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  • How to find an specific key/value (property list)

    - by Bob Rivers
    Hi, I'm learning cocoa/objective-c. Right now I'm dealing with key/value coding. After reading Aaron's book and other sources, I thought that I was able to left the simple examples and try a complex one... I'm trying read iTunes property list (iTunes Music Library.xml). I would like to retrieve the tracks held by an specific playlist. Probably everybody knows it, but bellow I put a piece of the xml: <plist version="1.0"> <dict> <key>Major Version</key><integer>1</integer> ... <key>Playlists</key> <array> <dict> <key>Name</key><string>Library</string> ... <key>Playlist Items</key> <array> <dict> <key>Track ID</key><integer>10281</integer> </dict> ... </array> </dict> <dict> ... </dict> </array> </dict> </plist> As you can see, the playlists are stored as dictionaries inside an array, and the key that identifies it is inside it, not as a <key> preceding it. The problem is that I'm not able to figure out how to search for a key that is inside another one. With the following code I can find the the array in which the playlists are stored, but how to find an specific <dict>? NSDictionary *rootDict = [[NSDictionary alloc] initWithContentsOfFile:file]; NSArray *playlists = [rootDict objectForKey:@"Playlists"]; Here at Stackoverflow I found this post, but I'm not sure if iterate over the array and test it is a good idea. I'm quite sure that I could use valueForKeyPath, but I'm unable to figure out how to do it. Any help is welcome. TIA, Bob

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  • Help using recursion in Java

    - by Mercer
    I have a class Group. In the class I have two fields, idGroup IdGroupGroup. Groups may be part of other groups. My class Group is defined in a HashMap<Integer,Integer>; the key is IdGroupGroup and value is idGroup. I want to search the map for a particular idGroup; can I use recursion to do this?

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  • How do I serve a large file using Pylons?

    - by Chris R
    I am writing a Pylons-based download gateway. The gateway's client will address files by ID: /file_gw/download/1 Internally, the file itself is accessed via HTTP from an internal file server: http://internal-srv/path/to/file_1.content The files may be quite large, so I want to stream the content. I store metadata about the file in a StoredFile model object: class StoredFile(Base): id = Column(Integer, primary_key=True) name = Column(String) size = Column(Integer) content_type = Column(String) url = Column(String) Given this, what's the best (ie: most architecturally-sound, performant, et al) way to write my file_gw controller?

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  • Floating point arithmetics restricted to integers

    - by user396672
    I use doubles for a uniform implementation of some arithmetic calculations. These calculations may be actually applied to integers too, but there are no C++-like templates in Java and I don't want to duplicate the implementation code, so I simply use "double" version for ints. Does JVM spec guarantees the correctness of integer operations such a <=,=, +, -, *, and / (in case of remainder==0) when the operations are emulated as corresponding floating point ops? (Any integer, of course, has reasonable size to be represented in double's mantissa)

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  • Iterating over hashmap injsp in struts application

    - by Rozer
    I have a Hashmap object that i am getting on a jsp page. HashMap<Integer,Gift_product> gift_hm = new HashMap<Integer,Gift_product>(); gift_hm.put(17,new Gift_product("doll",67)); now i need to iterate this and display content on jsp. as Gift_product class contains two fields name and price jsp output should be serial no. product name price 17 Doll 67 How can i achieve it..

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  • Regarding address operator C/C++

    - by iSight
    Hi, What does address operator mean. say in the method below. what should be passed in the method as parameter value of integer or the address of an integer variable. void func1(int&)// method declaration void func1(int& inNumber)//method definition { //some code }

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  • Runtime casting from String to other datatype

    - by Gengrlin
    Hi I have Map of String values. I want to cast this value at runtime. e.g. Map map = new HashMap(); map.put("key1","101"); map.put("key2","45.40"); Now runtime I know key1 is integer and key2 is double How can I cast this. I tried this: ("101").getClass().cast(Integer). Thanks.

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  • Python : Convert from C-Char to Int

    - by cuband
    I have a string read in from a binary file that is unpacked using struct.unpack as a string of length n. Each byte in the string is a single integer (1-byte) representing 0-255. So for each character in the string I want to convert it to an integer. I can't figure out how to do this. Using ord doesn't seem to be on the right track...

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  • where does main() return its value?

    - by piemesons
    I'm newly using CODE::BLOCKS+mingw compiler If I don't type return 0 at the end of program,I can see that main() is returning some integer,I learnt that main() returning 0 infers program executes successfully.I don't find any flaw in my code, why is it returning some integer? secondly any function returns its value to its function call, to where does main() return its value?

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  • How can i optimize this python code

    - by RandomVector
    def maxVote(nLabels): count = {} maxList = [] maxCount = 0 for nLabel in nLabels: if nLabel in count: count[nLabel] += 1 else: count[nLabel] = 1 #Check if the count is max if count[nLabel] > maxCount: maxCount = count[nLabel] maxList = [nLabel,] elif count[nLabel]==maxCount: maxList.append(nLabel) return random.choice(maxList) nLabels contains a list of integers. The above function returns the integer with highest frequency, if more than one have same frequency then a randomly selected integer from them is returned. E.g. maxVote([1,3,4,5,5,5,3,12,11]) is 5

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  • How do I make an object a property of a model in Ruby on Rails?

    - by iTake
    I have this in my schema: create_table "robots_matches", :force => true do |t| t.integer "robot_id" t.integer "match_id" and I think I want to be able to load a robot and match from within my robots_match model so I can do something like this: robots_match.find(:id).get_robot().Name My attempt in the robots_matches model was this: def get_robot Robot.find(this.id) end I am super new to rails, so feel free to correct my architectural decision here.

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  • Why is this condition never satisfied ?

    - by Patrick
    I don't know why this condition is never satisfied: I'm comparing two ArrayList values, and it is always false. if ( (Integer) results.get(rank) == (Integer) experts.get(j)) I'm debugging and I have exactly the same 2 values: 3043 and 3043 However it doesn't work. thanks

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  • typedef and operator overloading in C

    - by jocapco
    Suppose I typedef an integer or integer array or any known type: typedef int int2 Then I overload operator * for int2 pairs, now if I initialize variables a and b as int. Then will my * between a and b be the overloaded * ? How do I achieve overloading an int and yet also use * for int the way they are. Should I create a new type?

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  • How do i insert 1000 times in one statement? with SQLITE?

    - by acidzombie24
    I want to fill this table with 10000000 values but first i want only 1000. I tried this in sqlite database browser but 3 isnt inserted unless i drop everything after it. But more importantly i dont know how to have num go from 1 to 1000. create table if not exists test1(id integer primary key, val integer); insert into test1(val) select '3' as num where num between 1 and 1000

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  • Why are interfaces unusable in PHP?

