Search Results

Search found 24616 results on 985 pages for 'number'.

Page 114/985 | < Previous Page | 110 111 112 113 114 115 116 117 118 119 120 121  | Next Page >

  • ms-access: missing operator in query expression

    - by every_answer_gets_a_point
    i have this sql statement in access: SELECT * FROM (SELECT [Occurrence Number], [1 0 Preanalytical (Before Testing)], NULL, NULL,NULL FROM [Lab Occurrence Form] WHERE NOT ([1 0 Preanalytical (Before Testing)] IS NULL) UNION SELECT [Occurrence Number], NULL, [2 0 Analytical (Testing Phase)], NULL,NULL FROM [Lab Occurrence Form] WHERE NOT ([2 0 Analytical (Testing Phase)] IS NULL) UNION SELECT [Occurrence Number], NULL, NULL, [3 0 Postanalytical ( After Testing)],NULL FROM [Lab Occurrence Form] WHERE NOT ([3 0 Postanalytical ( After Testing)] IS NULL) UNION SELECT [Occurrence Number], NULL, NULL,NULL [4 0 Other] FROM [Lab Occurrence Form] WHERE NOT ([4 0 Other] IS NULL) ) AS mySubQuery ORDER BY mySubQuery.[Occurrence Number]; everything was fine until i added the last line: SELECT [Occurrence Number], NULL, NULL,NULL [4 0 Other] FROM [Lab Occurrence Form] WHERE NOT ([4 0 Other] IS NULL) i get this error: syntax error (missing operator) in query expression 'NULL [4 0 Other]' anyone have any clues why i am getting this error?

    Read the article

  • Best way to test instance methods without running __init__

    - by KenFar
    I've got a simple class that gets most of its arguments via init, which also runs a variety of private methods that do most of the work. Output is available either through access to object variables or public methods. Here's the problem - I'd like my unittest framework to directly call the private methods called by init with different data - without going through init. What's the best way to do this? So far, I've been refactoring these classes so that init does less and data is passed in separately. This makes testing easy, but I think the usability of the class suffers a little. EDIT: Example solution based on Ignacio's answer: import types class C(object): def __init__(self, number): new_number = self._foo(number) self._bar(new_number) def _foo(self, number): return number * 2 def _bar(self, number): print number * 10 #--- normal execution - should print 160: ------- MyC = C(8) #--- testing execution - should print 80 -------- MyC = object.__new__(C) MyC._bar(8)

    Read the article

  • How to use Scanner to accept only valid int as input

    - by John
    I'm trying to make a small program more robust and I need some help with that. Scanner kb = new Scanner(System.in); int num1; int num2 = 0; System.out.print("Enter number 1: "); num1 = kb.nextInt(); while(num2<num1) { System.out.print("Enter number 2: "); num2 = kb.nextInt(); } Number 2 has to be greater than number 1 Also I want the program to automatically check and ignore if the user enters a character instead of a number. Because right now when a user enters for example r instead of a number the program just exits.

    Read the article

  • Database design for a Fantasy league

    - by Samidh T
    Here's the basic schema for my database Table user{ userid numeber primary key, count number } Table player{ pid number primary key, } Table user-player{ userid number primary key foreign key(user), pid number primary key foreign key(player) } Table temp{ pid number primary key, points number } Here's what I intend to do... After every match the temp table is updated which holds the id of players that played the last match and the points they earned. Next run a procedure that will match the pid from temp table with every uid of user-player table having the same pid. add the points from temp table to the count of user table for every matching uid. empty temp table. My questions is considering 200 players and 10000 users,Will this method be efficient? I am going to be using mysql for this.

    Read the article

  • How to get jQuery animateNumber to pull value dynamically?

    - by mcography
    I am trying to implement jQuery animateNumber, and I have it working like the demo just fine. However, I want to modify it so that it pulls the number from the HTML, rather than setting it in the script. I tried the following, but it just shows "NAN." What am I doing wrong? <div class="stat-title animate-number">$16,309</div> <script> $('.animate-number').each(function(){ var value = new Number; // Grab contents of element and turn it into a number value = $(this).text(); value = parseInt(value); // Set the starting text to 0 $(this).text('0'); $(this).animateNumber( { number: value, }, 1000 ) }); </script>

    Read the article

  • Index for wildcard match of end of string

    - by Anders Abel
    I have a table of phone numbers, storing the phone number as varchar(20). I have a requirement to implement searching of both entire numbers, but also on only the last part of the number, so a typical query will be: SELECT * FROM PhoneNumbers WHERE Number LIKE '%1234' How can I put an index on the Number column to make those searchs efficient? Is there a way to create an index that sorts the records on the reversed string? Another option might be to reverse the numbers before storing them, which will give queries like: SELECT * FROM PhoneNumbers WHERE ReverseNumber LIKE '4321%' However that will require all users of the database to always reverse the string. It might be solved by storing both the normal and reversed number and having the reversed number being updated by a trigger on insert/update. But that kind of solution is not very elegant. Any other suggestions?

    Read the article

  • PHP - Why does my computation produce a different result when I assign it to a variable?

    - by David
    I want to round a number to a specific number of significant digits - basically I want the following function: round(12345.67, 2) -> 12000 round(8888, 3) -> 8890 I have the following, but there's a strange problem. function round_to_sf($number, $sf) { $mostsigplace = floor(log10(abs($number)))+1; $num = $number / pow(10, ($mostsigplace-$sf)); echo ($number / pow(10, ($mostsigplace-$sf))).' '.$num.'<BR>'; } round_to_sf(41918.522, 1); Produces the following output: 4.1918522 -0 How can the result of a computation be different when it's assigned to a variable?

