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  • What's an effective way to reuse ArrayLists in a for loop?

    - by Patrick
    hi, I'm reusing the same ArrayList in a for loop, and I use for loop results = new ArrayList<Integer>(); experts = new ArrayList<Integer>(); output = new ArrayList<String>(); .... to create new ones. I guess this is wrong, because I'm allocating new memory. Is this correct ? If yes, how can I empty them ? Added: another example I'm creating new variables each time I call this method. Is this good practice ? I mean to create new precision, relevantFound.. etc ? Or should I declare them in my class, outside the method to not allocate more and more memory ? public static void computeMAP(ArrayList<Integer> results, ArrayList<Integer> experts) { //compute MAP double precision = 0; int relevantFound = 0; double sumprecision = 0; thanks

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  • CentOS 5.4 NFS v4 client file permissions differ from original files & NFS Share file contents

    - by p4guru
    Having a strange problem with NFS share and file permissions on the 1 out of the 2 NFS clients, web1 has file permissions issues but web2 is fine. web1 and web2 are load balanced web servers. So questions are: how do I ensure NFS share file contents retain the same permissions for user/group as the original files on web1 server like they do on web2 server ? how do I reverse what I did on web1, i tried unmount command and said command not found ? Information: I'm using 3 dedicated server setup. All 3 servers CentOS 5.4 64bit based. servers are as follows: web1 - nfs client with file permissions issues web2 - nfs client file permissions are OKAY db1 - nfs share at /nfsroot web2 nfs client was setup by my web host, while web1 was setup by me. I did the following commands on web1 and it worked with updating db1 nfsroot share at /nfsroot/site_css with latest files on web1 but the file permissions don't stick even if i use tar with -p command to perserve file permissions ? cd /home/username/public_html/forums/script/ tar -zcp site_css/ > site_css.tar.gz mount -t nfs4 nfsshareipaddress:/site_css /home/username/public_html/forums/scripts/site_css/ -o rw,soft cd /home/username/public_html/forums/script/ tar -zxf site_css.tar.gz But checking on web1 file permissions no longer username user/group but owned by nobody ? but web2 file permissions correct ? This is only a problem for web1 while web2 is correct ? Looks like numeric ids aren't the same ? Not sure how to correct this ? web1 with incorrect user/group of nobody ls -alh /home/username/public_html/forums/scripts/site_css total 48K drwxrwxrwx 2 nobody nobody 4.0K Feb 22 02:37 ./ drwxr-xr-x 3 username username 4.0K Feb 22 02:43 ../ -rw-r--r-- 1 nobody nobody 1 Nov 30 2006 index.html -rw-r--r-- 1 nobody nobody 5.8K Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 nobody nobody 5.8K Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 nobody nobody 5.8K Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 nobody nobody 5.8K Feb 18 05:37 style-cc2f96c9-00011.css web1 numeric ids ls -n /home/username/public_html/forums/scripts/site_css total 48 drwxrwxrwx 2 99 99 4096 Feb 22 02:37 ./ drwxr-xr-x 3 503 500 4096 Feb 22 02:43 ../ -rw-r--r-- 1 99 99 1 Nov 30 2006 index.html -rw-r--r-- 1 99 99 5876 Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 99 99 5877 Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 99 99 5877 Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 99 99 5876 Feb 18 05:37 style-cc2f96c9-00011.css web2 correct username user/group permissions ls -alh /home/username/public_html/forums/scripts/site_css total 48K drwxrwxrwx 2 root root 4.0K Feb 22 02:37 ./ drwxr-xr-x 3 username username 4.0K Dec 2 14:51 ../ -rw-r--r-- 1 username username 1 Nov 30 2006 index.html -rw-r--r-- 1 username username 5.8K Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 username username 5.8K Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 username username 5.8K Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 username username 5.8K Feb 18 05:37 style-cc2f96c9-00011.css web2 numeric ids ls -n /home/username/public_html/forums/scripts/site_css total 48 drwxrwxrwx 2 503 500 4096 Feb 22 02:37 ./ drwxr-xr-x 3 503 500 4096 Dec 2 14:51 ../ -rw-r--r-- 1 503 500 1 Nov 30 2006 index.html -rw-r--r-- 1 503 500 5876 Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 503 500 5877 Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 503 500 5877 Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 503 500 5876 Feb 18 05:37 style-cc2f96c9-00011.css I checked db1 /nfsroot/site_css and user/group ownership was incorrect for newer files dated feb22 owned by root and not username ? on db1 originally incorrect root assigned user/group for new feb22 dated files ls -alh /nfsroot/site_css total 44K drwxrwxrwx 2 root root 4.0K Feb 22 02:37 . drwxr-xr-x 17 root root 4.0K Feb 17 12:06 .. -rw-r--r-- 1 root root 1 Nov 30 2006 index.html -rw-r--r-- 1 root root 5.8K Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 root root 5.8K Feb 22 02:37 style-95001864-00002.css -rw------- 1 username nfs 5.8K Feb 18 05:37 style-b1879ba7-00002.css -rw------- 1 username nfs 5.8K Feb 18 05:37 style-cc2f96c9-00011.css Then I chmod them all on db1 and chown to set to right ownership on db1 so it looks like below on db1 once corrected the newer feb22 dated files ls -alh /nfsroot/site_css total 44K drwxrwxrwx 2 root root 4.0K Feb 22 02:37 . drwxr-xr-x 17 root root 4.0K Feb 17 12:06 .. -rw-r--r-- 1 username username 1 Nov 30 2006 index.html -rw-r--r-- 1 username username 5.8K Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 username username 5.8K Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 username username 5.8K Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 username username 5.8K Feb 18 05:37 style-cc2f96c9-00011.css but still web1 shows owned by nobody ? while web2 shows correct permissions ? web1 still with incorrect user/group of nobody not matching what web2 and db1 are set to ? ls -alh /home/username/public_html/forums/scripts/site_css total 48K drwxrwxrwx 2 nobody nobody 4.0K Feb 22 02:37 ./ drwxr-xr-x 3 username username 4.0K Feb 22 02:43 ../ -rw-r--r-- 1 nobody nobody 1 Nov 30 2006 index.html -rw-r--r-- 1 nobody nobody 5.8K Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 nobody nobody 5.8K Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 nobody nobody 5.8K Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 nobody nobody 5.8K Feb 18 05:37 style-cc2f96c9-00011.css Just so confusing so any help is very very much appreciated! thanks

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  • CentOS 5.4 NFS v4 client file permissions differ from original files & NFS Share file contents

