does a series of things and prints some diagnostic output. I would also like to have a progress counter like this: Percentage done: 25% where the number increases "in place". If I use only string statements I can write separate numbers, but that would clutter the screen. Is there some way to achieve this, for example using some escape char for backspace in order to clear a number and write the next one? Thanks

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  • WPF Style Triggers: can I apply the one style for a variety of Properties?

    - by Matt H.
    It seems like there has to be a way to do this: I am applying an ItemContainerStyle in my Listbox, based on two property triggers. As you can see, I'm using the exact same set of trigger enter/exit actions, simply applied on two different properties. Is there something equivalent to a <Trigger Property="prop1" OR Property="prop2" ??? (Obviously wouldn't look like that, but that probably gets the point across.) <Style x:Key="ListBoxItemStyle" TargetType="ListBoxItem"> <Style.Triggers> <Trigger Property="IsKeyboardFocusWithin" Value="True"> <Trigger.EnterActions> <BeginStoryboard> <Storyboard> <DoubleAnimation Storyboard.TargetProperty="Height" To="50" Duration="0:0:.3"></DoubleAnimation> </Storyboard> </BeginStoryboard> </Trigger.EnterActions> <Trigger.ExitActions> <BeginStoryboard> <Storyboard> <DoubleAnimation Storyboard.TargetProperty="Height" To="25" Duration="0:0:.3" /> </Storyboard> </BeginStoryboard> </Trigger.ExitActions> </Trigger> </Style.Triggers> </Style> <Style x:Key="ListBoxItemStyle" TargetType="ListBoxItem"> <Style.Triggers> <Trigger Property="IsMouseOver" Value="True"> <Trigger.EnterActions> <BeginStoryboard> <Storyboard> <DoubleAnimation Storyboard.TargetProperty="Height" To="50" Duration="0:0:.3"></DoubleAnimation> </Storyboard> </BeginStoryboard> </Trigger.EnterActions> <Trigger.ExitActions> <BeginStoryboard> <Storyboard> <DoubleAnimation Storyboard.TargetProperty="Height" To="25" Duration="0:0:.3" /> </Storyboard> </BeginStoryboard> </Trigger.ExitActions> </Trigger> </Style.Triggers> </Style>

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  • I know I'm doing something wrong with RaiseCanExecuteChanged and CanExecute

    - by Cowman
    Well after fiddling with MVVM light to get my button to enable and disable when I want it to... I sort of mashed things together until it worked. However, I just know I'm doing something wrong here. I have RaiseCanExecuteChanged and CanExecute in the same area being called. Surely this is not how it's done? Here's my xaml <Button Margin="10, 25, 10, 25" VerticalAlignment="Center" HorizontalAlignment="Center" Width="50" Height="50" Grid.Column="1" Grid.Row="3" Content="Host"> <i:Interaction.Triggers> <i:EventTrigger EventName="Click"> <mvvmLight:EventToCommand Command="{Binding HostChat}" MustToggleIsEnabled="True" /> </i:EventTrigger> </i:Interaction.Triggers> </Button> And here's my code public override void InitializeViewAndViewModel() { view = UnityContainer.Resolve<LoginPromptView>(); viewModel = UnityContainer.Resolve<LoginPromptViewModel>(); view.DataContext = viewModel; InjectViewIntoRegion(RegionNames.PopUpRegion, view, true); viewModel.HostChat = new DelegateCommand(ExecuteHostChat, CanHostChat); viewModel.PropertyChanged += new System.ComponentModel.PropertyChangedEventHandler(ViewModelPropertyChanged); } void ViewModelPropertyChanged(object sender, System.ComponentModel.PropertyChangedEventArgs e) { if (e.PropertyName == "Name" || e.PropertyName == "Port" || e.PropertyName == "Address") { (viewModel.HostChat as DelegateCommand).RaiseCanExecuteChanged(); (viewModel.HostChat as DelegateCommand).CanExecute(); } } public void ExecuteHostChat() { } public bool CanHostChat() { if (String.IsNullOrEmpty(viewModel.Address) || String.IsNullOrEmpty(viewModel.Port) || String.IsNullOrEmpty(viewModel.Name)) { return false; } else return true; } See how these two are together? Surely that can't be right. I mean... it WORKS for me... but something seems wrong about it. Shouldn't RaiseCanExecuteChanged call CanExecute? It doesn't... and so if I don't have that CanExecute in there, my control never toggles its IsEnabled like I need it to. (viewModel.HostChat as DelegateCommand).RaiseCanExecuteChanged(); (viewModel.HostChat as DelegateCommand).CanExecute();