    Read the article

  • pl/sql creating a function with parameterized cursor with return date

    - by user3134365
    create or replace FUNCTION get_next_sch_date(cert_id VARCHAR2,test_id VARCHAR2) RETURN DATE AS CURSOR next_sch_date(pb_id number,test_no varchar2) IS SELECT Sch_Controls,PBY_FRQ,START_AFTER__CAL_DAYS,PBY_DUE_BY,PBY_NEXT_SCH_TEST_DATE FROM ms_cmp_plan_pby WHERE pby_id=pb_id AND test_plan_id=test_no; l_new_date DATE; l_new_sch number; sch_ctrl VARCHAR2(100); pb_frq VARCHAR2(100); start_days NUMBER; due_days NUMBER; test_date DATE; pb_id NUMBER; test_no NUMBER; BEGIN OPEN next_sch_date(pb_id,test_no); loop FETCH next_sch_date INTO sch_ctrl,pb_frq,start_days,due_days,test_date; SELECT DISTINCT pby_rec_id INTO l_new_sch FROM ms_cmp_assignment_log WHERE ASSIGNMENT_ID=cert_id AND PLAN_ID=test_id; exit; end loop; CLOSE next_sch_date; RETURN l_new_date; Exception WHEN others THEN RETURN NULL; end; this is my function but i dont getting excepted result

    Read the article

  • How do I implement an higher lower game algorithm?

    - by lazorde
    The computer will guess a player’s number between 1 and 100. After each guess the human player should respond “higher”, “lower” or “correct”. Your program should be able to guess the player’s number in no more than 7 tries. Begin by explaining the game to the player, telling him/her to think of a number between 1 and 100. Make the computer do what you would normally do to guess a number in a certain range. Allow the user to respond with “higher”, “lower”, or “correct” after each computer guess. Output the number of tries it took the computer to guess the number. Make the game as user friendly as you can.

    Read the article

  • Problem displaying contents of a class in Java

    - by LuckySlevin
    My problem is i have a class and in it there is a list of elements of another class. public class Branch { private ArrayList<Player> players = new ArrayList<Player>(); String brName; public Branch() {} public void setBr(String brName){this.brName = brName;} public String getBr(){return brName;} public ArrayList<Player> getPlayers() { return players; } public void setPlayers(ArrayList<Player> players) { this.players =new ArrayList<Player>(players); } } public class Player { private String name; private String pos; private Integer salary; private Integer number; public Player(String name, String pos, Integer salary, Integer number) { this.name = name; this.pos = pos; this.salary = salary; this.number = number; } public Player(){} public String getName() { return name; } public String getPos() { return pos; } public Integer getSalary() { return salary; } public Integer getNumber() { return number; } public void setName(String name) { this.name = name; } public void setPos(String pos) { this.pos = pos; } public void setSalary(Integer salary) { this.salary = salary; } public void setNumber(Integer number) { this.number = number; } } My problem is to print the players of a Branch with their name,pos,salary,number. For this i tried this simply : String p1,p2; int a1,a2; p1 = input.readLine(); p2 = input.readLine(); a1 = Integer.parseInt(input.readLine()); a2 = Integer.parseInt(input.readLine()); players[0].setName(p1); players[0].setPos(p2); players[0].setSalary(a1); players[0].setNumber(a2); ptmp.add(players[0]); myBranch[0].setPlayers(ptmp); System.out.println(myBranch[0].brName + " " + myBranch[0].getPlayers()); I wrote this just to try how to display. I created an array of Players, and Branches so they already defined. The problem is getPlayers() doesn't give me any result. What is the way to do this?

    Read the article

  • Remove parameter from link

    - by goordis
    I have many links with parameter number - value is numbers between 1-1000 http://mysite.com?one=2&two=4&number=2 http://mysite.com?one=2&two=4&four=4&number=124 http://mysite.com?one=2&three=4&number=9 http://mysite.com?two=4&number=242 http://mysite.com?one=2&two=4&number=52 How can i remove from this parameter and value with PHP? I would like receive: http://mysite.com?one=2&two=4 http://mysite.com?one=2&two=4&four=4 http://mysite.com?one=2&three=4 http://mysite.com?two=4 http://mysite.com?one=2&two=4

    Read the article

  • Is this the right way of handling command line arguments?

    - by shadyabhi
    ask_username = True ask_password = True ask_message = True ask_number = True def Usage(): print '\t-h, --help: View help' print '\t-u, --username: Username' print '\t-p, --password: Password' print '\t-n, --number: numbber to send the sms' print '\t-m, --message: Message to send' sys.exit(1) opts, args = getopt(sys.argv[1:], 'u:p:m:n:h',["username=","password=","message=","number=","help"]) print opts, args for o,v in opts: if o in ("-h", "--help"): Usage() elif o in ("-u", "--username"): username = v ask_username = False elif o in ("-p", "--password"): passwd = v ask_password = False elif o in ("-m", "--message"): message = v ask_message = False elif o in ("-n", "--number"): number = v ask_number = False #Credentials taken here if ask_username: username = raw_input("Enter USERNAME: ") if ask_password: passwd = getpass() if ask_message: message = raw_input("Enter Message: ") if ask_number: number = raw_input("Enter Mobile number: ") I dont think it is, because I am using 4 objects just for checking if command line argument was provided... Guide me with the best way of doing it..