    - by p4guru
    Having a strange problem with NFS share and file permissions on the 1 out of the 2 NFS clients, web1 has file permissions issues but web2 is fine. web1 and web2 are load balanced web servers. So questions are: how do I ensure NFS share file contents retain the same permissions for user/group as the original files on web1 server like they do on web2 server ? how do I reverse what I did on web1, i tried unmount command and said command not found ? Information: I'm using 3 dedicated server setup. All 3 servers CentOS 5.4 64bit based. servers are as follows: web1 - nfs client with file permissions issues web2 - nfs client file permissions are OKAY db1 - nfs share at /nfsroot web2 nfs client was setup by my web host, while web1 was setup by me. I did the following commands on web1 and it worked with updating db1 nfsroot share at /nfsroot/site_css with latest files on web1 but the file permissions don't stick even if i use tar with -p command to perserve file permissions ? cd /home/username/public_html/forums/script/ tar -zcp site_css/ > site_css.tar.gz mount -t nfs4 nfsshareipaddress:/site_css /home/username/public_html/forums/scripts/site_css/ -o rw,soft cd /home/username/public_html/forums/script/ tar -zxf site_css.tar.gz But checking on web1 file permissions no longer username user/group but owned by nobody ? but web2 file permissions correct ? This is only a problem for web1 while web2 is correct ? Looks like numeric ids aren't the same ? Not sure how to correct this ? web1 with incorrect user/group of nobody ls -alh /home/username/public_html/forums/scripts/site_css total 48K drwxrwxrwx 2 nobody nobody 4.0K Feb 22 02:37 ./ drwxr-xr-x 3 username username 4.0K Feb 22 02:43 ../ -rw-r--r-- 1 nobody nobody 1 Nov 30 2006 index.html -rw-r--r-- 1 nobody nobody 5.8K Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 nobody nobody 5.8K Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 nobody nobody 5.8K Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 nobody nobody 5.8K Feb 18 05:37 style-cc2f96c9-00011.css web1 numeric ids ls -n /home/username/public_html/forums/scripts/site_css total 48 drwxrwxrwx 2 99 99 4096 Feb 22 02:37 ./ drwxr-xr-x 3 503 500 4096 Feb 22 02:43 ../ -rw-r--r-- 1 99 99 1 Nov 30 2006 index.html -rw-r--r-- 1 99 99 5876 Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 99 99 5877 Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 99 99 5877 Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 99 99 5876 Feb 18 05:37 style-cc2f96c9-00011.css web2 correct username user/group permissions ls -alh /home/username/public_html/forums/scripts/site_css total 48K drwxrwxrwx 2 root root 4.0K Feb 22 02:37 ./ drwxr-xr-x 3 username username 4.0K Dec 2 14:51 ../ -rw-r--r-- 1 username username 1 Nov 30 2006 index.html -rw-r--r-- 1 username username 5.8K Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 username username 5.8K Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 username username 5.8K Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 username username 5.8K Feb 18 05:37 style-cc2f96c9-00011.css web2 numeric ids ls -n /home/username/public_html/forums/scripts/site_css total 48 drwxrwxrwx 2 503 500 4096 Feb 22 02:37 ./ drwxr-xr-x 3 503 500 4096 Dec 2 14:51 ../ -rw-r--r-- 1 503 500 1 Nov 30 2006 index.html -rw-r--r-- 1 503 500 5876 Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 503 500 5877 Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 503 500 5877 Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 503 500 5876 Feb 18 05:37 style-cc2f96c9-00011.css I checked db1 /nfsroot/site_css and user/group ownership was incorrect for newer files dated feb22 owned by root and not username ? on db1 originally incorrect root assigned user/group for new feb22 dated files ls -alh /nfsroot/site_css total 44K drwxrwxrwx 2 root root 4.0K Feb 22 02:37 . drwxr-xr-x 17 root root 4.0K Feb 17 12:06 .. -rw-r--r-- 1 root root 1 Nov 30 2006 index.html -rw-r--r-- 1 root root 5.8K Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 root root 5.8K Feb 22 02:37 style-95001864-00002.css -rw------- 1 username nfs 5.8K Feb 18 05:37 style-b1879ba7-00002.css -rw------- 1 username nfs 5.8K Feb 18 05:37 style-cc2f96c9-00011.css Then I chmod them all on db1 and chown to set to right ownership on db1 so it looks like below on db1 once corrected the newer feb22 dated files ls -alh /nfsroot/site_css total 44K drwxrwxrwx 2 root root 4.0K Feb 22 02:37 . drwxr-xr-x 17 root root 4.0K Feb 17 12:06 .. -rw-r--r-- 1 username username 1 Nov 30 2006 index.html -rw-r--r-- 1 username username 5.8K Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 username username 5.8K Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 username username 5.8K Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 username username 5.8K Feb 18 05:37 style-cc2f96c9-00011.css but still web1 shows owned by nobody ? while web2 shows correct permissions ? web1 still with incorrect user/group of nobody not matching what web2 and db1 are set to ? ls -alh /home/username/public_html/forums/scripts/site_css total 48K drwxrwxrwx 2 nobody nobody 4.0K Feb 22 02:37 ./ drwxr-xr-x 3 username username 4.0K Feb 22 02:43 ../ -rw-r--r-- 1 nobody nobody 1 Nov 30 2006 index.html -rw-r--r-- 1 nobody nobody 5.8K Feb 22 02:37 style-057c3df0-00011.css -rw-r--r-- 1 nobody nobody 5.8K Feb 22 02:37 style-95001864-00002.css -rw-r--r-- 1 nobody nobody 5.8K Feb 18 05:37 style-b1879ba7-00002.css -rw-r--r-- 1 nobody nobody 5.8K Feb 18 05:37 style-cc2f96c9-00011.css Just so confusing so any help is very very much appreciated! thanks

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  • C#/.NET Little Wonders: Fun With Enum Methods

