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  • To ref or not to ref

    - by nmarun
    So the question is what is the point of passing a reference type along with the ref keyword? I have an Employee class as below: 1: public class Employee 2: { 3: public string FirstName { get; set; } 4: public string LastName { get; set; } 5:  6: public override string ToString() 7: { 8: return string.Format("{0}-{1}", FirstName, LastName); 9: } 10: } In my calling class, I say: 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(Employee employee) 16: { 17: employee.FirstName = "Smith"; 18: employee.LastName = "Doe"; 19: } 20: }   After having a look at the code, you’ll probably say, Well, an instance of a class gets passed as a reference, so any changes to the instance inside the CallSomeMethod, actually modifies the original object. Hence the output will be ‘John-Doe’ on the first call and ‘Smith-Doe’ on the second. And you’re right: So the question is what’s the use of passing this Employee parameter as a ref? 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(ref employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(ref Employee employee) 16: { 17: employee.FirstName = "Smith"; 18: employee.LastName = "Doe"; 19: } 20: } The output is still the same: Ok, so is there really a need to pass a reference type using the ref keyword? I’ll remove the ‘ref’ keyword and make one more change to the CallSomeMethod method. 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(Employee employee) 16: { 17: employee = new Employee 18: { 19: FirstName = "Smith", 20: LastName = "John" 21: }; 22: } 23: } In line 17 you’ll see I’ve ‘new’d up the incoming Employee parameter and then set its properties to new values. The output tells me that the original instance of the Employee class does not change. Huh? But an instance of a class gets passed by reference, so why did the values not change on the original instance or how do I keep the two instances in-sync all the times? Aah, now here’s the answer. In order to keep the objects in sync, you pass them using the ‘ref’ keyword. 1: class Program 2: { 3: static void Main() 4: { 5: Employee employee = new Employee 6: { 7: FirstName = "John", 8: LastName = "Doe" 9: }; 10: Console.WriteLine(employee); 11: CallSomeMethod(ref employee); 12: Console.WriteLine(employee); 13: } 14:  15: private static void CallSomeMethod(ref Employee employee) 16: { 17: employee = new Employee 18: { 19: FirstName = "Smith", 20: LastName = "John" 21: }; 22: } 23: } Viola! Now, to prove it beyond doubt, I said, let me try with another reference type: string. 1: class Program 2: { 3: static void Main() 4: { 5: string name = "abc"; 6: Console.WriteLine(name); 7: CallSomeMethod(ref name); 8: Console.WriteLine(name); 9: } 10:  11: private static void CallSomeMethod(ref string name) 12: { 13: name = "def"; 14: } 15: } The output was as expected, first ‘abc’ and then ‘def’ - proves the 'ref' keyword works here as well. Now, what if I remove the ‘ref’ keyword? The output should still be the same as the above right, since string is a reference type? 1: class Program 2: { 3: static void Main() 4: { 5: string name = "abc"; 6: Console.WriteLine(name); 7: CallSomeMethod(name); 8: Console.WriteLine(name); 9: } 10:  11: private static void CallSomeMethod(string name) 12: { 13: name = "def"; 14: } 15: } Wrong, the output shows ‘abc’ printed twice. Wait a minute… now how could this be? This is because string is an immutable type. This means that any time you modify an instance of string, new memory address is allocated to the instance. The effect is similar to ‘new’ing up the Employee instance inside the CallSomeMethod in the absence of the ‘ref’ keyword. Verdict: ref key came to the rescue and saved the planet… again!

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  • SQL group and order

    - by John Lambert
    I have multiple users with multiple entries recording times they arrive at destinations Somehow, with my select query I would like to only show the most recent entries for each unique user name. Here is the code that doesn't work: SELECT * FROM $dbTable GROUP BY xNAME ORDER BY xDATETIME DESC This does the name grouping fine, but as far as showing ONLY their most recent entry, is just shows the first entry it sees in the SQL table. I guess my question is, is this possible? Here is my data sample: john 7:00 chris 7:30 greg 8:00 john 8:15 greg 8:30 chris 9:00 and my desired result should only be john 8:15 chris 9:00 greg 8:30

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  • C# - Determine if class initializaion causes infinite recursion?