    Read the article

  • where clausule on field defined by sub-query

    - by stUrb
    I have this query which displays some properties and count the number of references to it from an other table: SELECT p.id,p.propName ( SELECT COUNT(*) FROM propLoc WHERE propLoc.propID = p.id ) AS number FROM property as p WHERE p.category != 'natural' This generates a good table with all the information I want to filter: id | propName | number 3 | Name 1 | 3 4 | Name 2 | 1 5 | Name 3 | 0 6 | Name 4 | 10 etc etc I now want to filter out the properties with number <= 0 So I tried to add an AND number > 0 But it reacts with Unknown column 'number' in 'where clause' apparently you can't filter on a name specified by a subquery? How can I achieve my goal?

    Read the article

  • Criticize my code, please

    - by Micky
    Hey, I was applying for a position, and they asked me to complete a coding problem for them. I did so and submitted it, but I later found out I was rejected from the position. Anyways, I have an eclectic programming background so I'm not sure if my code is grossly wrong or if I just didn't have the best solution out there. I would like to post my code and get some feedback about it. Before I do, here's a description of a problem: You are given a sorted array of integers, say, {1, 2, 4, 4, 5, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 11, 13 }. Now you are supposed to write a program (in C or C++, but I chose C) that prompts the user for an element to search for. The program will then search for the element. If it is found, then it should return the first index the entry was found at and the number of instances of that element. If the element is not found, then it should return "not found" or something similar. Here's a simple run of it (with the array I just put up): Enter a number to search for: 4 4 was found at index 2. There are 2 instances for 4 in the array. Enter a number to search for: -4. -4 is not in the array. They made a comment that my code should scale well with large arrays (so I wrote up a binary search). Anyways, my code basically runs as follows: Prompts user for input. Then it checks if it is within bounds (bigger than a[0] in the array and smaller than the largest element of the array). If so, then I perform a binary search. If the element is found, then I wrote two while loops. One while loop will count to the left of the element found, and the second while loop will count to the right of the element found. The loops terminate when the adjacent elements do not match with the desired value. EX: 4, 4, 4, 4, 4 The bold 4 is the value the binary search landed on. One loop will check to the left of it, and another loop will check to the right of it. Their sum will be the total number of instances of the the number four. Anyways, I don't know if there are any advanced techniques that I am missing or if I just don't have the CS background and made a big error. Any constructive critiques would be appreciated! #include <stdio.h> #include <stdlib.h> #include <string.h> #include <stddef.h> /* function prototype */ int get_num_of_ints( const int* arr, size_t r, int N, size_t* first, size_t* count ); int main() { int N; /* input variable */ int arr[]={1,1,2,3,3,4,4,4,4,5,5,7,7,7,7,8,8,8,9,11,12,12}; /* array of sorted integers */ size_t r = sizeof(arr)/sizeof(arr[0]); /* right bound */ size_t first; /* first match index */ size_t count; /* total number of matches */ /* prompts the user to enter input */ printf( "\nPlease input the integer you would like to find.\n" ); scanf( "%d", &N ); int a = get_num_of_ints( arr, r, N, &first, &count ); /* If the function returns -1 then the value is not found. Else it is returned */ if( a == -1) printf( "%d has not been found.\n", N ); else if(a >= 0){ printf( "The first matching index is %d.\n", first ); printf( "The total number of instances is %d.\n", count ); } return 0; } /* function definition */ int get_num_of_ints( const int* arr, size_t r, int N, size_t* first, size_t* count ) { int lo=0; /* lower bound for search */ int m=0; /* middle value obtained */ int hi=r-1; /* upper bound for search */ int w=r-1; /* used as a fixed upper bound to calculate the number of right instances of a particular value. */ /* binary search to find if a value exists */ /* first check if the element is out of bounds */ if( N < arr[0] || arr[hi] < N ){ m = -1; } else{ /* binary search to find a value, if it exists, within given parameters */ while(lo <= hi){ m = (hi + lo)/2; if(arr[m] < N) lo = m+1; else if(arr[m] > N) hi = m-1; else if(arr[m]==N){ m=m; break; } } if (lo > hi) /* if it doesn't we assign it -1 */ m = -1; } /* If the value is found, then we compute the left and right instances of it */ if( m >= 0 ){ int j = m-1; /* starting with the first term to the left */ int L = 0; /* total number of left instances */ /* while loop computes total number of left instances */ while( j >= 0 && arr[j] == arr[m] ){ L++; j--; } /* There are six possible outcomes of this. Depending on the outcome, we must assign the first index variable accordingly */ if( j > 0 && L > 0 ) *first=j+1; else if( j==0 && L==0) *first=m; else if( j > 0 && L==0 ) *first=m; else if(j < 0 && L==0 ) *first=m; else if( j < 0 && L > 0 ) *first=0; else if( j=0 && L > 0 ) *first=j+1; int h = m + 1; /* starting with the first term to the right */ int R = 0; /* total number of right instances */ /* while loop computes total number of right instances */ /* we fixed w earlier so that it's value does not change */ while( arr[h]==arr[m] && h <= w ){ R++; h++; } *count = (R + L + 1); /* total number of instances stored as value of count */ return *first; /* first instance index stored here */ } /* if value does not exist, then we return a negative value */ else if( m==-1) return -1; }

    Read the article

  • The Most Common and Least Used 4-Digit PIN Numbers [Security Analysis Report]