    - by streetparade
    I mean an interface definition without defining the return type makes it unusable? This makes more Clear Interface run { public function getInteger(); } class MyString implements run { public function myNumber() { } public function getInteger() { return "Not a number"; } } In Java every Interface has a return type like Integer,String,Void I know that PHP is unfortunately a loosly typed Language but isnt there a Solution for that Problem? Is it Possible to defining a Interface with a Return type like Integer?

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  • Why cant i use the field user in SQL Server 8?

    - by acidzombie24
    Maybe not literally but the query below gets an error near user. If i change it to userZ it works. WHY can i not use that name? Is there a way to specific its a field instead of a keyword? (or whatever it is) create table Post2 ( id INTEGER PRIMARY KEY NOT NULL, title nvarchar(max) NOT NULL, body nvarchar(max) NOT NULL, user integer REFERENCES Post1(id));

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  • Else without if

    - by user2808951
    I'm trying to write a code for my computer programming class for a project due Monday, and I'm pretty new to Java, but I'm trying to write a program that will first determine if a number the user inputs is even or odd and then determine if the number is prime or not. I'm not sure if I did the algorithm right or not, so if anyone has any corrections on the program to my algorithm or anything else please say so, but my real issue is that the program is refusing to compile. Every time I try, it says it's having an else without if problem. Here's a link to my command box: http://s1341.photobucket.com/user/Emi_Nightshade/media/Capture_zps45f9a2ea.png.html Here's my code: import java.io.*; import java.util.*; public class Lesson9p1_ThuotteEmily { public static void main(String args[]) { Scanner kbReader0=new Scanner(System.in); System.out.print("\n\nPlease enter an integer. An integer is whole number, and it can be either negative or positive. Please enter your number: "); long num=kbReader0.nextLong(); if(num%2==0) //if and else with braces { System.out.println("Your integer " + num + " is even."); } else { System.out.println("Your integer " + num + " is odd."); } Scanner kbReader1=new Scanner(System.in); System.out.print("\n\nWould you like to know if your number is prime? Please enter yes or no: "); String yn=kbReader1.nextLine(); if(yn.equals.IgnoreCase("Yes")) { System.out.println("Okay. Give me a moment."); { if(num%2==0) { System.out.println("Your number isn't prime."); } else if(num==2) { System.out.println("Your number is 2, which is the only even prime number in existence. Cool, right?"); } for(int i=3;i*i<=n;i+=2) { if(n%1==0) { System.out.println("Your number isn't prime."); } } else { System.out.println("Your number is prime!"); } } } if(yn.equals.IgnoreCase("No")) { System.out.println("Okay."); } } } If anyone could help me out with this and also any problems I may have made elsewhere in the program, I'd be very grateful! Thanks.

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  • Know more about Enqueue Deadlock Detection