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  • JAVA String split on interval

    - by user2920611
    I would like to split my strings in JAVA based on a regular interval, not on regex. This is what I have to split: 1 x3.1.105.41 1 -10 2 x4.1.105.41 0 -10 3 x12.1.105.41 0 -10 4 y3.1.105.41.19 1 0 5 y4.1.105.41.21 0 0 6 y1.1.105.41.23 0 0 7 y12.1.105.41.25 0 0 I would like to seperate each column. Currently, I use the strLine.spli function Any help would be great!

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  • I need objective syntax for calculating bmi

    - by Umaid
    Actually I am working on bmi calculator. Where I would like to calculate bmi for height in inches and weight in lbs and also in need of correct formula for height in cm and weight in kgs. I have tried but couldn't calculate actual value coming withing the range as below. It exceeds the range. BMI Categories: * Underweight = <18.5 * Normal weight = 18.5-24.9 * Overweight = 25-29.9 * Obesity = BMI of 30 or greater

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  • How can I maintain a sorted hash in Perl?

    - by srk
    @aoh =( { 3 => 15, 4 => 8, 5 => 9, }, { 3 => 11, 4 => 25, 5 => 6, }, { 3 => 5, 4 => 18, 5 => 5, }, { 0 => 16, 1 => 11, 2 => 7, }, { 0 => 21, 1 => 13, 2 => 31, }, { 0 => 11, 1 => 14, 2 => 31, }, ); I want the hashes in each array index sorted in reverse order based on values.. @sorted = sort { ........... please fill this..........} @aoh; expected output @aoh =( { 4 => 8, 5 => 9, 3 => 15, }, { 5 => 6, 3 => 11, 4 => 25, }, { 5 => 5, 3 => 5, 4 => 18, }, { 2 => 7, 1 => 11, 0 => 16, }, { 1 => 13, 0 => 21, 2 => 31, }, { 0 => 11, 1 => 14, 2 => 31, }, ); Please help.. Thanks in advance.. Stating my request again: I only want the hashes in each array index to be sorted by values.. i dont want the array to be sorted..

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  • Database is very slow when creating an index

    - by kaliyaperumal M
    My database has around 25 core numbers, in that weekly basis I need to create an index and drop index. While creating the index it takes long time to complete, my log file also keeps on increasing, and when I delete some numbers from that table also taking too much time (because weekly basis I have to delete 30 to 50 lack numbers and add 30 to 40 lack new number also). Can u please give me the proper solution..

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  • dictionary/map/key-value pairs data structure in C

    - by morgancodes
    How does one construct and access a set of key-value pairs in C? To use a silly simple example, let's say I want to create a table which translates between an integer and its square root. If I were writing javascript, I could just do this: var squareRoots = { 4: 2, 9: 3, 16: 4, 25: 5 } and then access them like: var squareRootOf25 = squareRoots[5] What's the prettiest way to do this in C? What if I want to use one type of enum as the key and another type of enum as the value?

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  • Add an column to dataset in .net 1.1

    - by prince23
    hi, i have dataset(dataset name dsEmp) with an 25 columns , here i need to add an new column with columname as "EndDate" of data type as string now i need to import all the data from column18 whose datatype is Datetime to the newely created datacolumn "EndDate" once we import all the data from the column 18 to the newcolumn "EndDate" , we should remove the column18 hope my Question is clear, can any one please help me this.