    - by John M
    I am working on porting a VB6 application to C# (Winforms 3.5) and while doing so I'm trying to break up the functionality into various classes (ie database class, data validation class, string manipulation class). Right now when I attempt to run the program in Debug mode the program pauses and then crashes with a StackOverFlowException. VS 2008 suggests a infinite recursion cause. I have been trying to trace what might be causing this recursion and right now my only hypothesis is that class initializations (which I do in the header(?) of each class). My thought is this: mainForm initializes classA classA initializes classB classB initializes classA .... Does this make sense or should I be looking elsewhere? UPDATE1 (a code sample): mainForm namespace john { public partial class frmLogin : Form { stringCustom sc = new sc(); stringCustom namespace john { class stringCustom { retrieveValues rv = new retrieveValues(); retrieveValues namespace john { class retrieveValues { stringCustom sc = new stringCustom();

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  • struct assignment operator on arrays

    - by Django fan
    Suppose I defined a structure like this: struct person { char name [10]; int age; }; and declared two person variables: person Bob; person John; where Bob.name = "Bob", Bob.age = 30 and John.name = "John",John.age = 25. and I called Bob = John; struct person would do a Memberwise assignment and assign Johns's member values to Bob's. But arrays can't assign to arrays, so how does the assignment of the "name" array work?

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  • How do I average the difference between specific values in TSQL?

    - by jvenema
    Hey folks, sorry this is a bit of a longer question... I have a table with the following columns: [ChatID] [User] [LogID] [CreatedOn] [Text] What I need to find is the average response time for a given user id, to another specific user id. So, if my data looks like: [1] [john] [20] [1/1/11 3:00:00] [Hello] [1] [john] [21] [1/1/11 3:00:23] [Anyone there?] [1] [susan] [22] [1/1/11 3:00:43] [Hello!] [1] [susan] [23] [1/1/11 3:00:53] [What's up?] [1] [john] [24] [1/1/11 3:01:02] [Not much] [1] [susan] [25] [1/1/11 3:01:08] [Cool] ...then I need to see that Susan has an average response time of (20 + 6) / 2 = 13 seconds to John, and John has an average of (9 / 1) = 9 seconds to Susan. I'm not even sure this can be done in set-based logic, but if anyone has any ideas, they'd be much appreciated!

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  • Must I loop to search results for a specific value?

    - by tag
    I have a table in the database: name Opinion Tim Tim has an opinion John other random text Dan Dan's random text Al Al says something else I call this data and get it back in getRecords.lastResult To access John's opinion, I could use: getRecords.lastResult[1].opinion But that's only because I know that John is the second record (record 1), but this may change. So the right way is to search through the results to first find the record index for John, then access his opinion. My guess is I need some sort of a loop? Is there an easier way to search for John directly without a loop?

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  • Installing CUDA on Ubuntu 12.04 with nvidia driver 295.59