    - by James Michael Hare
    Once again lets dive into the Little Wonders of .NET, those small things in the .NET languages and BCL classes that make development easier by increasing readability, maintainability, and/or performance. So probably every one of us has used an enumerated type at one time or another in a C# program.  The enumerated types we create are a great way to represent that a value can be one of a set of discrete values (or a combination of those values in the case of bit flags). But the power of enum types go far beyond simple assignment and comparison, there are many methods in the Enum class (that all enum types “inherit” from) that can give you even more power when dealing with them. IsDefined() – check if a given value exists in the enum Are you reading a value for an enum from a data source, but are unsure if it is actually a valid value or not?  Casting won’t tell you this, and Parse() isn’t guaranteed to balk either if you give it an int or a combination of flags.  So what can we do? Let’s assume we have a small enum like this for result codes we want to return back from our business logic layer: 1: public enum ResultCode 2: { 3: Success, 4: Warning, 5: Error 6: } In this enum, Success will be zero (unless given another value explicitly), Warning will be one, and Error will be two. So what happens if we have code like this where perhaps we’re getting the result code from another data source (could be database, could be web service, etc)? 1: public ResultCode PerformAction() 2: { 3: // set up and call some method that returns an int. 4: int result = ResultCodeFromDataSource(); 5:  6: // this will suceed even if result is < 0 or > 2. 7: return (ResultCode) result; 8: } So what happens if result is –1 or 4?  Well, the cast does not fail, so what we end up with would be an instance of a ResultCode that would have a value that’s outside of the bounds of the enum constants we defined. This means if you had a block of code like: 1: switch (result) 2: { 3: case ResultType.Success: 4: // do success stuff 5: break; 6:  7: case ResultType.Warning: 8: // do warning stuff 9: break; 10:  11: case ResultType.Error: 12: // do error stuff 13: break; 14: } That you would hit none of these blocks (which is a good argument for always having a default in a switch by the way). So what can you do?  Well, there is a handy static method called IsDefined() on the Enum class which will tell you if an enum value is defined.  1: public ResultCode PerformAction() 2: { 3: int result = ResultCodeFromDataSource(); 4:  5: if (!Enum.IsDefined(typeof(ResultCode), result)) 6: { 7: throw new InvalidOperationException("Enum out of range."); 8: } 9:  10: return (ResultCode) result; 11: } In fact, this is often recommended after you Parse() or cast a value to an enum as there are ways for values to get past these methods that may not be defined. If you don’t like the syntax of passing in the type of the enum, you could clean it up a bit by creating an extension method instead that would allow you to call IsDefined() off any isntance of the enum: 1: public static class EnumExtensions 2: { 3: // helper method that tells you if an enum value is defined for it's enumeration 4: public static bool IsDefined(this Enum value) 5: { 6: return Enum.IsDefined(value.GetType(), value); 7: } 8: }   HasFlag() – an easier way to see if a bit (or bits) are set Most of us who came from the land of C programming have had to deal extensively with bit flags many times in our lives.  As such, using bit flags may be almost second nature (for a quick refresher on bit flags in enum types see one of my old posts here). However, in higher-level languages like C#, the need to manipulate individual bit flags is somewhat diminished, and the code to check for bit flag enum values may be obvious to an advanced developer but cryptic to a novice developer. For example, let’s say you have an enum for a messaging platform that contains bit flags: 1: // usually, we pluralize flags enum type names 2: [Flags] 3: public enum MessagingOptions 4: { 5: None = 0, 6: Buffered = 0x01, 7: Persistent = 0x02, 8: Durable = 0x04, 9: Broadcast = 0x08 10: } We can combine these bit flags using the bitwise OR operator (the ‘|’ pipe character): 1: // combine bit flags using 2: var myMessenger = new Messenger(MessagingOptions.Buffered | MessagingOptions.Broadcast); Now, if we wanted to check the flags, we’d have to test then using the bit-wise AND operator (the ‘&’ character): 1: if ((options & MessagingOptions.Buffered) == MessagingOptions.Buffered) 2: { 3: // do code to set up buffering... 4: // ... 5: } While the ‘|’ for combining flags is easy enough to read for advanced developers, the ‘&’ test tends to be easy for novice developers to get wrong.  First of all you have to AND the flag combination with the value, and then typically you should test against the flag combination itself (and not just for a non-zero)!  This is because the flag combination you are testing with may combine multiple bits, in which case if only one bit is set, the result will be non-zero but not necessarily all desired bits! Thanks goodness in .NET 4.0 they gave us the HasFlag() method.  This method can be called from an enum instance to test to see if a flag is set, and best of all you can avoid writing the bit wise logic yourself.  Not to mention it will be more readable to a novice developer as well: 1: if (options.HasFlag(MessagingOptions.Buffered)) 2: { 3: // do code to set up buffering... 4: // ... 5: } It is much more concise and unambiguous, thus increasing your maintainability and readability. It would be nice to have a corresponding SetFlag() method, but unfortunately generic types don’t allow you to specialize on Enum, which makes it a bit more difficult.  It can be done but you have to do some conversions to numeric and then back to the enum which makes it less of a payoff than having the HasFlag() method.  But if you want to create it for symmetry, it would look something like this: 1: public static T SetFlag<T>(this Enum value, T flags) 2: { 3: if (!value.GetType().IsEquivalentTo(typeof(T))) 4: { 5: throw new ArgumentException("Enum value and flags types don't match."); 6: } 7:  8: // yes this is ugly, but unfortunately we need to use an intermediate boxing cast 9: return (T)Enum.ToObject(typeof (T), Convert.ToUInt64(value) | Convert.ToUInt64(flags)); 10: } Note that since the enum types are value types, we need to assign the result to something (much like string.Trim()).  Also, you could chain several SetFlag() operations together or create one that takes a variable arg list if desired. Parse() and ToString() – transitioning from string to enum and back Sometimes, you may want to be able to parse an enum from a string or convert it to a string - Enum has methods built in to let you do this.  Now, many may already know this, but may not appreciate how much power are in these two methods. For example, if you want to parse a string as an enum, it’s easy and works just like you’d expect from the numeric types: 1: string optionsString = "Persistent"; 2:  3: // can use Enum.Parse, which throws if finds something it doesn't like... 4: var result = (MessagingOptions)Enum.Parse(typeof (MessagingOptions), optionsString); 5:  6: if (result == MessagingOptions.Persistent) 7: { 8: Console.WriteLine("It worked!"); 9: } Note that Enum.Parse() will throw if it finds a value it doesn’t like.  But the values it likes are fairly flexible!  You can pass in a single value, or a comma separated list of values for flags and it will parse them all and set all bits: 1: // for string values, can have one, or comma separated. 2: string optionsString = "Persistent, Buffered"; 3:  4: var result = (MessagingOptions)Enum.Parse(typeof (MessagingOptions), optionsString); 5:  6: if (result.HasFlag(MessagingOptions.Persistent) && result.HasFlag(MessagingOptions.Buffered)) 7: { 8: Console.WriteLine("It worked!"); 9: } Or you can parse in a string containing a number that represents a single value or combination of values to set: 1: // 3 is the combination of Buffered (0x01) and Persistent (0x02) 2: var optionsString = "3"; 3:  4: var result = (MessagingOptions) Enum.Parse(typeof (MessagingOptions), optionsString); 5:  6: if (result.HasFlag(MessagingOptions.Persistent) && result.HasFlag(MessagingOptions.Buffered)) 7: { 8: Console.WriteLine("It worked again!"); 9: } And, if you really aren’t sure if the parse will work, and don’t want to handle an exception, you can use TryParse() instead: 1: string optionsString = "Persistent, Buffered"; 2: MessagingOptions result; 3:  4: // try parse returns true if successful, and takes an out parm for the result 5: if (Enum.TryParse(optionsString, out result)) 6: { 7: if (result.HasFlag(MessagingOptions.Persistent) && result.HasFlag(MessagingOptions.Buffered)) 8: { 9: Console.WriteLine("It worked!"); 10: } 11: } So we covered parsing a string to an enum, what about reversing that and converting an enum to a string?  The ToString() method is the obvious and most basic choice for most of us, but did you know you can pass a format string for enum types that dictate how they are written as a string?: 1: MessagingOptions value = MessagingOptions.Buffered | MessagingOptions.Persistent; 2:  3: // general format, which is the default, 4: Console.WriteLine("Default : " + value); 5: Console.WriteLine("G (default): " + value.ToString("G")); 6:  7: // Flags format, even if type does not have Flags attribute. 8: Console.WriteLine("F (flags) : " + value.ToString("F")); 9:  10: // integer format, value as number. 11: Console.WriteLine("D (num) : " + value.ToString("D")); 12:  13: // hex format, value as hex 14: Console.WriteLine("X (hex) : " + value.ToString("X")); Which displays: 1: Default : Buffered, Persistent 2: G (default): Buffered, Persistent 3: F (flags) : Buffered, Persistent 4: D (num) : 3 5: X (hex) : 00000003 Now, you may not really see a difference here between G and F because I used a [Flags] enum, the difference is that the “F” option treats the enum as if it were flags even if the [Flags] attribute is not present.  Let’s take a non-flags enum like the ResultCode used earlier: 1: // yes, we can do this even if it is not [Flags] enum. 2: ResultCode value = ResultCode.Warning | ResultCode.Error; And if we run that through the same formats again we get: 1: Default : 3 2: G (default): 3 3: F (flags) : Warning, Error 4: D (num) : 3 5: X (hex) : 00000003 Notice that since we had multiple values combined, but it was not a [Flags] marked enum, the G and default format gave us a number instead of a value name.  This is because the value was not a valid single-value constant of the enum.  However, using the F flags format string, it broke out the value into its component flags even though it wasn’t marked [Flags]. So, if you want to get an enum to display appropriately for whether or not it has the [Flags] attribute, use G which is the default.  If you always want it to attempt to break down the flags, use F.  For numeric output, obviously D or  X are the best choice depending on whether you want decimal or hex. Summary Hopefully, you learned a couple of new tricks with using the Enum class today!  I’ll add more little wonders as I think of them and thanks for all the invaluable input!   Technorati Tags: C#,.NET,Little Wonders,Enum,BlackRabbitCoder