    - by Asian Angel
    How ‘secure’ is your 4-digit PIN number? Is your PIN number a far too common one or is it a bit more unique in comparison to others? The folks over at the Data Genetics blog have put together an interesting analysis report that looks at the most common and least used 4-digit PIN numbers chosen by people. Numerically based (0-9) 4-digit PIN numbers only allow for a total of 10,000 possible combinations, so it stands to reason that some combinations are going to be far more common than others. The question is whether or not your personal PIN number choices are among the commonly used ones or ‘stand out’ as being more unique. Note 1: Data Genetics used data condensed from released, exposed, & discovered password tables and security breaches to generate the analysis report. Note 2: The updates section at the bottom has some interesting tidbits concerning peoples’ use of dates and certain words for PIN number generation. The analysis makes for very interesting reading, so browse on over to get an idea of where you stand with regards to your personal PIN number choices. 8 Deadly Commands You Should Never Run on Linux 14 Special Google Searches That Show Instant Answers How To Create a Customized Windows 7 Installation Disc With Integrated Updates

    Read the article

  • code metrics for .net code

    - by user20358
    While the code metrics tool gives a pretty good analysis of the code being analyzed, I was wondering if there was any such benchmark on acceptable standards for the following as well: Maximum number of types per assembly Maximum number of such types that can be accessible Maximum number of parameters per method Acceptable RFC count Acceptable Afferent coupling count Acceptable Efferent coupling count Any other metrics to judge the quality of .Net code by? Thanks for your time.

    Read the article

  • Oracle Solaris Cluster 4.2 Event and its SNMP Interface

    - by user12609115
    Background The cluster event SNMP interface was first introduced in Oracle Solaris Cluster 3.2 release. The details of the SNMP interface are described in the Oracle Solaris Cluster System Administration Guide and the Cluster 3.2 SNMP blog. Prior to the Oracle Solaris Cluster 4.2 release, when the event SNMP interface was enabled, it would take effect on WARNING or higher severity events. The events with WARNING or higher severity are usually for the status change of a cluster component from ONLINE to OFFLINE. The interface worked like an alert/alarm interface when some components in the cluster were out of service (changed to OFFLINE). The consumers of this interface could not get notification for all status changes and configuration changes in the cluster. Cluster Event and its SNMP Interface in Oracle Solaris Cluster 4.2 The user model of the cluster event SNMP interface is the same as what was provided in the previous releases. The cluster event SNMP interface is not enabled by default on a freshly installed cluster; you can enable it by using the cluster event SNMP administration commands on any cluster nodes. Usually, you only need to enable it on one of the cluster nodes or a subset of the cluster nodes because all cluster nodes get the same cluster events. When it is enabled, it is responsible for two basic tasks. • Logs up to 100 most recent NOTICE or higher severity events to the MIB. • Sends SNMP traps to the hosts that are configured to receive the above events. The changes in the Oracle Solaris Cluster 4.2 release are1) Introduction of the NOTICE severity for the cluster configuration and status change events.The NOTICE severity is introduced for the cluster event in the 4.2 release. It is the severity between the INFO and WARNING severity. Now all severities for the cluster events are (from low to high) • INFO (not exposed to the SNMP interface) • NOTICE (newly introduced in the 4.2 release) • WARNING • ERROR • CRITICAL • FATAL In the 4.2 release, the cluster event system is enhanced to make sure at least one event with the NOTICE or a higher severity will be generated when there is a configuration or status change from a cluster component instance. In other words, the cluster events from a cluster with the NOTICE or higher severities will cover all status and configuration changes in the cluster (include all component instances). The cluster component instance here refers to an instance of the following cluster componentsnode, quorum, resource group, resource, network interface, device group, disk, zone cluster and geo cluster heartbeat. For example, pnode1 is an instance of the cluster node component, and oracleRG is an instance of the cluster resource group. With the introduction of the NOTICE severity event, when the cluster event SNMP interface is enabled, the consumers of the SNMP interface will get notification for all status and configuration changes in the cluster. A thrid-party system management platform with the cluster SNMP interface integration can generate alarms and clear alarms programmatically, because it can get notifications for the status change from ONLINE to OFFLINE and also from OFFLINE to ONLINE. 2) Customization for the cluster event SNMP interface • The number of events logged to the MIB is 100. When the number of events stored in the MIB reaches 100 and a new qualified event arrives, the oldest event will be removed before storing the new event to the MIB (FIFO, first in, first out). The 100 is the default and minimum value for the number of events stored in the MIB. It can be changed by setting the log_number property value using the clsnmpmib command. The maximum number that can be set for the property is 500. • The cluster event SNMP interface takes effect on the NOTICE or high severity events. The NOTICE severity is also the default and lowest event severity for the SNMP interface. The SNMP interface can be configured to take effect on other higher severity events, such as WARNING or higher severity events by setting the min_severity property to the WARNING. When the min_severity property is set to the WARNING, the cluster event SNMP interface would behave the same as the previous releases (prior to the 4.2 release). Examples, • Set the number of events stored in the MIB to 200 # clsnmpmib set -p log_number=200 event • Set the interface to take effect on WARNING or higher severity events. # clsnmpmib set -p min_severity=WARNING event Administering the Cluster Event SNMP Interface Oracle Solaris Cluster provides the following three commands to administer the SNMP interface. • clsnmpmib: administer the SNMP interface, and the MIB configuration. • clsnmphost: administer hosts for the SNMP traps • clsnmpuser: administer SNMP users (specific for SNMP v3 protocol) Only clsnmpmib is changed in the 4.2 release to support the aforementioned customization of the SNMP interface. Here are some simple examples using the commands. Examples: 1. Enable the cluster event SNMP interface on the local node # clsnmpmib enable event 2. Display the status of the cluster event SNMP interface on the local node # clsnmpmib show -v 3. Configure my_host to receive the cluster event SNMP traps. # clsnmphost add my_host Cluster Event SNMP Interface uses the common agent container SNMP adaptor, which is based on the JDMK SNMP implementation as its SNMP agent infrastructure. By default, the port number for the SNMP MIB is 11161, and the port number for the SNMP traps is 11162. The port numbers can be changed by using the cacaoadm. For example, # cacaoadm list-params Print all changeable parameters. The output includes the snmp-adaptor-port and snmp-adaptor-trap-port properties. # cacaoadm set-param snmp-adaptor-port=1161 Set the SNMP MIB port number to 1161. # cacaoadm set-param snmp-adaptor-trap-port=1162 Set the SNMP trap port number to 1162. The cluster event SNMP MIB is defined in sun-cluster-event-mib.mib, which is located in the /usr/cluster/lib/mibdirectory. Its OID is 1.3.6.1.4.1.42.2.80, that can be used to walk through the MIB data. Again, for more detail information about the cluster event SNMP interface, please see the Oracle Solaris Cluster 4.2 System Administration Guide. - Leland Chen 