    - by Liu Maclean(???)
    ??? ORACLE ALLSTAR???????????????????,??????? ???????enqueue lock?????????3 ??????,????????????????????????????ora-00060 dead lock??process???3s: SQL> select * from v$version; BANNER ---------------------------------------------------------------- Oracle Database 10g Enterprise Edition Release 10.2.0.5.0 - 64bi PL/SQL Release 10.2.0.5.0 - Production CORE 10.2.0.5.0 Production TNS for Linux: Version 10.2.0.5.0 - Production NLSRTL Version 10.2.0.5.0 - Production SQL> select * from global_name; GLOBAL_NAME -------------------------------------------------------------------------------- www.oracledatabase12g.com PROCESS A: set timing on; update maclean1 set t1=t1+1; PROCESS B: update maclean2 set t1=t1+1; PROCESS A: update maclean2 set t1=t1+1; PROCESS B: update maclean1 set t1=t1+1; ??3s? PROCESS A ?? ERROR at line 1: ORA-00060: deadlock detected while waiting for resource Elapsed: 00:00:03.02 ????Process A????????????? 3s,?????????????,??????? ?????????? ???????: SQL> col name for a30 SQL> col value for a5 SQL> col DESCRIB for a50 SQL> set linesize 140 pagesize 1400 SQL> SELECT x.ksppinm NAME, y.ksppstvl VALUE, x.ksppdesc describ 2 FROM SYS.x$ksppi x, SYS.x$ksppcv y 3 WHERE x.inst_id = USERENV ('Instance') 4 AND y.inst_id = USERENV ('Instance') 5 AND x.indx = y.indx 6 AND x.ksppinm='_enqueue_deadlock_scan_secs'; NAME VALUE DESCRIB ------------------------------ ----- -------------------------------------------------- _enqueue_deadlock_scan_secs 0 deadlock scan interval SQL> alter system set "_enqueue_deadlock_scan_secs"=18 scope=spfile; System altered. Elapsed: 00:00:00.01 SQL> startup force; ORACLE instance started. Total System Global Area 851443712 bytes Fixed Size 2100040 bytes Variable Size 738198712 bytes Database Buffers 104857600 bytes Redo Buffers 6287360 bytes Database mounted. Database opened. PROCESS A: SQL> set timing on; SQL> update maclean1 set t1=t1+1; 1 row updated. Elapsed: 00:00:00.06 Process B SQL> update maclean2 set t1=t1+1; 1 row updated. SQL> update maclean1 set t1=t1+1; Process A: SQL> SQL> alter session set events '10704 trace name context forever,level 10:10046 trace name context forever,level 8'; Session altered. SQL> update maclean2 set t1=t1+1; update maclean2 set t1=t1+1 * ERROR at line 1: ORA-00060: deadlock detected while waiting for resource  Elapsed: 00:00:18.05 ksqcmi: TX,90011,4a9 mode=6 timeout=21474836 WAIT #12: nam='enq: TX - row lock contention' ela= 2930070 name|mode=1415053318 usn<<16 | slot=589841 sequence=1193 obj#=56810 tim=1308114759849120 WAIT #12: nam='enq: TX - row lock contention' ela= 2930636 name|mode=1415053318 usn<<16 | slot=589841 sequence=1193 obj#=56810 tim=1308114762779801 WAIT #12: nam='enq: TX - row lock contention' ela= 2930439 name|mode=1415053318 usn<<16 | slot=589841 sequence=1193 obj#=56810 tim=1308114765710430 *** 2012-06-12 09:58:43.089 WAIT #12: nam='enq: TX - row lock contention' ela= 2931698 name|mode=1415053318 usn<<16 | slot=589841 sequence=1193 obj#=56810 tim=1308114768642192 WAIT #12: nam='enq: TX - row lock contention' ela= 2930428 name|mode=1415053318 usn<<16 | slot=589841 sequence=1193 obj#=56810 tim=1308114771572755 WAIT #12: nam='enq: TX - row lock contention' ela= 2931408 name|mode=1415053318 usn<<16 | slot=589841 sequence=1193 obj#=56810 tim=1308114774504207 DEADLOCK DETECTED ( ORA-00060 ) [Transaction Deadlock] The following deadlock is not an ORACLE error. It is a deadlock due to user error in the design of an application or from issuing incorrect ad-hoc SQL. The following information may aid in determining the deadlock: ??????Process A?’enq: TX – row lock contention’ ?????ORA-00060 deadlock detected????3s ??? 18s , ???hidden parameter “_enqueue_deadlock_scan_secs”?????,????????0? ??????????: SQL> alter system set "_enqueue_deadlock_scan_secs"=4 scope=spfile; System altered. Elapsed: 00:00:00.01 SQL> alter system set "_enqueue_deadlock_time_sec"=9 scope=spfile; System altered. Elapsed: 00:00:00.00 SQL> startup force; ORACLE instance started. Total System Global Area 851443712 bytes Fixed Size 2100040 bytes Variable Size 738198712 bytes Database Buffers 104857600 bytes Redo Buffers 6287360 bytes Database mounted. Database opened. SQL> set linesize 140 pagesize 1400 SQL> show parameter dead NAME TYPE VALUE ------------------------------------ -------------------------------- ------------------------------ _enqueue_deadlock_scan_secs integer 4 _enqueue_deadlock_time_sec integer 9 SQL> set timing on SQL> select * from maclean1 for update wait 8; T1 ---------- 11 Elapsed: 00:00:00.01 PROCESS B SQL> select * from maclean2 for update wait 8; T1 ---------- 3 SQL> select * from maclean1 for update wait 8; select * from maclean1 for update wait 8 PROCESS A SQL> select * from maclean2 for update wait 8; select * from maclean2 for update wait 8 * ERROR at line 1: ORA-30006: resource busy; acquire with WAIT timeout expired Elapsed: 00:00:08.00 ???????? ??? select for update wait?enqueue request timeout ?????8s? ,???????”_enqueue_deadlock_scan_secs”=4(deadlock scan interval),?4s???deadlock detected,????Process A????deadlock ???, ??????? ??Process A?????8s?raised??”ORA-30006: resource busy; acquire with WAIT timeout expired”??,??ORA-00060,?????process A???????? ????????”_enqueue_deadlock_time_sec”(requests with timeout <= this will not have deadlock detection)???,?enqueue request time < “_enqueue_deadlock_time_sec”?Server process?????dead lock detection,?????????enqueue request ??????timeout??????