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  • PLPGSQL array assignment not working, "array subscript in assignment must not be null"

    - by Koen Schmeets
    Hello there, When assigning mobilenumbers to a varchar[] in a loop through results it gives me the following error: "array subscript in assignment must not be null" Also, i think the query that joins member uuids, and group member uuids, into one, grouped on the user_id, i think it can be done better, or maybe this is even why it is going wrong in the first place! Any help is very appreciated.. Thank you very much! CREATE OR REPLACE FUNCTION create_membermessage(in_company_uuid uuid, in_user_uuid uuid, in_destinationmemberuuids uuid[], in_destinationgroupuuids uuid[], in_title character varying, in_messagecontents character varying, in_timedelta interval, in_messagecosts numeric, OUT out_status integer, OUT out_status_description character varying, OUT out_value VARCHAR[], OUT out_trigger uuid[]) RETURNS record LANGUAGE plpgsql AS $$ DECLARE temp_count INTEGER; temp_costs NUMERIC; temp_balance NUMERIC; temp_campaign_uuid UUID; temp_record RECORD; temp_mobilenumbers VARCHAR[]; temp_destination_uuids UUID[]; temp_iterator INTEGER; BEGIN out_status := NULL; out_status_description := NULL; out_value := NULL; out_trigger := NULL; SELECT INTO temp_count COUNT(*) FROM costs WHERE costtype = 'MEMBERMESSAGE' AND company_uuid = in_company_uuid AND startdatetime < NOW() AND (enddatetime > NOW() OR enddatetime IS NULL); IF temp_count > 1 THEN out_status := 1; out_status_description := 'Invalid rows in costs table!'; RETURN; ELSEIF temp_count = 1 THEN SELECT INTO temp_costs costs FROM costs WHERE costtype = 'MEMBERMESSAGE' AND company_uuid = in_company_uuid AND startdatetime < NOW() AND (enddatetime > NOW() OR enddatetime IS NULL); ELSE SELECT INTO temp_costs costs FROM costs WHERE costtype = 'MEMBERMESSAGE' AND company_uuid IS NULL AND startdatetime < NOW() AND (enddatetime > NOW() OR enddatetime IS NULL); END IF; IF temp_costs != in_messagecosts THEN out_status := 2; out_status_description := 'Message costs have changed during sending of the message'; RETURN; ELSE SELECT INTO temp_balance balance FROM companies WHERE company_uuid = in_company_uuid; SELECT INTO temp_count COUNT(*) FROM users WHERE (user_uuid = ANY(in_destinationmemberuuids)) OR (user_uuid IN (SELECT user_uuid FROM targetgroupusers WHERE targetgroup_uuid = ANY(in_destinationgroupuuids)) ) GROUP BY user_uuid; temp_campaign_uuid := generate_uuid('campaigns', 'campaign_uuid'); INSERT INTO campaigns (company_uuid, campaign_uuid, title, senddatetime, startdatetime, enddatetime, messagetype, state, message) VALUES (in_company_uuid, temp_campaign_uuid, in_title, NOW() + in_timedelta, NOW() + in_timedelta, NOW() + in_timedelta, 'MEMBERMESSAGE', 'DRAFT', in_messagecontents); IF in_timedelta > '00:00:00' THEN ELSE IF temp_balance < (temp_costs * temp_count) THEN UPDATE campaigns SET state = 'INACTIVE' WHERE campaign_uuid = temp_campaign_uuid; out_status := 2; out_status_description := 'Insufficient balance'; RETURN; ELSE UPDATE campaigns SET state = 'ACTIVE' WHERE campaign_uuid = temp_campaign_uuid; UPDATE companies SET balance = (temp_balance - (temp_costs * temp_count)) WHERE company_uuid = in_company_uuid; SELECT INTO temp_destination_uuids array_agg(DISTINCT(user_uuid)) FROM users WHERE (user_uuid = ANY(in_destinationmemberuuids)) OR (user_uuid IN(SELECT user_uuid FROM targetgroupusers WHERE targetgroup_uuid = ANY(in_destinationgroupuuids))); RAISE NOTICE 'Array is %', temp_destination_uuids; FOR temp_record IN (SELECT u.firstname, m.mobilenumber FROM users AS u LEFT JOIN mobilenumbers AS m ON m.user_uuid = u.user_uuid WHERE u.user_uuid = ANY(temp_destination_uuids)) LOOP IF temp_record.mobilenumber IS NOT NULL AND temp_record.mobilenumber != '' THEN --THIS IS WHERE IT GOES WRONG temp_mobilenumbers[temp_iterator] := ARRAY[temp_record.firstname::VARCHAR, temp_record.mobilenumber::VARCHAR]; temp_iterator := temp_iterator + 1; END IF; END LOOP; out_status := 0; out_status_description := 'Message created successfully'; out_value := temp_mobilenumbers; RETURN; END IF; END IF; END IF; END$$;

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  • Should we hire someone who writes C in Perl?