    Read the article

  • USB mouse does not work on boot

    - by Uku Loskit
    My problem is pretty much a duplicate of the one described in USB mouse late to load , but the solution there has not worked for me. I'm running the same OS and experiencing the exact same issue. It disappears after 10 seconds or so. Booting with the options specified in the other question did not fix it :/ Thanks in advance. sheepz@sheepz-desktop:~$ dmesg | egrep "hci|usb" [ 0.188000] usbcore: registered new interface driver usbfs [ 0.188000] usbcore: registered new interface driver hub [ 0.188000] usbcore: registered new device driver usb [ 0.358613] ehci_hcd: USB 2.0 'Enhanced' Host Controller (EHCI) Driver [ 0.358627] ohci_hcd: USB 1.1 'Open' Host Controller (OHCI) Driver [ 0.358637] uhci_hcd: USB Universal Host Controller Interface driver [ 0.358683] uhci_hcd 0000:00:1d.0: PCI INT A -> GSI 23 (level, low) -> IRQ 23 [ 0.358691] uhci_hcd 0000:00:1d.0: setting latency timer to 64 [ 0.358695] uhci_hcd 0000:00:1d.0: UHCI Host Controller [ 0.358726] uhci_hcd 0000:00:1d.0: new USB bus registered, assigned bus number 1 [ 0.358758] uhci_hcd 0000:00:1d.0: irq 23, io base 0x0000e100 [ 0.358927] uhci_hcd 0000:00:1d.1: PCI INT B -> GSI 19 (level, low) -> IRQ 19 [ 0.358932] uhci_hcd 0000:00:1d.1: setting latency timer to 64 [ 0.358935] uhci_hcd 0000:00:1d.1: UHCI Host Controller [ 0.358964] uhci_hcd 0000:00:1d.1: new USB bus registered, assigned bus number 2 [ 0.358991] uhci_hcd 0000:00:1d.1: irq 19, io base 0x0000e200 [ 0.359132] uhci_hcd 0000:00:1d.2: PCI INT C -> GSI 18 (level, low) -> IRQ 18 [ 0.359137] uhci_hcd 0000:00:1d.2: setting latency timer to 64 [ 0.359139] uhci_hcd 0000:00:1d.2: UHCI Host Controller [ 0.359165] uhci_hcd 0000:00:1d.2: new USB bus registered, assigned bus number 3 [ 0.359193] uhci_hcd 0000:00:1d.2: irq 18, io base 0x0000e300 [ 0.359327] uhci_hcd 0000:00:1d.3: PCI INT D -> GSI 16 (level, low) -> IRQ 16 [ 0.359332] uhci_hcd 0000:00:1d.3: setting latency timer to 64 [ 0.359334] uhci_hcd 0000:00:1d.3: UHCI Host Controller [ 0.359360] uhci_hcd 0000:00:1d.3: new USB bus registered, assigned bus number 4 [ 0.359387] uhci_hcd 0000:00:1d.3: irq 16, io base 0x0000e400 [ 0.731933] usb 1-1: new full speed USB device using uhci_hcd and address 2 [ 1.023859] usb 1-2: new full speed USB device using uhci_hcd and address 3 [ 16.136175] usb 1-2: device descriptor read/64, error -110 [ 31.352481] usb 1-2: device descriptor read/64, error -110 [ 31.568485] usb 1-2: new full speed USB device using uhci_hcd and address 4 [ 46.680794] usb 1-2: device descriptor read/64, error -110 [ 61.903555] usb 1-2: device descriptor read/64, error -110 [ 62.119671] usb 1-2: new full speed USB device using uhci_hcd and address 5 [ 72.541078] usb 1-2: device not accepting address 5, error -110 [ 72.653194] usb 1-2: new full speed USB device using uhci_hcd and address 6 [ 83.066637] usb 1-2: device not accepting address 6, error -110 [ 83.178615] usb 3-1: new low speed USB device using uhci_hcd and address 2 [ 83.562546] usbcore: registered new interface driver hiddev [ 83.578827] input: Logitech USB-PS/2 Optical Mouse as /devices/pci0000:00/0000:00:1d.2/usb3/3-1/3-1:1.0/input/input3 [ 83.579016] generic-usb 0003:046D:C01D.0001: input,hidraw0: USB HID v1.10 Mouse [Logitech USB-PS/2 Optical Mouse] on usb-0000:00:1d.2-1/input0 [ 83.579244] usbcore: registered new interface driver usbhid [ 83.579246] usbhid: USB HID core driver [114025.224407] usb 3-1: USB disconnect, address 2 sheepz@sheepz-desktop:~$ dmesg | egrep "hci|usb" [ 0.188000] usbcore: registered new interface driver usbfs [ 0.188000] usbcore: registered new interface driver hub [ 0.188000] usbcore: registered new device driver usb [ 0.358613] ehci_hcd: USB 2.0 'Enhanced' Host Controller (EHCI) Driver [ 0.358627] ohci_hcd: USB 1.1 'Open' Host Controller (OHCI) Driver [ 0.358637] uhci_hcd: USB Universal Host Controller Interface driver [ 0.358683] uhci_hcd 0000:00:1d.0: PCI INT A -> GSI 23 (level, low) -> IRQ 23 [ 0.358691] uhci_hcd 0000:00:1d.0: setting latency timer to 64 [ 0.358695] uhci_hcd 0000:00:1d.0: UHCI Host Controller [ 0.358726] uhci_hcd 0000:00:1d.0: new USB bus registered, assigned bus number 1 [ 0.358758] uhci_hcd 0000:00:1d.0: irq 23, io base 0x0000e100 [ 0.358927] uhci_hcd 0000:00:1d.1: PCI INT B -> GSI 19 (level, low) -> IRQ 19 [ 0.358932] uhci_hcd 0000:00:1d.1: setting latency timer to 64 [ 0.358935] uhci_hcd 0000:00:1d.1: UHCI Host Controller [ 0.358964] uhci_hcd 0000:00:1d.1: new USB bus registered, assigned bus number 2 [ 0.358991] uhci_hcd 0000:00:1d.1: irq 19, io base 0x0000e200 [ 0.359132] uhci_hcd 0000:00:1d.2: PCI INT C -> GSI 18 (level, low) -> IRQ 18 [ 0.359137] uhci_hcd 0000:00:1d.2: setting latency timer to 64 [ 0.359139] uhci_hcd 0000:00:1d.2: UHCI Host Controller [ 0.359165] uhci_hcd 0000:00:1d.2: new USB bus registered, assigned bus number 3 [ 0.359193] uhci_hcd 0000:00:1d.2: irq 18, io base 0x0000e300 [ 0.359327] uhci_hcd 0000:00:1d.3: PCI INT D -> GSI 16 (level, low) -> IRQ 16 [ 0.359332] uhci_hcd 0000:00:1d.3: setting latency timer to 64 [ 0.359334] uhci_hcd 0000:00:1d.3: UHCI Host Controller [ 0.359360] uhci_hcd 0000:00:1d.3: new USB bus registered, assigned bus number 4 [ 0.359387] uhci_hcd 0000:00:1d.3: irq 16, io base 0x0000e400 [ 0.731933] usb 1-1: new full speed USB device using uhci_hcd and address 2 [ 1.023859] usb 1-2: new full speed USB device using uhci_hcd and address 3 [ 16.136175] usb 1-2: device descriptor read/64, error -110 [ 31.352481] usb 1-2: device descriptor read/64, error -110 [ 31.568485] usb 1-2: new full speed USB device using uhci_hcd and address 4 [ 46.680794] usb 1-2: device descriptor read/64, error -110 [ 61.903555] usb 1-2: device descriptor read/64, error -110 [ 62.119671] usb 1-2: new full speed USB device using uhci_hcd and address 5 [ 72.541078] usb 1-2: device not accepting address 5, error -110 [ 72.653194] usb 1-2: new full speed USB device using uhci_hcd and address 6 [ 83.066637] usb 1-2: device not accepting address 6, error -110 [ 83.178615] usb 3-1: new low speed USB device using uhci_hcd and address 2 [ 83.562546] usbcore: registered new interface driver hiddev [ 83.578827] input: Logitech USB-PS/2 Optical Mouse as /devices/pci0000:00/0000:00:1d.2/usb3/3-1/3-1:1.0/input/input3 [ 83.579016] generic-usb 0003:046D:C01D.0001: input,hidraw0: USB HID v1.10 Mouse [Logitech USB-PS/2 Optical Mouse] on usb-0000:00:1d.2-1/input0 [ 83.579244] usbcore: registered new interface driver usbhid [ 83.579246] usbhid: USB HID core driver