(_enqueue_deadlock_time_sec????5,?timeout<5s),???????????????;??????timeout>”_enqueue_deadlock_time_sec”???,Oracle????????????????????? ??????????: SQL> show parameter dead NAME TYPE VALUE ------------------------------------ -------------------------------- ------------------------------ _enqueue_deadlock_scan_secs integer 4 _enqueue_deadlock_time_sec integer 9 Process A: SQL> set timing on; SQL> select * from maclean1 for update wait 10; T1 ---------- 11 Process B: SQL> select * from maclean2 for update wait 10; T1 ---------- 3 SQL> select * from maclean1 for update wait 10; PROCESS A: SQL> select * from maclean2 for update wait 10; select * from maclean2 for update wait 10 * ERROR at line 1: ORA-00060: deadlock detected while waiting for resource Elapsed: 00:00:06.02 ??????? select for update wait 10?10s??, ?? 10s?????_enqueue_deadlock_time_sec???(9s),??Process A???????? ???????????????6s ???????_enqueue_deadlock_scan_secs?4s ? ???????????,???????????_enqueue_deadlock_scan_secs?????????3???? ??: enqueue lock?????????????? 1. ?????????deadlock detection??3s????, ????????_enqueue_deadlock_scan_secs(deadlock scan interval)???,??????0,????????_enqueue_deadlock_scan_secs?????????3???, ?_enqueue_deadlock_scan_secs=0 ??3s??, ?_enqueue_deadlock_scan_secs=4??6s??,????? 2. ???????_enqueue_deadlock_time_sec(requests with timeout <= this will not have deadlock detection)???,?enqueue request timeout< _enqueue_deadlock_time_sec(????5),?Server process?????????enqueue request timeout>_enqueue_deadlock_time_sec ????_enqueue_deadlock_scan_secs???????, ??request timeout??????select for update wait [TIMEOUT]??? ??: ???10.2.0.1?????????2?hidden parameter , ???patchset 10.2.0.3????? _enqueue_deadlock_time_sec, ?patchset 10.2.0.5??????_enqueue_deadlock_scan_secs? ?????RAC???????????10s, ???????_lm_dd_interval(dd time interval in seconds) ,????????8.0.6???? ???????????????,??????,  ?10g???????60s,?11g???????10s?  ???????11g??_lm_dd_interval?????????????,?????11g??LMD????????????,??????????RAC?LMD?Deadlock Detection???????CPU,???11g?Oracle????Team???LMD????????CPU????: ????????11g?LMD???????,???????11g??? UTS TRACE ????? DD???: SQL> select * from v$version; BANNER -------------------------------------------------------------------------------- Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production PL/SQL Release 11.2.0.3.0 - Production CORE 11.2.0.3.0 Production TNS for Linux: Version 11.2.0.3.0 - Production NLSRTL Version 11.2.0.3.0 - Production SQL> SQL> select * from global_name 2 ; GLOBAL_NAME -------------------------------------------------------------------------------- www.oracledatabase12g.com SQL> alter system set "_lm_dd_interval"=20 scope=spfile; System altered. SQL> startup force; ORACLE instance started. Total System Global Area 1570009088 bytes Fixed Size 2228704 bytes Variable Size 1325403680 bytes Database Buffers 234881024 bytes Redo Buffers 7495680 bytes Database mounted. Database opened. SQL> set linesize 140 pagesize 1400 SQL> show parameter lm_dd NAME TYPE VALUE ------------------------------------ -------------------------------- ------------------------------ _lm_dd_interval integer 20 SQL> select count(*) from gv$instance; COUNT(*) ---------- 2 instance 1: SQL> oradebug setorapid 12 Oracle pid: 12, Unix process pid: 8608, image: [email protected] (LMD0) ? LMD0??? UTS TRACE??RAC???????????? SQL> oradebug event 10046 trace name context forever,level 8:10708 trace name context forever,level 103: trace[rac.*] disk high; Statement processed. Elapsed: 00:00:00.00 SQL> update maclean1 set t1=t1+1; 1 row updated. instance 2: SQL> update maclean2 set t1=t1+1; 1 row updated. SQL> update maclean1 set t1=t1+1; Instance 1: SQL> update maclean2 set t1=t1+1; update maclean2 set t1=t1+1 * ERROR at line 1: ORA-00060: deadlock detected while waiting for resource Elapsed: 00:00:20.51 LMD0???UTS TRACE 2012-06-12 22:27:00.929284 : [kjmpbmsg:process][type 22][msg 0x7fa620ac85a8][from 1][seq 8148.0][len 192] 2012-06-12 22:27:00.929346 : [kjmxmpm][type 22][seq 0.0][msg 0x7fa620ac85a8][from 1] *** 2012-06-12 22:27:00.929 * kjddind: received DDIND msg with subtype x6 * reqp->dd_master_inst_kjxmddi == 1 * kjddind: dump sgh: 2012-06-12 22:27:00.929346*: kjddind: req->timestamp [0.15], kjddt [0.13] 2012-06-12 22:27:00.929346*: >> DDmsg:KJX_DD_REMOTE,TS[0.15],Inst 1->2,ddxid[id1,id2,inst:2097153,31,1],ddlock[0x95023930,829],ddMasterInst 1 2012-06-12 22:27:00.929346*: lock [0x95023930,829], op = [mast] 2012-06-12 22:27:00.929346*: reqp->timestamp [0.15], kjddt [0.13] 2012-06-12 22:27:00.929346*: kjddind: updated local timestamp [0.15] * kjddind: case KJX_DD_REMOTE 2012-06-12 22:27:00.929346*: ADD IO NODE WFG: 0 frame pointer 2012-06-12 22:27:00.929346*: PUSH: type=res, enqueue(0xffffffff.0xffffffff)=0xbbb9af40, block=KJUSEREX, snode=1 2012-06-12 22:27:00.929346*: PROCESS: type=res, enqueue(0xffffffff.0xffffffff)=0xbbb9af40, block=KJUSEREX, snode=1 2012-06-12 22:27:00.929346*: POP: type=res, enqueue(0xffffffff.0xffffffff)=0xbbb9af40, block=KJUSEREX, snode=1 2012-06-12 22:27:00.929346*: kjddopr[TX 0xe000c.0x32][ext 0x5,0x0]: blocking lock 0xbbb9a800, owner 2097154 of inst 2 2012-06-12 22:27:00.929346*: PUSH: type=txn, enqueue(0xffffffff.0xffffffff)=0xbbb9a800, block=KJUSEREX, snode=1 2012-06-12 22:27:00.929346*: PROCESS: type=txn, enqueue(0xffffffff.0xffffffff)=0xbbb9a800, block=KJUSEREX, snode=1 2012-06-12 22:27:00.929346*: ADD NODE TO WFG: type=txn, enqueue(0xffffffff.0xffffffff)=0xbbb9a800, block=KJUSEREX, snode=1 2012-06-12 22:27:00.929346*: POP: type=txn, enqueue(0xffffffff.0xffffffff)=0xbbb9a800, block=KJUSEREX, snode=1 2012-06-12 22:27:00.929346*: