Search Results

Search found 33443 results on 1338 pages for 'reply list'.

Page 30/1338 | < Previous Page | 26 27 28 29 30 31 32 33 34 35 36 37  | Next Page >

• Should xml represent a set or a list?

  - by sixtyfootersdude

  I always think of xml like a set data structure. Ie: <class> <person>john</person> <person>sarah</person> </class> Is equivalent to: <class> <person>sarah</person> <person>john</person> </class> Question One: Are these two things logicly equivalant? Are you allowed to make things like this in xml? <methodCall> <param>happy</param> <param>sad</param> </methodCall> Or do you need to do it like this: <methodCall> <param arg="1">happy</param> <param arg="2">sad</param> </methodCall> Question Two: Are these two things logically equivalent? Question Three: Is xml usually treated like a set or a list?

  Read the article

• adding elements in to the doubly linked list

  - by user329820

  Hi this is my code for main class and doubly linked class and node class but when I run the program ,in the concole will show this"datastructureproject.DoublyLinkedList@19ee1ac" instead of the random numbers .please help me thanks! main class: public class Main { public static int getRandomNumber(double min, double max) { Random random = new Random(); return (int) (random.nextDouble() * (max - min) + min); } public static void main(String[] args) { int j; int i = 0; i = getRandomNumber(10, 10000); DoublyLinkedList listOne = new DoublyLinkedList(); for (j = 0; j <= i / 2; j++) { listOne.add(getRandomNumber(10, 10000)); } System.out.println(listOne); } } doubly linked list class: public class DoublyLinkedList { private Node head ; private Node tail; private long size = 0; public DoublyLinkedList() { head= new Node(0, null, null); tail = new Node(0, head, null); } public void add(int i){ head.setValue(i); Node newNode = new Node(); head.setNext(newNode); newNode.setPrev(head); newNode = head; } } and the node class is like the class that you have seen before (Node prev,Node next,int value)

  Read the article

• Linked List Inserting strings in alphabetical order

  - by user69514

  I have a linked list where each node contains a string and a count. my insert method needs to inset a new node in alphabetical order based on the string. if there is a node with the same string, then i increment the count. the problem is that my method is not inserting in alphabetical order public Node findIsertionPoint(Node head, Node node){ if( head == null) return null; Node curr = head; while( curr != null){ if( curr.getValue().compareTo(node.getValue()) == 0) return curr; else if( curr.getNext() == null || curr.getNext().getValue().compareTo(node.getValue()) > 0) return curr; else curr = curr.getNext(); } return null; } public void insert(Node node){ Node newNode = node; Node insertPoint = this.findIsertionPoint(this.head, node); if( insertPoint == null) this.head = newNode; else{ if( insertPoint.getValue().compareTo(node.getValue()) == 0) insertPoint.getItem().incrementCount(); else{ newNode.setNext(insertPoint.getNext()); insertPoint.setNext(newNode); } } count++; }

  Read the article

• Deleting first and last element of a linked list in C

  - by LuckySlevin

  struct person { int age; char name[100]; struct person *next; }; void delfirst(struct person **p)// For deleting the beginning { struct person *tmp,*m; m = (*p); tmp = (*p)->next; free(m); return; } void delend(struct person **p)// For deleting the end { struct person *tmp,*m; tmp=*p; while(tmp->next!=NULL) { tmp=tmp->next; } m->next=tmp; free(tmp); m->next = NULL; return; } I'm looking for two seperate functions to delete the first and last elements of a linked list. Here is what i tried. What do you suggest? Especially deleting first is so problematic for me.