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  • Is there anyway that we can get a label value to a Sql

    - by Pradeep
    SELECT COUNT(*) AS Expr1 FROM Book INNER JOIN Temp_Order ON Book.Book_ID = Temp_Order.Book_ID WHERE (Temp_Order.User_ID = 25) AND (CONVERT (nvarchar, Temp_Order.OrderDate, 111) = CONVERT (nvarchar, GETDATE(), 111)) In here i want to change my User_ID to get from a label.Text this Sql Statement is in a DataView. so in the Wizard it not accepting a text box values or anything. can someone please help me to solve this

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  • how to validate

    - by kumar
    I have this Calender Control I am using..user can select any date from the calender.. I need to validate the dates like Saturday and Sundays and 1/1 and 1/25.. if they select these days I need to show them Popup message not valid date? can anybody help me out.. thanks

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  • apache eat up too many ram per child

    - by mrc4r7m4n
    Hello to everyone. I've got fallowing problem: Apache eat to many ram per child. The fallowing comments shows: cat /etc/redhat-release -- Fedora release 8 (Werewolf) free -m: total used free shared buffers cached Mem: 3566 3136 429 0 339 1907 -/+ buffers/cache: 889 2676 Swap: 4322 0 4322 I know that you will say that there is nothing to worry about because swap is not use, but i think it's not use for now. 3.httpd -v: Server version: Apache/2.2.14 (Unix) 4.httpd -l: Compiled in modules: core.c mod_authn_file.c mod_authn_default.c mod_authz_host.c mod_authz_groupfile.c mod_authz_user.c mod_authz_default.c mod_auth_basic.c mod_include.c mod_filter.c mod_log_config.c mod_env.c mod_setenvif.c mod_version.c mod_ssl.c prefork.c http_core.c mod_mime.c mod_status.c mod_autoindex.c mod_asis.c mod_cgi.c mod_negotiation.c mod_dir.c mod_actions.c mod_userdir.c mod_alias.c mod_rewrite.c mod_so.c 5.List of loaded dynamic modules: LoadModule authz_host_module modules/mod_authz_host.so LoadModule include_module modules/mod_include.so LoadModule log_config_module modules/mod_log_config.so LoadModule setenvif_module modules/mod_setenvif.so LoadModule mime_module modules/mod_mime.so LoadModule autoindex_module modules/mod_autoindex.so LoadModule vhost_alias_module modules/mod_vhost_alias.so LoadModule negotiation_module modules/mod_negotiation.so LoadModule dir_module modules/mod_dir.so LoadModule alias_module modules/mod_alias.so LoadModule rewrite_module modules/mod_rewrite.so LoadModule proxy_module modules/mod_proxy.so LoadModule cgi_module modules/mod_cgi.so 6.My prefrok directive <IfModule prefork.c> StartServers 8 MinSpareServers 5 MaxSpareServers 25 ServerLimit 80 MaxClients 80 MaxRequestsPerChild 4000 </IfModule> KeepAliveTimeout 6 MaxKeepAliveRequests 100 KeepAlive On 7.top -u apache: ctrl+ M top - 09:19:42 up 2 days, 19 min, 2 users, load average: 0.85, 0.87, 0.80 Tasks: 113 total, 1 running, 112 sleeping, 0 stopped, 0 zombie Cpu(s): 7.3%us, 15.7%sy, 0.0%ni, 75.7%id, 0.0%wa, 0.7%hi, 0.7%si, 0.0%st Mem: 3652120k total, 3149964k used, 502156k free, 348048k buffers Swap: 4425896k total, 0k used, 4425896k free, 1944952k cached PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 16956 apache 20 0 700m 135m 100m S 0.0 3.8 2:16.78 httpd 16953 apache 20 0 565m 130m 96m S 0.0 3.7 1:57.26 httpd 16957 apache 20 0 587m 129m 102m S 0.0 3.6 1:47.41 httpd 16955 apache 20 0 567m 126m 93m S 0.0 3.6 1:43.60 httpd 17494 apache 20 0 626m 125m 96m S 0.0 3.5 1:58.77 httpd 17515 apache 20 0 540m 120m 88m S 0.0 3.4 1:45.57 httpd 17516 apache 20 0 573m 120m 88m S 0.0 3.4 1:50.51 httpd 16954 apache 20 0 551m 120m 88m S 0.0 3.4 1:52.47 httpd 17493 apache 20 0 586m 120m 94m S 0.0 3.4 1:51.02 httpd 17279 apache 20 0 568m 117m 87m S 16.0 3.3 1:51.87 httpd 17302 apache 20 0 560m 116m 90m S 0.3 3.3 1:59.06 httpd 17495 apache 20 0 551m 116m 89m S 0.0 3.3 1:47.51 httpd 17277 apache 20 0 476m 114m 81m S 0.0 3.2 1:37.14 httpd 30097 apache 20 0 536m 113m 83m S 0.0 3.2 1:47.38 httpd 30112 apache 20 0 530m 112m 81m S 0.0 3.2 1:40.15 httpd 17513 apache 20 0 516m 112m 85m S 0.0 3.1 1:43.92 httpd 16958 apache 20 0 554m 111m 82m S 0.0 3.1 1:44.18 httpd 1617 apache 20 0 487m 111m 85m S 0.0 3.1 1:31.67 httpd 16952 apache 20 0 461m 107m 75m S 0.0 3.0 1:13.71 httpd 16951 apache 20 0 462m 103m 76m S 0.0 2.9 1:28.05 httpd 17278 apache 20 0 497m 103m 76m S 0.0 2.9 1:31.25 httpd 17403 apache 20 0 537m 102m 79m S 0.0 2.9 1:52.24 httpd 25081 apache 20 0 412m 101m 70m S 0.0 2.8 1:01.74 httpd I guess thats all information needed to help me solve this problem. I think the virt memory is to big, the same res. The consumption of ram is increasing all the time. Maybe it's memory leak because i see there is so many static modules compiled. Could someone help me with this issue? Thank you in advance.