kjddopt: converting lock 0xbbce92f8 on 'TX' 0x80016.0x5d4,txid [2097154,34]of inst 2 2012-06-12 22:27:00.929346*: PUSH: type=res, enqueue(0xffffffff.0xffffffff)=0xbbce92f8, block=KJUSEREX, snode=1 2012-06-12 22:27:00.929346*: PROCESS: type=res, enqueue(0xffffffff.0xffffffff)=0xbbce92f8, block=KJUSEREX, snode=1 2012-06-12 22:27:00.929346*: ADD NODE TO WFG: type=res, enqueue(0xffffffff.0xffffffff)=0xbbce92f8, block=KJUSEREX, snode=1 2012-06-12 22:27:00.929855 : GSIPC:AMBUF: rcv buff 0x7fa620aa8cd8, pool rcvbuf, rqlen 1102 2012-06-12 22:27:00.929878 : GSIPC:GPBMSG: new bmsg 0x7fa620aa8d48 mb 0x7fa620aa8cd8 msg 0x7fa620aa8d68 mlen 192 dest x100 flushsz -1 2012-06-12 22:27:00.929878*: << DDmsg:KJX_DD_REMOTE,TS[0.15],Inst 2->1,ddxid[id1,id2,inst:2097153,31,1],ddlock[0x95023930,829],ddMasterInst 1 2012-06-12 22:27:00.929878*: lock [0xbbce92f8,287], op = [mast] 2012-06-12 22:27:00.929878*: ADD IO NODE WFG: 0 frame pointer 2012-06-12 22:27:00.929923 : [kjmpbmsg:compl][msg 0x7fa620ac8588][typ p][nmsgs 1][qtime 0][ptime 0] 2012-06-12 22:27:00.929947 : GSIPC:PBAT: flush start. flag 0x79 end 0 inc 4.4 2012-06-12 22:27:00.929963 : GSIPC:PBAT: send bmsg 0x7fa620aa8d48 blen 224 dest 1.0 2012-06-12 22:27:00.929979 : GSIPC:SNDQ: enq msg 0x7fa620aa8d48, type 65521 seq 8325, inst 1, receiver 0, queued 1 012-06-12 22:27:00.929979 : GSIPC:SNDQ: enq msg 0x7fa620aa8d48, type 65521 seq 8325, inst 1, receiver 0, queued 1 2012-06-12 22:27:00.929996 : GSIPC:BSEND: flushing sndq 0xb491dd28, id 0, dcx 0xbc517770, inst 1, rcvr 0 qlen 0 1 2012-06-12 22:27:00.930014 : GSIPC:BSEND: no batch1 msg 0x7fa620aa8d48 type 65521 len 224 dest (1:0) 2012-06-12 22:27:00.930088 : kjbsentscn[0x0.3f72dc][to 1] 2012-06-12 22:27:00.930144 : GSIPC:SENDM: send msg 0x7fa620aa8d48 dest x10000 seq 8325 type 65521 tkts x1 mlen xe00110 2012-06-12 22:27:00.930531 : GSIPC:KSXPCB: msg 0x7fa620aa8d48 status 30, type 65521, dest 1, rcvr 0 WAIT #0: nam='ges remote message' ela= 1372 waittime=80 loop=0 p3=74 obj#=-1 tim=1339554420931640 2012-06-12 22:27:00.931728 : GSIPC:RCVD: ksxp msg 0x7fa620af6490 sndr 1 seq 0.8149 type 65521 tkts 1 2012-06-12 22:27:00.931746 : GSIPC:RCVD: watq msg 0x7fa620af6490 sndr 1, seq 8149, type 65521, tkts 1 2012-06-12 22:27:00.931763 : GSIPC:RCVD: seq update (0.8148)->(0.8149) tp -15 fg 0x4 from 1 pbattr 0x0 2012-06-12 22:27:00.931779 : GSIPC:TKT: collect msg 0x7fa620af6490 from 1 for rcvr 0, tickets 1 2012-06-12 22:27:00.931794 : kjbrcvdscn[0x0.3f72dc][from 1][idx 2012-06-12 22:27:00.931810 : kjbrcvdscn[no bscn dd_master_inst_kjxmddi == 1 * kjddind: dump sgh: NXTIN (nil) 0 wq 0 cvtops x0 0x0.0x0(ext 0x0,0x0)[0000-0000-00000000] inst 1 BLOCKER 0xbbb9a800 5 wq 1 cvtops x28 TX 0xe000c.0x32(ext 0x5,0x0)[20000-0002-00000022] inst 2 BLOCKED 0xbbce92f8 5 wq 2 cvtops x1 TX 0x80016.0x5d4(ext 0x2,0x0)[20000-0002-00000022] inst 2 NXTOUT (nil) 0 wq 0 cvtops x0 0x0.0x0(ext 0x0,0x0)[0000-0000-00000000] inst 1 2012-06-12 22:27:00.932058*: kjddind: req->timestamp [0.15], kjddt [0.15] 2012-06-12 22:27:00.932058*: >> DDmsg:KJX_DD_VALIDATE,TS[0.15],Inst 1->2,ddxid[id1,id2,inst:2097153,31,1],ddlock[0x95023930,829],ddMasterInst 1 2012-06-12 22:27:00.932058*: lock [(nil),0], op = [vald_dd] 2012-06-12 22:27:00.932058*: kjddind: updated local timestamp [0.15] * kjddind: case KJX_DD_VALIDATE *** 2012-06-12 22:27:00.932 * kjddvald called: kjxmddi stuff: * cont_lockp (nil) * dd_lockp 0x95023930 * dd_inst 1 * dd_master_inst 1 * sgh graph: NXTIN (nil) 0 wq 0 cvtops x0 0x0.0x0(ext 0x0,0x0)[0000-0000-00000000] inst 1 BLOCKER 0xbbb9a800 5 wq 1 cvtops x28 TX 0xe000c.0x32(ext 0x5,0x0)[20000-0002-00000022] inst 2 BLOCKED 0xbbce92f8 5 wq 2 cvtops x1 TX 0x80016.0x5d4(ext 0x2,0x0)[20000-0002-00000022] inst 2 NXTOUT (nil) 0 wq 0 cvtops x0 0x0.0x0(ext 0x0,0x0)[0000-0000-00000000] inst 1 POP WFG NODE: lock=(nil) * kjddvald: dump the PRQ: BLOCKER 0xbbb9a800 5 wq 1 cvtops x28 TX 0xe000c.0x32(ext 0x5,0x0)[20000-0002-00000022] inst 2 BLOCKED 0xbbce92f8 5 wq 2 cvtops x1 TX 0x80016.0x5d4(ext 0x2,0x0)[20000-0002-00000022] inst 2 * kjddvald: KJDD_NXTONOD ->node_kjddsg.dinst_kjddnd =1 * kjddvald: ... which is not my node, my subgraph is validated but the cycle is not complete Global blockers dump start:--------------------------------- DUMP LOCAL BLOCKER/HOLDER: block level 5 res [0x80016][0x5d4],[TX][ext 0x2,0x0] ??dead lock!!! ???????11.2.0.3???? RAC LMD???????????”_lm_dd_interval”????????????20s?  ???????10g?_lm_dd_interval???60s,??????Processes?????????????????,????????????Server Process????????60s??????11g?????(??????LMD???????)???????,???????????10s??? Enqueue Deadlock Detection? ?11g??? RAC?LMD???????hidden parameter ????”_lm_dd_interval”???,RAC????????????????,???????????: SQL> col name for a50 SQL> col describ for a60 SQL> col value for a20 SQL> set linesize 140 pagesize 1400 SQL> SELECT x.ksppinm NAME, y.ksppstvl VALUE, x.ksppdesc describ 2 FROM SYS.x$ksppi x, SYS.x$ksppcv y 3 WHERE x.inst_id = USERENV ('Instance') 4 AND y.inst_id = USERENV ('Instance') 5 AND x.indx = y.indx 6 AND x.ksppinm like '_lm_dd%'; NAME VALUE DESCRIB -------------------------------------------------- -------------------- ------------------------------------------------------------ _lm_dd_interval 20 dd time interval in seconds _lm_dd_scan_interval 5 dd scan interval in seconds _lm_dd_search_cnt 3 number of dd search per token get _lm_dd_max_search_time 180 max dd search time per token _lm_dd_maxdump 50 max number of locks to be dumped during dd validation _lm_dd_ignore_nodd FALSE if TRUE nodeadlockwait/nodeadlockblock options are ignored 6 rows selected.