    - by johnmcd
    I have been trying to get cuda to run on a nvidia gt 650m based laptop. I am running Ubuntu 12.04 with the nvidia 295.59 driver. Also, my laptop uses Optimus so I have install the driver via bumblebee. Bumblebee is not working correctly yet -- however I believe it is possible to install CUDA independently. To install CUDA I have followed the instructions detailed here: How can I get nVidia CUDA or OpenCL working on a laptop with nVidia discrete card/Intel Integrated Graphics? However I am still running into problem building the sdk. I made the changes specified at the above link in common.mk, but I got the following (snippet) from the build process: make[2]: Entering directory `/home/john/NVIDIA_GPU_Computing_SDK/C/src/fluidsGL' /usr/bin/ld: warning: libnvidia-tls.so.302.17, needed by /usr/lib/nvidia-current/libGL.so, not found (try using -rpath or -rpath-link) /usr/bin/ld: warning: libnvidia-glcore.so.302.17, needed by /usr/lib/nvidia-current/libGL.so, not found (try using -rpath or -rpath-link) /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv018tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv012glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv017glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv012tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv015tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv019tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv000glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv017tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv013tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv013glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv018glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv022tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv007tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv009tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv020tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv014glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv015glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv016tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv001glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv006tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv021tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv011tls' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv020glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv019glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv002glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv021glcore' /usr/lib/nvidia-current/libGL.so: undefined reference to `_nv014tls' collect2: ld returned 1 exit status make[2]: *** [../../bin/linux/release/fluidsGL] Error 1 make[2]: Leaving directory `/home/john/NVIDIA_GPU_Computing_SDK/C/src/fluidsGL' make[1]: *** [src/fluidsGL/Makefile.ph_build] Error 2 make[1]: Leaving directory `/home/john/NVIDIA_GPU_Computing_SDK/C' make: *** [all] Error 2 The libraries that ld warns about are on my system and are installed on the system: $ locate libnvidia-tls.so.302.17 libnvidia-glcore.so.302.17 /usr/lib/nvidia-current/libnvidia-glcore.so.302.17 /usr/lib/nvidia-current/libnvidia-tls.so.302.17 /usr/lib/nvidia-current/tls/libnvidia-tls.so.302.17 /usr/lib32/nvidia-current/libnvidia-glcore.so.302.17 /usr/lib32/nvidia-current/libnvidia-tls.so.302.17 /usr/lib32/nvidia-current/tls/libnvidia-tls.so.302.17 however /usr/lib/nvidia-current and /usr/lib32/nvidia-current are not being picked up by ldconfig. I have tried adding them by adding a file to /etc/ld.so.conf.d/ which gets past this error, however now I am getting the following error: make[2]: Entering directory `/home/john/NVIDIA_GPU_Computing_SDK/C/src/deviceQueryDrv' cc1plus: warning: command line option ‘-Wimplicit’ is valid for C/ObjC but not for C++ [enabled by default] obj/x86_64/release/deviceQueryDrv.cpp.o: In function `main': deviceQueryDrv.cpp:(.text.startup+0x5f): undefined reference to `cuInit' deviceQueryDrv.cpp:(.text.startup+0x99): undefined reference to `cuDeviceGetCount' deviceQueryDrv.cpp:(.text.startup+0x10b): undefined reference to `cuDeviceComputeCapability' deviceQueryDrv.cpp:(.text.startup+0x127): undefined reference to `cuDeviceGetName' deviceQueryDrv.cpp:(.text.startup+0x16a): undefined reference to `cuDriverGetVersion' deviceQueryDrv.cpp:(.text.startup+0x1f0): undefined reference to `cuDeviceTotalMem_v2' deviceQueryDrv.cpp:(.text.startup+0x262): undefined reference to `cuDeviceGetAttribute' deviceQueryDrv.cpp:(.text.startup+0x457): undefined reference to `cuDeviceGetAttribute' deviceQueryDrv.cpp:(.text.startup+0x4bc): undefined reference to `cuDeviceGetAttribute' deviceQueryDrv.cpp:(.text.startup+0x502): undefined reference to `cuDeviceGetAttribute' deviceQueryDrv.cpp:(.text.startup+0x533): undefined reference to `cuDeviceGetAttribute' obj/x86_64/release/deviceQueryDrv.cpp.o:deviceQueryDrv.cpp:(.text.startup+0x55e): more undefined references to `cuDeviceGetAttribute' follow collect2: ld returned 1 exit status make[2]: *** [../../bin/linux/release/deviceQueryDrv] Error 1 make[2]: Leaving directory `/home/john/NVIDIA_GPU_Computing_SDK/C/src/deviceQueryDrv' make[1]: *** [src/deviceQueryDrv/Makefile.ph_build] Error 2 make[1]: Leaving directory `/home/john/NVIDIA_GPU_Computing_SDK/C' make: *** [all] Error 2 I would appreciate any help that anyone can provide me with. If I can provide any further information please let me know. Thanks.