    Read the article

  • 3D Dice using Maya - for integration with iOS game app

    - by Anil
    My designer is building a 3D design of a dice using Maya. I want to integrate this in my iOS app so that the user can spin the dice and get a number. Then they play the game using that number. So, I have two questions: 1) How can I make the dice spin and stop at a random position so that a number is presented to the user? and 2) Once it stops spinning how can I detect the number that is displayed to the user (programmatically)? Many thanks. -Anil

    Read the article

  • choosing Database and Its Design for Rails

    - by Gaurav Shah
    I am having a difficulty in deciding the database & its structure. Let us say the problem is like this. For my product I have various customers( each is an educational institute) Each customer have their own sub-clients ( Institution have students) Each student record will have some basic information like "name" & "Number" . There are also additional information that a customer(institution) might want to ask sub-client(student) like "email" or "semester" I have come up with two solutions : 1. Mysql _insititution__ id-|- Description| __Student__ id-|-instituition_id-|-Name-|-Number| __student_additional_details__ student_id -|- field_name -|- Value Student_additional_details will have multiple records for each student depending upon number of questions asked from institution. 2.MongoDb _insititution___ id-|- Description| _Student__ id-|-instituition_id-|-Name-|-Number|-otherfield1 -|- otherfield2 with mongo the structure itself can be dynamic so student table seems really good in mongo . But the problem comes when I have to relate student with institution . So which one is a better design ? Or some other idea ?