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  • Windows 8 Apps with HTML5 and JavaScript

    - by Stephen.Walther
    Last week, I finished writing Windows 8 Apps with HTML5 and JavaScript – Yikes! That is a long title. This book is all about writing apps for Windows 8 which can be added to the Windows Store. The book focuses on building apps using HTML5 and JavaScript. If you are already comfortable building websites, then building Windows Store apps is not a huge leap.  I explain how you can create productivity apps, like a Task List app, and games, like a simple arcade game. I also explain how you can publish your app to the Windows Store and make money. To celebrate the release of Windows 8, my publisher is offering a huge 40% discount on the book until November 30, 2012. If you want to take advantage of this discount, follow the link below and enter the discount code WINDEV40 during checkout. http://www.informit.com/promotions/promotion.aspx?promo=139036&walther So what’s in the book?  Here’s an overview of each of the chapters: Chapter 1 – Building Windows Store Apps Contains a walkthrough of creating a super simple Windows app for taking pictures from your webcam. Explains how to publish your app to the Windows Store. Chapter 2 – WinJS Fundamentals Provides an overview of the Windows Library for JavaScript which is the Microsoft library for creating Windows Store apps with JavaScript. Chapter 3 – Observables, Bindings, and Templates You learn how to display a list of items using a template. For example, you learn how to create a template which can be used to display a list of products. Chapter 4 – Using WinJS Controls Overview of the core set of JavaScript controls included with the WinJS library. You learn how to use the Tooltip, ToggleSwitch, Rating, DatePicker, TimePicker, and FlipView controls. Chapter 5 – Creating Forms This chapter explains how to take advantage of HTML5 forms to display specialized keyboards and perform form validation. Chapter 6 – Menus and Flyouts You learn how to display popups, menus, and toolbars using the JavaScript controls included with the WinJS library. Chapter 7 – Using the ListView Control This entire chapter is devoted to the ListView control which is the most important control in the WinJS library. You can use the ListView control to display, sort, filter, and edit a list of items. Chapter 8 – Creating Data Sources Learn how to use a ListView control to display data from the file system, a web service, and IndexedDB. Chapter 9 – App Events and States This chapter explains the standard application events which are raised in a Windows Store app such as the activated and checkpoint events. You also learn how to build apps which adapt automatically to different view states such as portrait and landscape. Chapter 10 – Page Fragments and Navigation This chapter discusses two subjects: You learn how to create custom WinJS controls with Page Controls and you learn how to build apps with multiple pages.  Chapter 11 – Using the Live Connect API Learn how to use Windows Live Services to authenticate users, interact with SkyDrive, and retrieve user profile information (such as a user’s birthday or profile picture). Chapter 12 – Graphics and Games This chapter is devoted to building the Brain Eaters app which is a simple arcade game. Navigate a maze and eat all of the food pellets while avoiding the brain-eating zombies to win the game. Learn how to create the game using HTML5 Canvas.   If you want to buy the book, remember to use the magic discount code WINDEV40 and visit the following link: http://www.informit.com/promotions/promotion.aspx?promo=139036&walther