  Read the article

• Cleaner method for list comprehension clean-up

  - by Dan McGrath

  This relates to my previous question: Converting from nested lists to a delimited string I have an external service that sends data to us in a delimited string format. It is lists of items, up to 3 levels deep. Level 1 is delimited by '|'. Level 2 is delimited by ';' and level 3 is delimited by ','. Each level or element can have 0 or more items. An simplified example is: a,b;c,d|e||f,g|h;; We have a function that converts this to nested lists which is how it is manipulated in Python. def dyn_to_lists(dyn): return [[[c for c in b.split(',')] for b in a.split(';')] for a in dyn.split('|')] For the example above, this function results in the following: >>> dyn = "a,b;c,d|e||f,g|h;;" >>> print (dyn_to_lists(dyn)) [[['a', 'b'], ['c', 'd']], [['e']], [['']], [['f', 'g']], [['h'], [''], ['']]] For lists, at any level, with only one item, we want it as a scalar rather than a 1 item list. For lists that are empty, we want them as just an empty string. I've came up with this function, which does work: def dyn_to_min_lists(dyn): def compress(x): return "" if len(x) == 0 else x if len(x) != 1 else x[0] return compress([compress([compress([item for item in mv.split(',')]) for mv in attr.split(';')]) for attr in dyn.split('|')]) Using this function and using the example above, it returns: [[['a', 'b'], ['c', 'd']], 'e', '', ['f', 'g'], ['h', '', '']] Being new to Python, I'm not confident this is the best way to do it. Are there any cleaner ways to handle this? This will potentially have large amounts of data passing through it, are there any more efficient/scalable ways to achieve this?

  Read the article

• Python - creating a list with 2 characteristics bug

  - by user2733911

  The goal is to create a list of 99 elements. All elements must be 1s or 0s. The first element must be a 1. There must be 7 1s in total. import random import math import time # constants determined through testing generation_constant = 0.96 def generate_candidate(): coin_vector = [] coin_vector.append(1) for i in range(0, 99): random_value = random.random() if (random_value > generation_constant): coin_vector.append(1) else: coin_vector.append(0) return coin_vector def validate_candidate(vector): vector_sum = sum(vector) sum_test = False if (vector_sum == 7): sum_test = True first_slot = vector[0] first_test = False if (first_slot == 1): first_test = True return (sum_test and first_test) vector1 = generate_candidate() while (validate_candidate(vector1) == False): vector1 = generate_candidate() print vector1, sum(vector1), validate_candidate(vector1) Most of the time, the output is correct, saying something like [1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0] 7 True but sometimes, the output is: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 2 False What exactly am I doing wrong?

  Read the article

• Apache gives empty reply

  - by Jorge Bernal

  It happens randomly, and only on moodle installations. Apache don't add a line in the logs when this happens, and I don't know where to look. koke@escher:~/Code/eboxhq/moodle[master]$ curl -I http://training.ebox-technologies.com/login/signup.php?course=WNA001 curl: (52) Empty reply from server koke@escher:~/Code/eboxhq/moodle[master]$ curl -I http://training.ebox-technologies.com/login/signup.php?course=WNA001 HTTP/1.1 200 OK The apache conf is quite straightforward and works perfectly in the other vhosts <VirtualHost *:80> ServerAdmin [email protected] DocumentRoot /srv/apache/training.ebox-technologies.com/htdocs ServerName training.eboxhq.com ErrorLog /var/log/apache2/training.ebox-technologies.com-error.log CustomLog /var/log/apache2/training.ebox-technologies.com-access.log combined <FilesMatch "\.(ico|gif|jpe?g|png|js|css)$"> ExpiresActive On ExpiresDefault "access plus 1 week" Header add Cache-Control public </FilesMatch> </VirtualHost> Using apache 2.2.9 php 5.2.6 and moodle 1.9.5+ (Build: 20090722) Any ideas welcome :)

  Read the article

• Domain of sender address does not resolve (in reply to MAIL FROM command)

  - by horen

  When sending out emails with postfix I sometimes get this error: 451 #4.1.8 Domain of sender address <[email protected]> does not resolve (in reply to MAIL FROM command) The domain mydomain.tld is resolvable though, meaning A, MX, PTR records are set properly. However, the sending server does have a different domain anotherdomain.tld but it is allowed to send emails from mydomain.tld since I set the MX records of mydomain.tld to anotherdomain.tld. The envelope from of the problematic emails is [email protected]. Is there some other dns entry I have to set? Or how else could I solve the problem? (I would like to keep the server structure though)