    - by paxdiablo
    One of my colleagues recently interviewed some candidates for a job and one said they had very good Perl experience. Since my colleague didn't know Perl, he asked me for a critique of some code written (off-site) by that potential hire, so I had a look and told him my concerns (the main one was that it originally had no comments and it's not like we gave them enough time). However, the code works so I'm loathe to say no-go without some more input. Another concern is that this code basically looks exactly how I'd code it in C. It's been a while since I did Perl (and I didn't do a lot, I'm more a Python bod for quick scripts) but I seem to recall that it was a much more expressive language than what this guy used. I'm looking for input from real Perl coders, and suggestions for how it could be improved (and why a Perl coder should know that method of improvement). You can also wax lyrical about whether people who write one language in a totally different language should (or shouldn't be hired). I'm interested in your arguments but this question is primarily for a critique of the code. The spec was to successfully process a CSV file as follows and output the individual fields: User ID,Name , Level,Numeric ID pax, Pax Morgan ,admin,0 gt," Turner, George" rubbish,user,1 ms,"Mark \"X-Men\" Spencer","guest user",2 ab,, "user","3" The output was to be something like this (the potential hire's code actually output this): User ID,Name , Level,Numeric ID: [User ID] [Name] [Level] [Numeric ID] pax, Pax Morgan ,admin,0: [pax] [Pax Morgan] [admin] [0] gt," Turner, George " rubbish,user,1: [gt] [ Turner, George ] [user] [1] ms,"Mark \"X-Men\" Spencer","guest user",2: [ms] [Mark "X-Men" Spencer] [guest user] [2] ab,, "user","3": [ab] [] [user] [3] Here is the code they submitted: #!/usr/bin/perl # Open file. open (IN, "qq.in") || die "Cannot open qq.in"; # Process every line. while (<IN>) { chomp; $line = $_; print "$line:\n"; # Process every field in line. while ($line ne "") { # Skip spaces and start with empty field. if (substr ($line,0,1) eq " ") { $line = substr ($line,1); next; } $field = ""; $minlen = 0; # Detect quoted field or otherwise. if (substr ($line,0,1) eq "\"") { $line = substr ($line,1); $pastquote = 0; while ($line ne "") { # Special handling for quotes (\\ and \"). if (length ($line) >= 2) { if (substr ($line,0,2) eq "\\\"") { $field = $field . "\""; $line = substr ($line,2); next; } if (substr ($line,0,2) eq "\\\\") { $field = $field . "\\"; $line = substr ($line,2); next; } } # Detect closing quote. if (($pastquote == 0) && (substr ($line,0,1) eq "\"")) { $pastquote = 1; $line = substr ($line,1); $minlen = length ($field); next; } # Only worry about comma if past closing quote. if (($pastquote == 1) && (substr ($line,0,1) eq ",")) { $line = substr ($line,1); last; } $field = $field . substr ($line,0,1); $line = substr ($line,1); } } else { while ($line ne "") { if (substr ($line,0,1) eq ",") { $line = substr ($line,1); last; } if ($pastquote == 0) { $field = $field . substr ($line,0,1); } $line = substr ($line,1); } } # Strip trailing space. while ($field ne "") { if (length ($field) == $minlen) { last; } if (substr ($field,length ($field)-1,1) eq " ") { $field = substr ($field,0, length ($field)-1); next; } last; } print " [$field]\n"; } } close (IN);

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  • Need help to debug application using .Net and MySQL

    - by Tim Murphy
    What advice can you give me on how to track down a bug I have while inserting data into MySQL database with .Net? The error message is: MySql.Data.MySqlClient.MySqlException: Duplicate entry '26012' for key 'StockNumber_Number_UNIQUE' Reviewing of the log proves that StockNumber_Number of 26012 has not been inserted yet. Products in use. Visual Studio 2008. mysql.data.dll 6.0.4.0. Windows 7 Ultimate 64 bit and Windows 2003 32 bit. Custom built ORM framework (have source code). Importing data from Access 2003 database. The code works fine for 3000 - 5000 imports. The record being imported that causes the problem in a full run works fine if just importing by itself. I've also seen the error on other records if I sort the data to be imported a different way. Have tried import with and without transactions. Have logged the hell out of the system. The SQL command to create the table: CREATE TABLE `RareItems_RareItems` ( `RareItemKey` CHAR(36) NOT NULL PRIMARY KEY, `StockNumber_Text` VARCHAR(7) NOT NULL, `StockNumber_Number` INT NOT NULL AUTO_INCREMENT, UNIQUE INDEX `StockNumber_Number_UNIQUE` (`StockNumber_Number` ASC), `OurPercentage` NUMERIC , `SellPrice` NUMERIC(19, 2) , `Author` VARCHAR(250) , `CatchWord` VARCHAR(250) , `Title` TEXT , `Publisher` VARCHAR(250) , `InternalNote` VARCHAR(250) , `DateOfPublishing` VARCHAR(250) , `ExternalNote` LONGTEXT , `Description` LONGTEXT , `Scrap` LONGTEXT , `SuppressionKey` CHAR(36) NOT NULL, `TypeKey` CHAR(36) NOT NULL, `CatalogueStatusKey` CHAR(36) NOT NULL, `CatalogueRevisedDate` DATETIME , `CatalogueRevisedByKey` CHAR(36) NOT NULL, `CatalogueToBeRevisedByKey` CHAR(36) NOT NULL, `DontInsure` BIT NOT NULL, `ExtraCosts` NUMERIC(19, 2) , `IsWebReady` BIT NOT NULL, `LocationKey` CHAR(36) NOT NULL, `LanguageKey` CHAR(36) NOT NULL, `CatalogueDescription` VARCHAR(250) , `PlacePublished` VARCHAR(250) , `ToDo` LONGTEXT , `Headline` VARCHAR(250) , `DepartmentKey` CHAR(36) NOT NULL, `Temp1` INT , `Temp2` INT , `Temp3` VARCHAR(250) , `Temp4` VARCHAR(250) , `InternetStatusKey` CHAR(36) NOT NULL, `InternetStatusInfo` LONGTEXT , `PurchaseKey` CHAR(36) NOT NULL, `ConsignmentKey` CHAR(36) , `IsSold` BIT NOT NULL, `RowCreated` DATETIME NOT NULL, `RowModified` DATETIME NOT NULL ); The SQL command and parameters to insert the record: INSERT INTO `RareItems_RareItems` (`RareItemKey`, `StockNumber_Text`, `StockNumber_Number`, `OurPercentage`, `SellPrice`, `Author`, `CatchWord`, `Title`, `Publisher`, `InternalNote`, `DateOfPublishing`, `ExternalNote`, `Description`, `Scrap`, `SuppressionKey`, `TypeKey`, `CatalogueStatusKey`, `CatalogueRevisedDate`, `CatalogueRevisedByKey`, `CatalogueToBeRevisedByKey`, `DontInsure`, `ExtraCosts`, `IsWebReady`, `LocationKey`, `LanguageKey`, `CatalogueDescription`, `PlacePublished`, `ToDo`, `Headline`, `DepartmentKey`, `Temp1`, `Temp2`, `Temp3`, `Temp4`, `InternetStatusKey`, `InternetStatusInfo`, `PurchaseKey`, `ConsignmentKey`, `IsSold`, `RowCreated`, `RowModified`) VALUES (@RareItemKey, @StockNumber_Text, @StockNumber_Number, @OurPercentage, @SellPrice, @Author, @CatchWord, @Title, @Publisher, @InternalNote, @DateOfPublishing, @ExternalNote, @Description, @Scrap, @SuppressionKey, @TypeKey, @CatalogueStatusKey, @CatalogueRevisedDate, @CatalogueRevisedByKey, @CatalogueToBeRevisedByKey, @DontInsure, @ExtraCosts, @IsWebReady, @LocationKey, @LanguageKey, @CatalogueDescription, @PlacePublished, @ToDo, @Headline, @DepartmentKey, @Temp1, @Temp2, @Temp3, @Temp4, @InternetStatusKey, @InternetStatusInfo, @PurchaseKey, @ConsignmentKey, @IsSold, @RowCreated, @RowModified) @RareItemKey = 0b625bd6-776d-43d6-9405-e97159d172a6 @StockNumber_Text = 199305 @StockNumber_Number = 26012 @OurPercentage = 22.5 @SellPrice = 1250 @Author = SPARRMAN, Anders. @CatchWord = COOK: SECOND VOYAGE @Title = A Voyage Round the World with Captain James Cook in H.M.S. Resolution… Introduction and notes by Owen Rutter, wood engravings by Peter Barker-Mill. @Publisher = @InternalNote = @DateOfPublishing = 1944 @ExternalNote = The first English translation of Sparrman’s narrative, which had originally been published in Sweden in 1802-1818, and the only complete version of his account to appear in English. The eighteenth-century translation had appeared some time before the Swedish publication of the final sections of his account. Sparrman’s observant and well-written narrative of the second voyage contains much that appears nowhere else, emphasising naturally his interests in medicine, health, and natural history.&lt;br&gt;&lt;br&gt;One of 350 numbered copies: a handsomely produced and beautifully illustrated work. @Description = Small folio, wood-engravings in the text; original olive glazed cloth, top edges gilt, a very good copy. London, Golden Cockerel Press, 1944. @Scrap = @SuppressionKey = 00000000-0000-0000-0000-000000000000 @TypeKey = 93f58155-7471-46ad-84c5-262ab9dd37e8 @CatalogueStatusKey = 00000000-0000-0000-0000-000000000003 @CatalogueRevisedDate = @CatalogueRevisedByKey = c4f6fc06-956d-44c4-b393-0d5462cbffec @CatalogueToBeRevisedByKey = 00000000-0000-0000-0000-000000000000 @DontInsure = False @ExtraCosts = @IsWebReady = False @LocationKey = 00000000-0000-0000-0000-000000000000 @LanguageKey = 00000000-0000-0000-0000-000000000000 @CatalogueDescription = @PlacePublished = Golden Cockerel Press @ToDo = @Headline = @DepartmentKey = 529578a3-9189-40de-b656-eef9039d00b8 @Temp1 = @Temp2 = @Temp3 = @Temp4 = v @InternetStatusKey = 00000000-0000-0000-0000-000000000000 @InternetStatusInfo = @PurchaseKey = 00000000-0000-0000-0000-000000000000 @ConsignmentKey = @IsSold = True @RowCreated = 8/04/2010 8:49:16 PM @RowModified = 8/04/2010 8:49:16 PM Suggestions on what is causing the error and/or how to track down what is causing the problem?