    Read the article

  • SQL SERVER – Removing Leading Zeros From Column in Table

    - by pinaldave
    Some questions surprises me and make me write code which I have never explored before. Today was similar experience as well. I have always received the question regarding how to reserve leading zeroes in SQL Server while displaying them on the SSMS or another application. I have written articles on this subject over here. SQL SERVER – Pad Ride Side of Number with 0 – Fixed Width Number Display SQL SERVER – UDF – Pad Ride Side of Number with 0 – Fixed Width Number Display SQL SERVER – Preserve Leading Zero While Coping to Excel from SSMS Today I received a very different question where the user wanted to remove leading zero and white space. I am using the same sample sent by user in this example. USE tempdb GO -- Create sample table CREATE TABLE Table1 (Col1 VARCHAR(100)) INSERT INTO Table1 (Col1) SELECT '0001' UNION ALL SELECT '000100' UNION ALL SELECT '100100' UNION ALL SELECT '000 0001' UNION ALL SELECT '00.001' UNION ALL SELECT '01.001' GO -- Original data SELECT * FROM Table1 GO -- Remove leading zeros SELECT SUBSTRING(Col1, PATINDEX('%[^0 ]%', Col1 + ' '), LEN(Col1)) FROM Table1 GO -- Clean up DROP TABLE Table1 GO Here is the resultset of above script. It will remove any leading zero or space and will display the number accordingly. This problem is a very generic problem and I am confident there are alternate solutions to this problem as well. If you have an alternate solution or can suggest a sample data which does not satisfy the SUBSTRING solution proposed, I will be glad to include them in follow up blog post with due credit. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Function, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

    Read the article

  • TakeWhile and SkipWhile method in LINQ

    - by vik20000in
     In my last post I talked about how to use the take and the Skip keyword to filter out the number of records that we are fetching. But there is only problem with the take and skip statement. The problem lies in the dependency where by the number of records to be fetched has to be passed to it. Many a times the number of records to be fetched is also based on the query itself. For example if we want to continue fetching records till a certain condition is met on the record set. Let’s say we want to fetch records from the array of number till we get 7. For this kind of query LINQ has exposed the TakeWhile Method.     int[] numbers = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };     var firstNumbersLessThan6 = numbers.TakeWhile(n => n < 7);   In the same way we can also use the SkipWhile statement. The skip while statement will skip all the records that do not match certain condition provided. In the example below we are skiping all those number which are not divisible by 3. Remember we could have done this with where clause also, but SkipWhile method can be useful in many other situation and hence the example and the keyword.     int[] numbers = { 5, 4, 1, 3, 9, 8, 6, 7, 2, 0 };     var allButFirst3Numbers = numbers.SkipWhile(n => n % 3 != 0); Vikram

    Read the article

  • Login failed for user 'sa' because the account is currently locked out. The system administrator can

    - by cabhilash
    Login failed for user 'sa' because the account is currently locked out. The system administrator can unlock it. (Microsoft SQL Server, Error: 18486) SQL server has local password policies. If policy is enabled which locks down the account after X number of failed attempts then the account is automatically locked down.This error with 'sa' account is very common. sa is default administartor login available with SQL server. So there are chances that an ousider has tried to bruteforce your system. (This can cause even if a legitimate tries to access the account with wrong password.Sometimes a user would have changed the password without informing others. So the other users would try to lo) You can unlock the account with the following options (use another admin account or connect via windows authentication) Alter account & unlock ALTER LOGIN sa WITH PASSWORD='password' UNLOCK Use another account Almost everyone is aware of the sa account. This can be the potential security risk. Even if you provide strong password hackers can lock the account by providing the wrong password. ( You can provide extra security by installing firewall or changing the default port but these measures are not always practical). As a best practice you can disable the sa account and use another account with same privileges.ALTER LOGIN sa DISABLE You can edit the lock-ot options using gpedit.msc( in command prompt type gpedit.msc and press enter). Navigate to Account Lokout policy as shown in the figure The Following options are available Account lockout threshold This security setting determines the number of failed logon attempts that causes a user account to be locked out. A locked-out account cannot be used until it is reset by an administrator or until the lockout duration for the account has expired. You can set a value between 0 and 999 failed logon attempts. If you set the value to 0, the account will never be locked out. Failed password attempts against workstations or member servers that have been locked using either CTRL+ALT+DELETE or password-protected screen savers count as failed logon attempts. Account lockout duration This security setting determines the number of minutes a locked-out account remains locked out before automatically becoming unlocked. The available range is from 0 minutes through 99,999 minutes. If you set the account lockout duration to 0, the account will be locked out until an administrator explicitly unlocks it. If an account lockout threshold is defined, the account lockout duration must be greater than or equal to the reset time. Default: None, because this policy setting only has meaning when an Account lockout threshold is specified. Reset account lockout counter after This security setting determines the number of minutes that must elapse after a failed logon attempt before the failed logon attempt counter is reset to 0 bad logon attempts. The available range is 1 minute to 99,999 minutes. If an account lockout threshold is defined, this reset time must be less than or equal to the Account lockout duration. Default: None, because this policy setting only has meaning when an Account lockout threshold is specified.When creating SQL user you can set CHECK_POLICY=on which will enforce the windows password policy on the account. The following policies will be applied Define the Enforce password history policy setting so that several previous passwords are remembered. With this policy setting, users cannot use the same password when their password expires.  Define the Maximum password age policy setting so that passwords expire as often as necessary for your environment, typically, every 30 to 90 days. With this policy setting, if an attacker cracks a password, the attacker only has access to the network until the password expires.  Define the Minimum password age policy setting so that passwords cannot be changed until they are more than a certain number of days old. This policy setting works in combination with the Enforce password historypolicy setting. If a minimum password age is defined, users cannot repeatedly change their passwords to get around the Enforce password history policy setting and then use their original password. Users must wait the specified number of days to change their passwords.  Define a Minimum password length policy setting so that passwords must consist of at least a specified number of characters. Long passwords--seven or more characters--are usually stronger than short ones. With this policy setting, users cannot use blank passwords, and they have to create passwords that are a certain number of characters long.  Enable the Password must meet complexity requirements policy setting. This policy setting checks all new passwords to ensure that they meet basic strong password requirements.  Password must meet the following complexity requirement, when they are changed or created: Not contain the user's entire Account Name or entire Full Name. The Account Name and Full Name are parsed for delimiters: commas, periods, dashes or hyphens, underscores, spaces, pound signs, and tabs. If any of these delimiters are found, the Account Name or Full Name are split and all sections are verified not to be included in the password. There is no check for any character or any three characters in succession. Contain characters from three of the following five categories:  English uppercase characters (A through Z) English lowercase characters (a through z) Base 10 digits (0 through 9) Non-alphabetic characters (for example, !, $, #, %) A catch-all category of any Unicode character that does not fall under the previous four categories. This fifth category can be regionally specific.