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  • Checking data of all same class elements

    - by Tiffani
    I need the code to check the data-name value of all instances of .account-select. Right now it just checks the first .account-select element and not any subsequent ones. The function right now is on click of an element such as John Smith, it checks the data-name of the .account-select lis. If the data-names are the same, it does not create a new li with the John Smith data. If no data-names are equal to John Smith, then it adds an li with John Smith. This is the JS-Fiddle I made for it so you can see what I am referring to: http://jsfiddle.net/rsxavior/vDCNy/22/ Any help would be greatly appreciated. This is the Jquery Code I am using right now. $('.account').click(function () { var acc = $(this).data("name"); var sel = $('.account-select').data("name"); if (acc === sel) { } else { $('.account-hidden-li').append('<li class="account-select" data-name="'+ $(this).data("name") +'">' + $(this).data("name") + '<a class="close bcn-close" data-dismiss="alert" href="#">&times;</a></li>'); } }); And the HTML: <ul> <li><a class="account" data-name="All" href="#">All</a></li> <li><a class="account" data-name="John Smith" href="#">John Smith</a></li> </ul> <ul class="account-hidden-li"> <ul>

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  • Pattern/Matcher in Java?

    - by user1007059
    I have a certain text in Java, and I want to use pattern and matcher to extract something from it. This is my program: public String getItemsByType(String text, String start, String end) { String patternHolder; StringBuffer itemLines = new StringBuffer(); patternHolder = start + ".*" + end; Pattern pattern = Pattern.compile(patternHolder); Matcher matcher = pattern.matcher(text); while (matcher.find()) { itemLines.append(text.substring(matcher.start(), matcher.end()) + "\n"); } return itemLines.toString(); } This code works fully WHEN the searched text is on the same line, for instance: String text = "My name is John and I am 18 years Old"; getItemsByType(text, "My", "John"); immediately grabs the text "My name is John" out of the text. However, when my text looks like this: String text = "My name\nis John\nand I'm\n18 years\nold"; getItemsByType(text, "My", "John"); It doesn't grab anything, since "My" and "John" are on different lines. How do I solve this?

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  • Parsing back to 'messy' API strcuture

    - by Eric Fail
    I'm fetching data from an online database (REDcap) via API and the data gets delivered in as comma separated string like this, RAW.API <- structure("id,event_arm,name,dob,pushed_text,pushed_calc,complete\n\"01\",\"event_1_arm_1\",\"John\",\"1979-05-01\",\"\",\"\",2\n\"01\",\"event_2_arm_1\",\"John\",\"2012-09-02\",\"abc\",\"123\",1\n\"01\",\"event_3_arm_1\",\"John\",\"2012-09-10\",\"\",\"\",2\n\"02\",\"event_1_arm_1\",\"Mary\",\"1951-09-10\",\"def\",\"456\",2\n\"02\",\"event_2_arm_1\",\"Mary\",\"1978-09-12\",\"\",\"\",2\n", "`Content-Type`" = structure(c("text/html", "utf-8"), .Names = c("", "charset"))) I have this script that nicely parses it into a data frame, (df <- read.table(file = textConnection(RAW.API), header = TRUE, sep = ",", na.strings = "", stringsAsFactors = FALSE)) id event_arm name dob pushed_text pushed_calc complete 1 1 event_1_arm_1 John 1979-05-01 <NA> NA 2 2 1 event_2_arm_1 John 2012-09-02 abc 123 1 3 1 event_3_arm_1 John 2012-09-10 <NA> NA 2 4 2 event_1_arm_1 Mary 1951-09-10 def 456 2 5 2 event_2_arm_1 Mary 1978-09-12 <NA> NA 2 I then do some calculations and write them to pushed_text and pushed_calc whereafter I need to format the data back to the messy comma separated structure it came in. I imagine something like this, API.back <- `some magic command`(df, ...) identical(RAW.API, API.back) [1] TRUE Some command that can format my data from the data frame I made, df, back to the structure that the raw API-object came in, RAW.API. Any help would be very appreciated.