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  • Debian, 6rd tunnel, and connection troubles

    - by Chris B
    Long story short I am having issues with IPv6 using a 6rd tunnel with my ISP, charter business. They offer a 6rd tunnel that I think I have properly set up, but the server doesn’t reply to every ipv6 request. When the server has the network interfaces idle with no traffic for about 10 minutes, then IPv6 stops accepting inbound connections. to re-allow it, I must go into the server, and make it do a outbound ipv6 connection (normally a ping) to start it back up. Whats weird though i that if I run iptraf when its not working, it still shows a inbound ipv6 packet… the server is just not replying, and I can’t figure out why. Also, if I try to access my server over IPv6 from a house about 1 mile away on the same ISP, it is never able to connect. it always times out, but again the iptraf shows a ipv6 inbound packet. Again, it just does not reply. To test if my server is accessible through IPv6 I always have to use my vzw 4g phone (they use IPv6) or ipv6proxy dot net. Here is all of the configuration information my ISP gives on there tunnel server: 6rd Prefix = 2602:100::/32 Border Relay Address = 68.114.165.1 6rd prefix length = 32 IPv4 mask length = 0 Here is my /etc/network/interfaces for ipv6 (used x's to block real addresses) auto charterv6 iface charterv6 inet6 v4tunnel address 2602:100:189f:xxxx::1 netmask 32 ttl 64 gateway ::68.114.165.1 endpoint 68.114.165.1 local 24.159.218.xxx up ip link set mtu 1280 dev charterv6 here is my iptables config filter :INPUT DROP [0:0] :fail2ban-ssh – [0:0] :OUTPUT ACCEPT [0:0] :FORWARD DROP [0:0] :hold – [0:0] -A INPUT -p tcp -m tcp —dport 22 -j fail2ban-ssh -A INPUT -m state —state RELATED,ESTABLISHED -j ACCEPT -A INPUT -p tcp -m multiport -j ACCEPT —dports 80,443,25,465,110,995,143,993,587,465,22 -A INPUT -i lo -j ACCEPT -A INPUT -p tcp -m tcp —dport 10000 -j ACCEPT -A INPUT -p tcp -m tcp —dport 5900:5910 -j ACCEPT -A fail2ban-ssh -j RETURN -A INPUT -p icmp -j ACCEPT COMMIT and last here is my ip6tables firewall config filter :INPUT DROP [1653:339023] :FORWARD DROP [0:0] :OUTPUT ACCEPT [60141:13757903] :hold – [0:0] -A INPUT -m state —state RELATED,ESTABLISHED -j ACCEPT -A INPUT -p tcp -m multiport —dports 80,443,25,465,110,995,143,993,587,465,22 -j ACCEPT -A INPUT -i lo -j ACCEPT -A INPUT -p tcp -m tcp —dport 10000 -j ACCEPT -A INPUT -p tcp -m tcp —dport 5900:5910 -j ACCEPT -A INPUT -p ipv6-icmp -j ACCEPT COMMIT So Summary: 1.iptraf always shows IPv6 traffic, so its always making it to the server 2.server stops replying on ipv6 after no traffic for awhile (10 minutesish) until a outbound connection is made, then the process repeats. 3.server is NEVER accessable vi same ISP (yet iptraf still shows ipv6 request) Notes: When I try to access it from the same ISP from across town, even with iptables and ip6tables allowing ALL inbound traffic, this is what iptraf shows. IPv6 (92 bytes) from 97.92.18.xxx to 24.159.218.xxx on eth0 ICMP dest unrch (port) (120 bytes) from 24.159.218.xxx to 97.92.18.xxx on eth1 its strange, like its trying to forward to LAN? (eth1 is LAN, eth0 is WAN) even with the IPv6 address being set in the hosts file to the servers domain name. With iptables set up normally with the above configurations it only says this: IPv6 (100 bytes) from 97.92.18.xxx to 24.159.218.xxx on eth0 Im REALLY stuck on this, and any help would be GREATLY appreciated.