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  • DataTable to JSON

    - by Joel Coehoorn
    I recently needed to serialize a datatable to JSON. Where I'm at we're still on .Net 2.0, so I can't use the JSON serializer in .Net 3.5. I figured this must have been done before, so I went looking online and found a number of different options. Some of them depend on an additional library, which I would have a hard time pushing through here. Others require first converting to List<Dictionary<>>, which seemed a little awkward and needless. Another treated all values like a string. For one reason or another I couldn't really get behind any of them, so I decided to roll my own, which is posted below. As you can see from reading the //TODO comments, it's incomplete in a few places. This code is already in production here, so it does "work" in the basic sense. The places where it's incomplete are places where we know our production data won't currently hit it (no timespans or byte arrays in the db). The reason I'm posting here is that I feel like this can be a little better, and I'd like help finishing and improving this code. Any input welcome. public static class JSONHelper { public static string FromDataTable(DataTable dt) { string rowDelimiter = ""; StringBuilder result = new StringBuilder("["); foreach (DataRow row in dt.Rows) { result.Append(rowDelimiter); result.Append(FromDataRow(row)); rowDelimiter = ","; } result.Append("]"); return result.ToString(); } public static string FromDataRow(DataRow row) { DataColumnCollection cols = row.Table.Columns; string colDelimiter = ""; StringBuilder result = new StringBuilder("{"); for (int i = 0; i < cols.Count; i++) { // use index rather than foreach, so we can use the index for both the row and cols collection result.Append(colDelimiter).Append("\"") .Append(cols[i].ColumnName).Append("\":") .Append(JSONValueFromDataRowObject(row[i], cols[i].DataType)); colDelimiter = ","; } result.Append("}"); return result.ToString(); } // possible types: // http://msdn.microsoft.com/en-us/library/system.data.datacolumn.datatype(VS.80).aspx private static Type[] numeric = new Type[] {typeof(byte), typeof(decimal), typeof(double), typeof(Int16), typeof(Int32), typeof(SByte), typeof(Single), typeof(UInt16), typeof(UInt32), typeof(UInt64)}; // I don't want to rebuild this value for every date cell in the table private static long EpochTicks = new DateTime(1970, 1, 1).Ticks; private static string JSONValueFromDataRowObject(object value, Type DataType) { // null if (value == DBNull.Value) return "null"; // numeric if (Array.IndexOf(numeric, DataType) > -1) return value.ToString(); // TODO: eventually want to use a stricter format // boolean if (DataType == typeof(bool)) return ((bool)value) ? "true" : "false"; // date -- see http://weblogs.asp.net/bleroy/archive/2008/01/18/dates-and-json.aspx if (DataType == typeof(DateTime)) return "\"\\/Date(" + new TimeSpan(((DateTime)value).ToUniversalTime().Ticks - EpochTicks).TotalMilliseconds.ToString() + ")\\/\""; // TODO: add Timespan support // TODO: add Byte[] support //TODO: this would be _much_ faster with a state machine // string/char return "\"" + value.ToString().Replace(@"\", @"\\").Replace(Environment.NewLine, @"\n").Replace("\"", @"\""") + "\""; } }

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  • sql perfomance on new server