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  • Rewriting from headers in Postfix

    - by inxilpro
    I want to configure Postfix to replace the 'From' header in all forwarded/aliased messages with a custom email address, and the 'Reply-To' header with the original sender's address. Is that something that can be done with a simple configuration change, or am I looking at a more complex problem? For example: Original Message: From: "John Smith" <[email protected]> To: "Jane Rice" <[email protected]> Would get translated to: From: "My Email Forwarding Service" <[email protected]> Reply-To: "John Smith" <[email protected]> To: "Jane Rice" <[email protected]> Ideally, I would also have it rewrite the message body (adding something about how the message was forwarded for them), but I know that's much more difficult. We have a number of email aliases, and everytime someone reports spam they received through their alias, our server gets flagged. I'm trying to minimize that damage as much as possible. Any help is greatly appreciated!

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  • ubuntu apache subdomains pointing to main domain

    - by Suhail Thakur
    i have a ubuntu server with apache setup, the main domain on the server is a subdomain app.example.com, which is working fine, now if i setup john.app.example.com, then that also is displaying the web page of app.example.com, the DocumentRoot for john.app.example.com is different, still it shows the web page of app.example.com. how can i resolve this, so john.app.example.com displays the pages that are there in its DocumentRoot.

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  • Multiple contacts with shared information

    - by Keith Thompson
    Background: I currently have several hundred contacts, synchronized between a Microsoft Exchange server and several mobile devices. I also save exported copies of the contacts in .vcf format. Is there a good way (application, file format, whatever) to maintain contacts with shared information? A very common scenario is that I have contacts for two or more people who live in the same house, for example: John Doe 123 Main Street, Anytown USA Home: 555-555-1111 Work: 555-555-2222 Mobile: 555-555-3333 E-mail: John[email protected] Jane Doe 123 Main Street, Anytown USA Home: 555-555-1111 Work: 555-555-4444 Mobile: 555-555-5555 E-mail: [email protected] As you can see, both contacts have the same home address and phone number, but distinct names and work and mobile phone numbers. (Other information might also be either shared or distinct.) The applications and file formats I'm familiar with don't seem to have a good way to deal with this. If I use a single "John & Jane Doe" contact for both, it's difficult to distinguish the distinct information (if I want to call Jane's mobile phone rather than John's). If I use a separate contact for each, I have to remember to update both of them (or all of them for N 2) when they move or change their home phone number. An ideal solution would let me create a record containing information for their household, and have each of their contact records contain a reference to the household record, so that when I view John's contact record I see both shared and distinct information. Is there anything out there that has good support this kind of thing? (I would think there would be, since it's a very common scenario.) (I suppose I could roll my own system that generates merged .vcf files from some extended format, but that wouldn't play well with synchronizing across multiple devices.)

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  • Thunderbird: how to move mails into correct thread? (mailing lists)

    - by unor
    I'm subscribed to some mailing lists and every day people reply to a wrong mail (or they don't reply at all), so that their mail lands in the wrong (or a new) thread. I set the mail display in "View ? Sort by" to "Threaded". Example: mailing list "Foobar": [Foobar] random topic Re: [Foobar] random topic Re: [Foobar] random topic Re: [Foobar] random topic Re: [Foobar] random topic [Foobar] I'm John Doe Re: [Foobar] I'm John Doe Re: [Foobar] Welcome, John Re: [Foobar] random topic There are two discussions, one about "random topic", one about "John Doe". The subject line changes in the discussion about John Doe, which is fine (no problem here). But the last mail should be pigeonholed in the first thread. Instead it is at the top-level. Now, how can I move that last mail into the correct thread? I tried to drag&drop it at the mail I think it should be a reply to, but this doesn't work. I think theoretically it should be possible by fiddling with the mail headers after receiving the mail, but this doesn't seem to be a comfortable way.

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  • I recieved an email but the to address is not mine

    - by user35072
    As per title, my email address is [email protected] and received an email from [email protected]. But in my Web Client i see: From: [email protected] To: [email protected] I received this on my [email protected] account so how did i get this email in my inbox? I have no affiliation whatsoever with [email protected]. Actually i have received a few emails from [email protected] where the TO address differs. What's going on?