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  • Seeking on a Heap, and Two Useful DMVs

    - by Paul White
    So far in this mini-series on seeks and scans, we have seen that a simple ‘seek’ operation can be much more complex than it first appears.  A seek can contain one or more seek predicates – each of which can either identify at most one row in a unique index (a singleton lookup) or a range of values (a range scan).  When looking at a query plan, we will often need to look at the details of the seek operator in the Properties window to see how many operations it is performing, and what type of operation each one is.  As you saw in the first post in this series, the number of hidden seeking operations can have an appreciable impact on performance. Measuring Seeks and Scans I mentioned in my last post that there is no way to tell from a graphical query plan whether you are seeing a singleton lookup or a range scan.  You can work it out – if you happen to know that the index is defined as unique and the seek predicate is an equality comparison, but there’s no separate property that says ‘singleton lookup’ or ‘range scan’.  This is a shame, and if I had my way, the query plan would show different icons for range scans and singleton lookups – perhaps also indicating whether the operation was one or more of those operations underneath the covers. In light of all that, you might be wondering if there is another way to measure how many seeks of either type are occurring in your system, or for a particular query.  As is often the case, the answer is yes – we can use a couple of dynamic management views (DMVs): sys.dm_db_index_usage_stats and sys.dm_db_index_operational_stats. Index Usage Stats The index usage stats DMV contains counts of index operations from the perspective of the Query Executor (QE) – the SQL Server component that is responsible for executing the query plan.  It has three columns that are of particular interest to us: user_seeks – the number of times an Index Seek operator appears in an executed plan user_scans – the number of times a Table Scan or Index Scan operator appears in an executed plan user_lookups – the number of times an RID or Key Lookup operator appears in an executed plan An operator is counted once per execution (generating an estimated plan does not affect the totals), so an Index Seek that executes 10,000 times in a single plan execution adds 1 to the count of user seeks.  Even less intuitively, an operator is also counted once per execution even if it is not executed at all.  I will show you a demonstration of each of these things later in this post. Index Operational Stats The index operational stats DMV contains counts of index and table operations from the perspective of the Storage Engine (SE).  It contains a wealth of interesting information, but the two columns of interest to us right now are: range_scan_count – the number of range scans (including unrestricted full scans) on a heap or index structure singleton_lookup_count – the number of singleton lookups in a heap or index structure This DMV counts each SE operation, so 10,000 singleton lookups will add 10,000 to the singleton lookup count column, and a table scan that is executed 5 times will add 5 to the range scan count. The Test Rig To explore the behaviour of seeks and scans in detail, we will need to create a test environment.  The scripts presented here are best run on SQL Server 2008 Developer Edition, but the majority of the tests will work just fine on SQL Server 2005.  A couple of tests use partitioning, but these will be skipped if you are not running an Enterprise-equivalent SKU.  Ok, first up we need a database: USE master; GO IF DB_ID('ScansAndSeeks') IS NOT NULL DROP DATABASE ScansAndSeeks; GO CREATE DATABASE ScansAndSeeks; GO USE ScansAndSeeks; GO ALTER DATABASE ScansAndSeeks SET ALLOW_SNAPSHOT_ISOLATION OFF ; ALTER DATABASE ScansAndSeeks SET AUTO_CLOSE OFF, AUTO_SHRINK OFF, AUTO_CREATE_STATISTICS OFF, AUTO_UPDATE_STATISTICS OFF, PARAMETERIZATION SIMPLE, READ_COMMITTED_SNAPSHOT OFF, RESTRICTED_USER ; Notice that several database options are set in particular ways to ensure we get meaningful and reproducible results from the DMVs.  In particular, the options to auto-create and update statistics are disabled.  There are also three stored procedures, the first of which creates a test table (which may or may not be partitioned).  The table is pretty much the same one we used yesterday: The table has 100 rows, and both the key_col and data columns contain the same values – the integers from 1 to 100 inclusive.  The table is a heap, with a non-clustered primary key on key_col, and a non-clustered non-unique index on the data column.  The only reason I have used a heap here, rather than a clustered table, is so I can demonstrate a seek on a heap later on.  The table has an extra column (not shown because I am too lazy to update the diagram from yesterday) called padding – a CHAR(100) column that just contains 100 spaces in every row.  It’s just there to discourage SQL Server from choosing table scan over an index + RID lookup in one of the tests. The first stored procedure is called ResetTest: CREATE PROCEDURE dbo.ResetTest @Partitioned BIT = 'false' AS BEGIN SET NOCOUNT ON ; IF OBJECT_ID(N'dbo.Example', N'U') IS NOT NULL BEGIN DROP TABLE dbo.Example; END ; -- Test table is a heap -- Non-clustered primary key on 'key_col' CREATE TABLE dbo.Example ( key_col INTEGER NOT NULL, data INTEGER NOT NULL, padding CHAR(100) NOT NULL DEFAULT SPACE(100), CONSTRAINT [PK dbo.Example key_col] PRIMARY KEY NONCLUSTERED (key_col) ) ; IF @Partitioned = 'true' BEGIN -- Enterprise, Trial, or Developer -- required for partitioning tests IF SERVERPROPERTY('EngineEdition') = 3 BEGIN EXECUTE (' DROP TABLE dbo.Example ; IF EXISTS ( SELECT 1 FROM sys.partition_schemes WHERE name = N''PS'' ) DROP PARTITION SCHEME PS ; IF EXISTS ( SELECT 1 FROM sys.partition_functions WHERE name = N''PF'' ) DROP PARTITION FUNCTION PF ; CREATE PARTITION FUNCTION PF (INTEGER) AS RANGE RIGHT FOR VALUES (20, 40, 60, 80, 100) ; CREATE PARTITION SCHEME PS AS PARTITION PF ALL TO ([PRIMARY]) ; CREATE TABLE dbo.Example ( key_col INTEGER NOT NULL, data INTEGER NOT NULL, padding CHAR(100) NOT NULL DEFAULT SPACE(100), CONSTRAINT [PK dbo.Example key_col] PRIMARY KEY NONCLUSTERED (key_col) ) ON PS (key_col); '); END ELSE BEGIN RAISERROR('Invalid SKU for partition test', 16, 1); RETURN; END; END ; -- Non-unique non-clustered index on the 'data' column CREATE NONCLUSTERED INDEX [IX dbo.Example data] ON dbo.Example (data) ; -- Add 100 rows INSERT dbo.Example WITH (TABLOCKX) ( key_col, data ) SELECT key_col = V.number, data = V.number FROM master.dbo.spt_values AS V WHERE V.[type] = N'P' AND V.number BETWEEN 1 AND 100 ; END; GO The second stored procedure, ShowStats, displays information from the Index Usage Stats and Index Operational Stats DMVs: CREATE PROCEDURE dbo.ShowStats @Partitioned BIT = 'false' AS BEGIN -- Index Usage Stats DMV (QE) SELECT index_name = ISNULL(I.name, I.type_desc), scans = IUS.user_scans, seeks = IUS.user_seeks, lookups = IUS.user_lookups FROM sys.dm_db_index_usage_stats AS IUS JOIN sys.indexes AS I ON I.object_id = IUS.object_id AND I.index_id = IUS.index_id WHERE IUS.database_id = DB_ID(N'ScansAndSeeks') AND IUS.object_id = OBJECT_ID(N'dbo.Example', N'U') ORDER BY I.index_id ; -- Index Operational Stats DMV (SE) IF @Partitioned = 'true' SELECT index_name = ISNULL(I.name, I.type_desc), partitions = COUNT(IOS.partition_number), range_scans = SUM(IOS.range_scan_count), single_lookups = SUM(IOS.singleton_lookup_count) FROM sys.dm_db_index_operational_stats ( DB_ID(N'ScansAndSeeks'), OBJECT_ID(N'dbo.Example', N'U'), NULL, NULL ) AS IOS JOIN sys.indexes AS I ON I.object_id = IOS.object_id AND I.index_id = IOS.index_id GROUP BY I.index_id, -- Key I.name, I.type_desc ORDER BY I.index_id; ELSE SELECT index_name = ISNULL(I.name, I.type_desc), range_scans = SUM(IOS.range_scan_count), single_lookups = SUM(IOS.singleton_lookup_count) FROM sys.dm_db_index_operational_stats ( DB_ID(N'ScansAndSeeks'), OBJECT_ID(N'dbo.Example', N'U'), NULL, NULL ) AS IOS JOIN sys.indexes AS I ON I.object_id = IOS.object_id AND I.index_id = IOS.index_id GROUP BY I.index_id, -- Key I.name, I.type_desc ORDER BY I.index_id; END; The final stored procedure, RunTest, executes a query written against the example table: CREATE PROCEDURE dbo.RunTest @SQL VARCHAR(8000), @Partitioned BIT = 'false' AS BEGIN -- No execution plan yet SET STATISTICS XML OFF ; -- Reset the test environment EXECUTE dbo.ResetTest @Partitioned ; -- Previous call will throw an error if a partitioned -- test was requested, but SKU does not support it IF @@ERROR = 0 BEGIN -- IO statistics and plan on SET STATISTICS XML, IO ON ; -- Test statement EXECUTE (@SQL) ; -- Plan and IO statistics off SET STATISTICS XML, IO OFF ; EXECUTE dbo.ShowStats @Partitioned; END; END; The Tests The first test is a simple scan of the heap table: EXECUTE dbo.RunTest @SQL = 'SELECT * FROM Example'; The top result set comes from the Index Usage Stats DMV, so it is the Query Executor’s (QE) view.  The lower result is from Index Operational Stats, which shows statistics derived from the actions taken by the Storage Engine (SE).  We see that QE performed 1 scan operation on the heap, and SE performed a single range scan.  Let’s try a single-value equality seek on a unique index next: EXECUTE dbo.RunTest @SQL = 'SELECT key_col FROM Example WHERE key_col = 32'; This time we see a single seek on the non-clustered primary key from QE, and one singleton lookup on the same index by the SE.  Now for a single-value seek on the non-unique non-clustered index: EXECUTE dbo.RunTest @SQL = 'SELECT data FROM Example WHERE data = 32'; QE shows a single seek on the non-clustered non-unique index, but SE shows a single range scan on that index – not the singleton lookup we saw in the previous test.  That makes sense because we know that only a single-value seek into a unique index is a singleton seek.  A single-value seek into a non-unique index might retrieve any number of rows, if you think about it.  The next query is equivalent to the IN list example seen in the first post in this series, but it is written using OR (just for variety, you understand): EXECUTE dbo.RunTest @SQL = 'SELECT data FROM Example WHERE data = 32 OR data = 33'; The plan looks the same, and there’s no difference in the stats recorded by QE, but the SE shows two range scans.  Again, these are range scans because we are looking for two values in the data column, which is covered by a non-unique index.  I’ve added a snippet from the Properties window to show that the query plan does show two seek predicates, not just one.  Now let’s rewrite the query using BETWEEN: EXECUTE dbo.RunTest @SQL = 'SELECT data FROM Example WHERE data BETWEEN 32 AND 33'; Notice the seek operator only has one predicate now – it’s just a single range scan from 32 to 33 in the index – as the SE output shows.  For the next test, we will look up four values in the key_col column: EXECUTE dbo.RunTest @SQL = 'SELECT key_col FROM Example WHERE key_col IN (2,4,6,8)'; Just a single seek on the PK from the Query Executor, but four singleton lookups reported by the Storage Engine – and four seek predicates in the Properties window.  On to a more complex example: EXECUTE dbo.RunTest @SQL = 'SELECT * FROM Example WITH (INDEX([PK dbo.Example key_col])) WHERE key_col BETWEEN 1 AND 8'; This time we are forcing use of the non-clustered primary key to return eight rows.  The index is not covering for this query, so the query plan includes an RID lookup into the heap to fetch the data and padding columns.  The QE reports a seek on the PK and a lookup on the heap.  The SE reports a single range scan on the PK (to find key_col values between 1 and 8), and eight singleton lookups on the heap.  Remember that a bookmark lookup (RID or Key) is a seek to a single value in a ‘unique index’ – it finds a row in the heap or cluster from a unique RID or clustering key – so that’s why lookups are always singleton lookups, not range scans. Our next example shows what happens when a query plan operator is not executed at all: EXECUTE dbo.RunTest @SQL = 'SELECT key_col FROM Example WHERE key_col = 8 AND @@TRANCOUNT < 0'; The Filter has a start-up predicate which is always false (if your @@TRANCOUNT is less than zero, call CSS immediately).  The index seek is never executed, but QE still records a single seek against the PK because the operator appears once in an executed plan.  The SE output shows no activity at all.  This next example is 2008 and above only, I’m afraid: EXECUTE dbo.RunTest @SQL = 'SELECT * FROM Example WHERE key_col BETWEEN 1 AND 30', @Partitioned = 'true'; This is the first example to use a partitioned table.  QE reports a single seek on the heap (yes – a seek on a heap), and the SE reports two range scans on the heap.  SQL Server knows (from the partitioning definition) that it only needs to look at partitions 1 and 2 to find all the rows where key_col is between 1 and 30 – the engine seeks to find the two partitions, and performs a range scan seek on each partition. The final example for today is another seek on a heap – try to work out the output of the query before running it! EXECUTE dbo.RunTest @SQL = 'SELECT TOP (2) WITH TIES * FROM Example WHERE key_col BETWEEN 1 AND 50 ORDER BY $PARTITION.PF(key_col) DESC', @Partitioned = 'true'; Notice the lack of an explicit Sort operator in the query plan to enforce the ORDER BY clause, and the backward range scan. © 2011 Paul White email: [email protected] twitter: @SQL_Kiwi