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  • sudo in Debian squeeze inside linux-vserver always wants password

    - by mark
    Every since I upgraded all my linux-vserver Debian guests from Lenny to Squeeze I've the apparent problem that whenever I want to use sudo it asks me for my password. Every time. I've configured sudo to have a timeout of 30 minutes: Defaults timestamp_timeout=30 . This has been configured when it was still Lenny (note: as suggested by EightBitTony I've also tried without this setting - no change). I've a hard time figuring out what the problem here is, since I think my configuration is right. I thought about it being a problem with the file used to record the timestamp, maybe a permission issue, but was unlucky to find any hard evidence. I've compared the contents of /var/lib/sudo/ between a working and a non-working system but couldn't spot any difference. The version of sudo used in both environments is 1.7.4p4-2.squeeze.3. My non-working system(s): find /var/lib/sudo/ -ls 17319289 4 drwx------ 4 root root 4096 Jan 1 1985 /var/lib/sudo/ 17319286 4 drwx------ 2 root mark 4096 Jan 1 1985 /var/lib/sudo/mark 17319312 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/6 17319361 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/9 17319490 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/10 17319326 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/4 17319491 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/2 A working system: find /var/lib/sudo -ls 2598921 4 drwx------ 5 root root 4096 Jan 1 1985 /var/lib/sudo 1999522 4 drwx------ 2 root mark 4096 Jan 1 1985 /var/lib/sudo/mark 2000781 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/8 1998998 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/17 1999459 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/26 1998930 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/24 2000771 4 -rw------- 1 root mark 40 Jun 25 11:39 /var/lib/sudo/mark/4 2000773 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/5 1999223 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/0 1998908 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/14 2000769 4 -rw------- 1 root mark 40 Jul 9 13:30 /var/lib/sudo/mark/2 2000770 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/3 2000782 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/9 2000778 4 -rw------- 1 root mark 40 Jul 8 00:11 /var/lib/sudo/mark/7 1998892 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/19 1999264 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/23 2000789 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/12 1999093 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/25 1998880 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/18 1998853 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/20 2000790 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/15 1998878 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/16 1998874 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/13 2000774 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/6 2000786 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/11 1998893 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/22 2000783 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/10 1998949 4 -rw------- 1 root mark 40 Jan 1 1985 /var/lib/sudo/mark/1 Despite the obvious (some up2date timestamps on the working system) I don't see anything wrong here, so it could be as well be a wrong track. Here's my current /etc/sudoers: # /etc/sudoers # # This file MUST be edited with the 'visudo' command as root. # # See the man page for details on how to write a sudoers file. # Defaults env_reset # Host alias specification # User alias specification User_Alias FULLADMIN = user1, user2, user3 # Cmnd alias specification # User privilege specification root ALL=(ALL) ALL FULLADMIN ALL = (ALL) ALL # Allow members of group sudo to execute any command # (Note that later entries override this, so you might need to move # it further down) %sudo ALL=(ALL) ALL # #includedir /etc/sudoers.d #Defaults always_set_home,timestamp_timeout=30

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