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  • ubuntu apache subdomains pointing to main domain

    - by Suhail Thakur
    I have a ubuntu server with apache setup, the main domain on the server is a subdomain app.example.com, which is working fine. Now if I setup john.app.example.com, then that also is displaying the web page of app.example.com, the DocumentRoot for john.app.example.com is different, still it shows the web page of app.example.com. how can I resolve this, so john.app.example.com displays the pages that are there in its DocumentRoot.

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  • How to add LDAP user to existing local group in RHEL?

    - by Highway of Life
    I'm attempting to add some of our LDAP users to a locally defined group on our RHEL server, however I get an error stating that the LDAP user is not found in /etc/passwd. What would be the best way to allow LDAP users to be added to local groups? My feeling is that this must be done manually. I could edit: /etc/group and add the LDAP group to the list. Would that be ideal? [server]# id apache uid=409(apache) gid=409(apache) groups=409(apache) context=user_u:system_r:unconfined_t:s0 [server]# id john.doe uid=11389(john.doe) gid=6097(ABC_Corporate_US) groups=6097(ABC_Corporate_US) context=user_u:system_r:unconfined_t:s0 [server]# /usr/sbin/usermod -a -G apache john.doe usermod: john.doe not found in /etc/passwd OS: RHEL (Red Hat Enterprise Linux Server release 5.3 (Tikanga)) Note: Updating the OS on this machine is not an option.

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  • Archiving mails with postfix: how to filter mails?

    - by Tronic
    i wanto to implement the following scenario: we use a postfix mailserver. to archive all old and new mails, i want to setup a second postfix on our fileserver and create a single mailbox "archive". then every mail gets forwarded as bcc to this mailbox automatically. now, i want to create different folders in a maildir structure and let the server move each mail to the right subfolder of the mailbox based on its sender or receiver. e.g. when we get a mail to one of our employees named "John Doe" at john[email protected], the mail should be moved to "Inbox/John Doe Incoming". the same applies when john doe sends a mail, folder would be "Inbox/John Doe Outgoing". how can i implement this filter behaviour. i heard of Procmail and Maildrop. Which of the two would you prefer? Which is more easy to configure? Any out-of-box solutions here? thanks in advance!

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  • Managing Linux Directory Permissions & SFTP

    - by Dizzle
    Good morning; I have a RHEL 5.7 web server configured to allow SSH/SFTP only by specific groups. I'd like for content managers to upload content to their respective directories and have that content inherit the user/group ownership of the directory regardless of upload method or application. For example: John is in group "web" for SSH/SFTP rights and "finance" for directory permissions, and uploads to directory "webstuff" via SFTP. Directory "webstuff" has permissions of "2760" (rwxrws---), and ownership of "apache:finance". If John uploads an update to an existing file in "webstuff", the ownership of the file stays at "apache:finance". If John uploads a new file to "webstuff", the ownership of the file is "john:finance". My desire is to have any file from John uploaded to "webstuff" to change to the directory's owner. I've tried with setuid and setgid both set, but the user-ownership didn't take. I've seen mentions on ServerFault of using ACL's, or a chrooted jail for SFTP but I have yet to configure and test them, and I don't know if they're a viable solution (they could be, I just don't know because I've never done either). Any thoughts and assistance would be greatly appreciated.

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  • [Ubuntu]df Total size is not correct compared with the size of the disk

    - by John John
    I'm running Ubuntu Squeeze and on one of the partitions df is showing the Total size as 335G: Filesystem Size Used Avail Use% Mounted on /dev/sdb 335G 225G 94G 71% /mnt However in the past it was showing as 360GB (which is the actual size): fdisk -l /dev/sdb Disk /dev/sdb: 365.0 GB, 365041287168 bytes lsof +L1 does not return anything (and anyway if this would be the case the Total space should not be affected.) On this partition I'm writing (and deleting) a lot of files and this happened before in the past, but problem solved by itself.

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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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