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  • Focus on Social Relationship Management at Oracle OpenWorld

    - by Pat Ma
    v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} w\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} 0 0 1 422 2408 involver 20 5 2825 14.0 Normal 0 false false false false EN-US JA X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin:0in; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:Cambria; mso-ascii-font-family:Cambria; mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Cambria; mso-hansi-theme-font:minor-latin;} Greetings from Oracle OpenWorld 2012. Today, we’re going to focus on Social Relationship Management at Oracle OpenWorld.?Social networking is touching all businesses today.  Customers are speaking about your brand right now on social media sites. Your employees are speaking to one another on social media sites. In an Oracle survey, 40% of consumers factor in Facebook recommendations when making purchasing decisions. Despite the rise of social networking, 70% of marketers report having little understanding of social media conversations happening around their brand. Oracle has invested in technologies that will help companies leverage social media technologies for their enterprise. Our suite of social products is collectively known as Social Relationship Management. Customers are using Social Relationship Management to get analytics to social media conversations around their brand, manage multiple social media channels while keeping their brand consistent, optimize internal workflows and processes, and create better customer relationships and experiences. In this example, using Social Relationship Management, a high-end national grocery chain is able to see that “Coconut Water” is trending in San Francisco. They are now able to send a $2-off coconut water coupon to shoppers who have checked into their San Francisco locations. This promotion further drives sales of coconut water in San Francisco. In another example, using Social Relationship Management, a technology company creates multiple Facebook pages and runs campaigns on them. These social campaigns are now integrated and tracked as another marketing channel in Oracle Fusion CRM. The technology company can now track and respond to a particular customer as he moves across multiple channels – without having to restart the conversation each time the customer contacts the company. Furthermore, the technology company can see in one interface what marketing channels – including social – is performing best for each promotion. Besides being a Software-as-a-Service solution, social is also a Platform-as-a-Service solution. The benefit here is that customers can extend the functionality of our social applications to suit their particular needs or create their own social application from scratch. During the Social Developer track, developers are learning how to use Java and other industry-standard programming languages to plug in social functionality to enterprise applications. To see how Social Relationship Management can help your business build better relationships and experience with customers, visit us on the web at oracle.com/social. There are a lot more social-oriented sessions left at OpenWorld. To view a schedule of the upcoming social-oriented sessions, go here.

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  • Nominations now open for the Oracle FMW Excellence Awards 2014

    - by Greg Jensen
    2014 Oracle Excellence Award NominationsWho Is the Innovative Leader for Identity Management? •    Is your organization leveraging one of Oracle’s Identity and Access Management solutions in your production environment?•    Are you a leading edge organization that has adopted a forward thinking approach to Identity and Access Management processes across the organization?•    Are you ready to promote and highlight the success of your deployment to your peers? •    Would you a chance to win FREE registration to Oracle OpenWorld 2014? Oracle is pleased to announce the call for nominations for the 2014 Oracle Excellence Awards: Oracle Fusion Middleware Innovation.  The Oracle Excellence Awards for Oracle Fusion Middleware Innovation honor organizations using Oracle Fusion Middleware to deliver unique business value.  This year, the awards will recognize customers across nine distinct categories, including Identity and Access Management.  Oracle customers, who feel they are pioneers in their implementation of at least one of the Oracle Identity and Access Management offerings in a production environment or active deployment, should submit a nomination.  If submitted by June 20th, 2014, you will have a chance to win a FREE registration to Oracle OpenWorld 2014 (September 28 - October 2) in San Francisco, CA.  Top customers will be showcased at Oracle OpenWorld and featured in Oracle publications.   The  Identity and Access Management Nomination Form Additional benefits to nomineesNominating your organization opens additional opportunities to partner with Oracle such as:•    Promotion of your Customer Success StoriesProvides a platform for you to share the success of your initiatives and programs to peer groups raising the overall visibility of your team and your organization as a leader in security•    Social Media promotion (Video, Blog & Podcast)Reach the masses of Oracle’s customers through sharing of success stories, or customer created blog content that highlights the advanced thought leadership role in security with co-authored articles on Oracle Blog page that reaches close to 100,000 subscribers. There are numerous options to promote activities on Facebook, Twitter and co-branded activities using Video and Audio. •    Live speaking opportunities to your peersAs a technology leader within your organization, you can represent your organization at Oracle sponsored events (online, in person or webcasts) to help share the success of your organizations efforts building out your team/organization brand and success. •    Invitation to the IDM Architect ForumOracle is able to invite the right customers into the IDM Architect Forum which is an invite only group of customers that meet monthly to hear technology driven presentations from their own peers (not from Oracle) on today’s trends.  If you want to hear privately what some of the most successful companies in every industry are doing about security, this is the forum to be in. All presentations are private and remain within the forum, and only members can see take advantage of the lessons gained from these meetings.  To date, there are 125 members. There are many more advantages to partnering with Oracle, however, it can start with the simple nomination form for Identity and Access Management category of the 2014 Oracle Excellence Award Normal 0 false false false EN-US X-NONE X-NONE /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;}

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