    - by Rapunzo
    My database is running on a pc (AMD Phenom x6, intel ssd disk, 8GB DDR3 RAM and windows 7 OS + sql server 2008 R2 sp3 ) and it started working hard, timeout problems and up to 30 seconds long queries after 200 mb of database And I also have an old server pc (IBM x-series 266: 72*3 15k rpm scsi discs with raid5, 4 gb ram and windows server 2003 + sql server 2008 R2 sp3 ) and same query start to give results in 100 seconds.. I tried query analyser tool for tuning my indexed. but not so much improvements. its a big dissapointment for me. because I thought even its an old server pc it should be more powerfull with 15k rpm discs with raid5. what should I do. do I need $10.000 new server to get a good performance for my sql server? cant I use that IBM server? Extra information: there is 50 sql users and its an ERP program. There is my query ALTER FUNCTION [dbo].[fnDispoTerbiye] ( ) RETURNS TABLE AS RETURN ( SELECT MD.dispoNo, SV.sevkNo, M1.musteriAdi AS musteri, SD.tipTurId, TT.tipTur, SD.tipNo, SD.desenNo, SD.varyantNo, SUM(T.topMetre) AS toplamSevkMetre, MD.dispoMetresi, DT.gelisMetresi, ISNULL(DT.fire, 0) AS fire, SV.sevkTarihi, DT.gelisTarihi, SP.mamulTermin, SD.miktar AS siparisMiktari, M.musteriAdi AS boyahane, MD.akisNotu AS islemler, --dbo.fnAkisIslemleri(MD.dispoNo) DT.partiNo, DT.iplikBoyaId, B.tanimAd AS BoyaTuru, MAX(HD.hamEn) AS hamEn, MAX(HD.hamGramaj) AS hamGramaj, TS.mamulEn, TS.mamulGramaj, DT.atkiCekmesi, DT.cozguCekmesi, DT.fiyat, DV.dovizCins, DT.dovizId, (SELECT CASE WHEN DT.dovizId = 2 THEN CAST(round(SUM(T .topMetre) * DT.fiyat * (SELECT TOP 1 satis FROM tblKur WHERE dovizId = 2 ORDER BY tarih DESC), 2) AS numeric(18, 2)) WHEN DT.dovizId = 3 THEN CAST(round(SUM(T .topMetre) * DT.fiyat * (SELECT TOP 1 satis FROM tblKur WHERE dovizId = 3 ORDER BY tarih DESC), 2) AS numeric(18, 2)) WHEN DT.dovizId = 1 THEN CAST(round(SUM(T .topMetre) * DT.fiyat * (SELECT TOP 1 satis FROM tblKur WHERE dovizId = 1 ORDER BY tarih DESC), 2) AS numeric(18, 2)) END AS Expr1) AS ToplamTLfiyat, DT.aciklama, MD.dispoNotu, SD.siparisId, SD.siparisDetayId, DT.sqlUserName, DT.kayitTarihi, O.orguAd, 'Çözgü=(' + (SELECT dbo.fnTipIplikler(SD.tipTurId, SD.tipNo, SD.desenNo, SD.varyantNo, 1) AS Expr1) + ')' + ' Atki=(' + (SELECT dbo.fnTipIplikler(SD.tipTurId, SD.tipNo, SD.desenNo, SD.varyantNo, 2) AS Expr1) + ')' AS iplikAciklama, DT.prosesOk, dbo.[fnYikamaTalimat](SP.siparisId) yikamaTalimati FROM tblDoviz AS DV WITH(NOLOCK) INNER JOIN tblDispoTerbiye AS DT WITH(NOLOCK) INNER JOIN tblTanimlar AS B WITH(NOLOCK) ON DT.iplikBoyaId = B.tanimId AND B.tanimTurId = 2 ON DV.id = DT.dovizId RIGHT OUTER JOIN tblMusteri AS M1 WITH(NOLOCK) INNER JOIN tblSiparisDetay AS SD WITH(NOLOCK) INNER JOIN tblDispo AS MD WITH(NOLOCK) ON SD.siparisDetayId = MD.siparisDetayId INNER JOIN tblTipTur AS TT WITH(NOLOCK) ON SD.tipTurId = TT.tipTurId INNER JOIN tblSiparis AS SP WITH(NOLOCK) ON SD.siparisId = SP.siparisId ON M1.musteriNo = SP.musteriNo INNER JOIN tblTip AS TP WITH(NOLOCK) ON SD.tipTurId = TP.tipTurId AND SD.tipNo = TP.tipNo AND SD.desenNo = TP.desen AND SD.varyantNo = TP.varyant INNER JOIN tblOrgu AS O WITH(NOLOCK) ON TP.orguId = O.orguId INNER JOIN tblMusteri AS M WITH(NOLOCK) INNER JOIN tblSevkiyat AS SV WITH(NOLOCK) ON M.musteriNo = SV.musteriNo INNER JOIN tblSevkDetay AS SVD WITH(NOLOCK) ON SV.sevkNo = SVD.sevkNo ON MD.mamulDispoHamSevkno = SV.sevkNo LEFT OUTER JOIN tblTop AS T WITH(NOLOCK) INNER JOIN tblDispo AS HD WITH(NOLOCK) ON T.dispoNo = HD.dispoNo AND T.dispoTuruId = HD.dispoTuruId ON SVD.dispoTuruId = T.dispoTuruId AND SVD.dispoNo = T.dispoNo AND SVD.topNo = T.topNo AND MD.siparisDetayId = HD.siparisDetayId ON DT.dispoTuruId = MD.dispoTuruId AND DT.dispoNo = MD.dispoNo LEFT OUTER JOIN tblDispoTerbiyeTest AS TS WITH(NOLOCK) ON DT.dispoTuruId = TS.dispoTuruId AND DT.dispoNo = TS.dispoNo --WHERE DT.gelisTarihi IS NULL -- OR DT.gelisTarihi > GETDATE()-30 GROUP BY MD.dispoNo, DT.partiNo, DT.iplikBoyaId, TS.mamulEn, TS.mamulGramaj, DT.gelisMetresi, DT.gelisTarihi, DT.atkiCekmesi, DT.cozguCekmesi, DT.fire, DT.fiyat, DT.aciklama, DT.sqlUserName, DT.kayitTarihi, SD.tipTurId, TT.tipTur, SD.tipNo, SD.desenNo, SD.varyantNo, SD.siparisId, SD.siparisDetayId, B.tanimAd, M.musteriAdi, M.musteriAdi, M1.musteriAdi, O.orguAd, TP.iplikAciklama, SD.miktar, MD.dispoNotu, SP.mamulTermin, DT.dovizId, DV.dovizCins, MD.dispoMetresi, MD.akisNotu, SV.sevkNo, SV.sevkTarihi, DT.prosesOk,SP.siparisId )

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  • Gmail - error adding pop3 account from my mail server (postfix+courier)

    - by Lucas Lobosque
    I use courier to add pop3/imap support to my mail server, and I get this when I try to add a new pop3 account in gmail: Server returned error: "Missing +OK response upon connecting to the server: * OK [CAPABILITY IMAP4rev1 UIDPLUS CHILDREN NAMESPACE THREAD=ORDEREDSUBJECT THREAD=REFERENCES SORT QUOTA IDLE ACL ACL2=UNION STARTTLS] Courier-IMAP ready. Copyright 1998-2011 Double Precision, Inc. See COPYING for distribution information." Any help on how to fix this would be appreciated.

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  • Xenserver 5.5 doesn't see RAID volume

    - by Roy Chan
    Hi Gurus, I am trying to install Xenserver on a Dell precision 490 workstation. After booting into the install wizard and next-ed a few times, On the disk step, it only shows physical harddrive but not the RAID (RAID-10) volume that I set up on the Dell RAID. Is there a special option that I have to set on the boot? or do I need a special driver for this? Please Advise Thanks

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  • Can I run 8 to 12 Monitors from one Laptop?