    Read the article

  • Oracle Linux and Oracle VM pricing guide

    - by wcoekaer
    A few days ago someone showed me a pricing guide from a Linux vendor and I was a bit surprised at the complexity of it. Especially when you look at larger servers (4 or 8 sockets) and when adding virtual machine use into the mix. I think we have a very compelling and simple pricing model for both Oracle Linux and Oracle VM. Let me see if I can explain it in 1 page, not 10 pages. This pricing information is publicly available on the Oracle store, I am using the current public list prices. Also keep in mind that this is for customers using non-oracle x86 servers. When a customer purchases an Oracle x86 server, the annual systems support includes full use (all you can eat) of Oracle Linux, Oracle VM and Oracle Solaris (no matter how many VMs you run on that server, in case you deploy guests on a hypervisor). This support level is the equivalent of premier support in the list below. Let's start with Oracle VM (x86) : Oracle VM support subscriptions are per physical server on which you deploy the Oracle VM Server product. (1) Oracle VM Premier Limited - 1- or 2 socket server : $599 per server per year (2) Oracle VM Premier - more than 2 socket server (4, or 8 or whatever more) : $1199 per server per year The above includes the use of Oracle VM Manager and Oracle Enterprise Manager Cloud Control's Virtualization management pack (including self service cloud portal, etc..) 24x7 support, access to bugfixes, updates and new releases. It also includes all options, live migrate, dynamic resource scheduling, high availability, dynamic power management, etc If you want to play with the product, or even use the product without access to support services, the product is freely downloadable from edelivery. Next, Oracle Linux : Oracle Linux support subscriptions are per physical server. If you plan to run Oracle Linux as a guest on Oracle VM, VMWare or Hyper-v, you only have to pay for a single subscription per system, we do not charge per guest or per number of guests. In other words, you can run any number of Oracle Linux guests per physical server and count it as just a single subscription. (1) Oracle Linux Network Support - any number of sockets per server : $119 per server per year Network support does not offer support services. It provides access to the Unbreakable Linux Network and also offers full indemnification for Oracle Linux. (2) Oracle Linux Basic Limited Support - 1- or 2 socket servers : $499 per server per year This subscription provides 24x7 support services, access to the Unbreakable Linux Network and the Oracle Support portal, indemnification, use of Oracle Clusterware for Linux HA and use of Oracle Enterprise Manager Cloud control for Linux OS management. It includes ocfs2 as a clustered filesystem. (3) Oracle Linux Basic Support - more than 2 socket server (4, or 8 or more) : $1199 per server per year This subscription provides 24x7 support services, access to the Unbreakable Linux Network and the Oracle Support portal, indemnification, use of Oracle Clusterware for Linux HA and use of Oracle Enterprise Manager Cloud control for Linux OS management. It includes ocfs2 as a clustered filesystem (4) Oracle Linux Premier Limited Support - 1- or 2 socket servers : $1399 per server per year This subscription provides 24x7 support services, access to the Unbreakable Linux Network and the Oracle Support portal, indemnification, use of Oracle Clusterware for Linux HA and use of Oracle Enterprise Manager Cloud control for Linux OS management, XFS filesystem support. It also offers Oracle Lifetime support, backporting of patches for critical customers in previous versions of package and ksplice zero-downtime updates. (5) Oracle Linux Premier Support - more than 2 socket servers : $2299 per server per year This subscription provides 24x7 support services, access to the Unbreakable Linux Network and the Oracle Support portal, indemnification, use of Oracle Clusterware for Linux HA and use of Oracle Enterprise Manager Cloud control for Linux OS management, XFS filesystem support. It also offers Oracle Lifetime support, backporting of patches for critical customers in previous versions of package and ksplice zero-downtime updates. (6) Freely available Oracle Linux - any number of sockets You can freely download Oracle Linux, install it on any number of servers and use it for any reason, without support, without right to use of these extra features like Oracle Clusterware or ksplice, without indemnification. However, you do have full access to all errata as well. Need support? then use options (1)..(5) So that's it. Count number of 2 socket boxes, more than 2 socket boxes, decide on basic or premier support level and you are done. You don't have to worry about different levels based on how many virtual instance you deploy or want to deploy. A very simple menu of choices. We offer, inclusive, Linux OS clusterware, Linux OS Management, provisioning and monitoring, cluster filesystem (ocfs), high performance filesystem (xfs), dtrace, ksplice, ofed (infiniband stack for high performance networking). No separate add-on menus. NOTE : socket/cpu can have any number of cores. So whether you have a 4,6,8,10 or 12 core CPU doesn't matter, we count the number of physical CPUs.

    Read the article

< Previous Page | 110 111 112 113 114 115 116 117 118 119 120 121  | Next Page >