    - by Dan
    I would like to run at least 8 monitors & possibly 12, from a Dell Precision M6800 mobile Laptop Workstation. Monitors I want to run are Dell U2412M ( 1920x1200 ) Ultra Sharp Monitors. The specs for the Laptop are : Intel Core i7-4700MQ Processor (17.3") UltraSharp FHD(1920x1080) 8GB (2x4GB) 1600MHz DDR3L AMD FirePro M6100 w/2GB GDDR5 2.5 inch 500GB Solid State Hybrid Drive Is it possible to do it & if yes, what other hardware, software do I need besides monitors ? Thanks

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  • PCMCIA Does Not Work on Dell Latitude under Windows XP

    - by Rich
    PCMCIA cards do not work on Dell Latitude e5510 under Windows XP. Works fine on Windows 7. Cards are recognized by Device Manager but do not work correctly. For example, CF adapter does not recognize when a card is inserted and network card is unable to obtain an IP address. Other systems that may be affected: Latitude E4310 Latitude E5410 Latitude E5510 Latitude E6410 Latitude E6410 ATG Latitude E6510 Dell Precision Mobile WorkStation M4500

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  • XP shared folders not accessible after BIOS changed

    - by stijn
    Here's what worked for over a year: PC A runs Windows 7, PC B runs Windows XP. Both are on the same subnet behind a router. A uses user account X, but logs in to PC B using the Administrator account. PC B is a Dell Precision 470. A known problem with these is that sometimes when plugging in their power cable they somehow loses all BIOS settings. This happened yesterday. After this happens Windows won't boot, because the default BIOS setting is 'RAID ON' while there is no RAID configured. No problem though, changing the BIOS settings to 'RAID OFF' makes it boot without problems. Note that in the meantime, nothing config-related was changed on machine A. It wasn't even on. Indeed after doing this, everything is fine. Everything includes all normal operations, remote desktop from PC A to PC B, running Synergy between A and B, accessing shared folders from B to A. But accessing the shared folders on B from A does not work any more. I tried pretty much everything I found via Google (fiddling with policies/registry kes/...) but no avail. > ping -a 192.168.2.2 Pinging A [192.168.2.2] with 32 bytes of data: Reply from 192.168.2.2: bytes=32 time<1ms TTL=128 > net view \\192.168.2.2 System error 5 has occurred. Access is denied. > net use /persistent:no K: \\A\myshare /user:A\USERNAME PASSWORD > net use /persistent:no K: \\192.168.2.2\myshare /user:192.168.2.2\USERNAME PASSWORD > net use /persistent:no K: \\192.168.2.2\myshare /user:USERNAME PASSWORD System error 86 has occurred. The specified network password is not correct. A solution to this would be great: I haven't been able to do any work since yesterday ;] update after taking the hard drive out of B and putting it in another Precision 470 with almost exactly the same hardware (at first sight, only the video card differs) the shared folders work.. Putting the disk back into A, same problem remains. Why does this depend on hardware, and more important, on which hardware?

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  • How to set my Bamboo Fun to work only on ONE of my dual screens?

    - by Przemyslaw Piekarski
    I need dual screen for my web developer job but when I do illustrations I prefer to work on a single screen to avoid the stretching of the workspace which affects tablet's precision. Is there a way to make my tablet work only on my primary screen and, at the same time, use mouse for both screens? I've looked into my tablet's preferences and haven't found it. I use Windows XP, Bamboo Fun A5, ATI Radeon X 1050. Thanks in advance.

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  • Change the delay before the Taskbar expands

    - by Will
    I like to keep my Taskbar collapsed on my laptop. Unfortunately, this plus the lack of precision control that the touchpad brings, I am constantly accidentally expanding the taskbar. What I'd like is to lengthen the delay time between when the cursor nears the taskbar and when it expands. Extra points for controlling the behavior differently for mouse input and for touch input.

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  • RHEL6 kick start fails with kernel panic - not syncing error

    - by mrchampe
    On a brand new (just unboxed) Dell Precision 7500 computer, I began a kickstart using a red hat boot disk and a kickstart file hosted at an http address. There is a picture of the full error below with trace information I believe this is different than the other questions relating to kernel panic errors. What I would like to know is What causes the kernel panic How does one fix this such that the kickstart finishes Additional information can be provided as necessary.

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  • Trackpad and USB mouse - different settings

    - by soupagain
    On my laptop I use both the trackpad and an external USB mouse. I'd like different settings for each device. e.g. the mouse is high resolution so I prefer to use a slow mouse settings, and for the trackpad I prefer to use the "Enhance pointer precision" option. But I can only seem to set these option for both devices, not individually. Can this be done? [Windows 7]

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  • Page layout software that allows mixed visual and programatic editing

    - by Justin Love
    I'd like to use a programming model for custom graphics and precision placement, and an interactive visual mode for large scale layout and less precise placements. I've used tools (PostScript, various vector drawing programs) that do one of these modes well, but leave me pining for the other model. Which tools should I be investigating? I'm currently on OS X. Examples: Creating diagrams with precise spacing, sets of cards, either likely drawing from some sort of data.

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  • Three monitor setup not working on Server 2012

    - by maxp
    Using an ATI firepro 4800 card, with three monitors connected (1 dvi, 2 displayport) on Server 2008 R2 worked fine. Ive now moved to a new machine, although identical spec, with a fresh Server 2012 install, and cannot get a three display output - only two. When I try to extend the display on to the third monitor, I get the message "The display settings could not be saved. Please try a different combination of display settings" The machine is a Dell Precision T1600. Any help appreciated.

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  • Why is the screen resolution different between Windows 7 Ultimate and Enterprise editions?

    - by IDispose
    I installed Windows 7 Enterprise 64 bit edition on my Dell Precision M6400 with an nVidia Quadro Fx 2700M card and I see that even though the screen resolution is set to 1920 X 1200, its not the same as the Windows 7 Ultimate 64 bit installed on a second hard drive. Both OSes show that the screen resolution is set at 1920 X 1200 but Ultimate shows more pixels than Enterprise. I also reinstalled the display driver but in vain. Any ideas? TIA

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  • SQL-Server 2008 on a windows vista doesn't start.

    - by Ice
    Hi, i have installed a SQL2008 Server developer-Edition on my Dell Precision M90 Notebook with windows Vista, but the service dosen't start. SQL Server Configuration Manager shows MSSQLSERVER as stopped and an attempt to start this service fails. No entry in eventviewer... where to look? What might be the reason?

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  • Oracle error when logging into database

    - by Bryan
    When I try to log into my db with a specific user I get this message. Below is from the alert log. I can login as system just fine. Anyone know how to figure out what is causing this? Thanks in advance for the help. ----- Error Stack Dump ----- ORA-00604: error occurred at recursive SQL level 1 ORA-01438: value larger than specified precision allowed for this column ORA-06512: at line 2 Oracle 10g OEL 5.5

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