  Read the article

• NSD reply from unexpected source

  - by Ximik

  I have server with NSD. There are MAIN_IP and ADD_IP. When I try to get IP of my site from server I have right output dig @localhost my_site.com But when I try to make this from my PC, I have dig @my_ns_server.com my_site.com ;; reply from unexpected source: MAIN_IP#53, expected ADD_IP#53 (ADD_IP is IP of my_ns_server.com) What should I do? UPD: My interfaces conf auto eth2 allow-hotplug eth2 iface eth2 inet static address xxx.xxx.xxx.234 netmask 255.255.255.252 network xxx.xxx.xxx.232 broadcast xxx.xxx.xxx.235 gateway xxx.xxx.xxx.233 dns-nameservers MY_ISP_IP dns-search MY_ISP_DOMAIN auto eth2:0 iface eth2:0 inet static address xxx.xxx.xxx.124 netmask 255.255.255.0 xxx.xxx.xxx is the same for all IPs

  Read the article

• Making Thunderbird auto-add SMTP identities whenever I reply

  - by 0xC0000022L

  How can I teach Thunderbird to automatically add an SMTP identity whenever I reply to an email directed to <whatever>@<mydomain>? So if an SMTP is configured for <mydomain> but no identity exists for <whatever>@<mydomain>, how can I make Thunderbird dynamically recognize this and add it. Currently I have to manually add the identity every single time, but I would prefer it to to be added ad-hoc. As long as Thunderbird was configured to know about the SMTP serving <mydomain> this should be trivial, but I couldn't find an option. An add-on or something like a catch-all/wildcard identity would also do as long as it doesn't require manually setting up a new identity every time.

  Read the article

• print a linear linked list into a table

  - by user1796970

  I am attempting to print some values i have stored into a LLL into a readable table. The data i have stored is the following : DEBBIE STARR F 3 W 1000.00 JOAN JACOBUS F 9 W 925.00 DAVID RENN M 3 H 4.75 ALBERT CAHANA M 3 H 18.75 DOUGLAS SHEER M 5 W 250.00 SHARI BUCHMAN F 9 W 325.00 SARA JONES F 1 H 7.50 RICKY MOFSEN M 6 H 12.50 JEAN BRENNAN F 6 H 5.40 JAMIE MICHAELS F 8 W 150.00 i have stored each firstname, lastname, gender, tenure, payrate, and salary into their own List. And would like to be able to print them out in the same format that they are viewed on the text file i read them in from. i have messed around with a few methods that allow me to traverse and print the Lists, but i end up with ugly output. . . here is my code for the storage of the text file and the format i would like to print out: public class Payroll { private LineWriter lw; private ObjectList output; ListNode input; private ObjectList firstname, lastname, gender, tenure, rate, salary; public Payroll(LineWriter lw) { this.lw = lw; this.firstname = new ObjectList(); this.lastname = new ObjectList(); this.gender = new ObjectList(); this.tenure = new ObjectList(); this.rate = new ObjectList(); this.salary = new ObjectList(); this.output = new ObjectList(); this.input = new ListNode(); } public void readfile() { File file = new File("payfile.txt"); try{ Scanner scanner = new Scanner(file); while(scanner.hasNextLine()) { String line = scanner.nextLine(); Scanner lineScanner = new Scanner(line); lineScanner.useDelimiter("\\s+"); while(lineScanner.hasNext()) { firstname.insert1(lineScanner.next()); lastname.insert1(lineScanner.next()); gender.insert1(lineScanner.next()); tenure.insert1(lineScanner.next()); rate.insert1(lineScanner.next()); salary.insert1(lineScanner.next()); } } }catch(FileNotFoundException e) {e.printStackTrace();} } public void printer(LineWriter lw) { String msg = " FirstName " + " LastName " + " Gender " + " Tenure " + " Pay Rate " + " Salary "; output.insert1(msg); System.out.println(output.getFirst()); System.out.println(" " + firstname.getFirst() + " " + lastname.getFirst() + "\t" + gender.getFirst() + "\t" + tenure.getFirst() + "\t" + rate.getFirst() + "\t" + salary.getFirst()); } }

  Read the article

• WPF, C# - Making Intellisense/Autocomplete list, fastest way to filter list of strings

  - by user559548

  Hello everyone, I'm writing an Intellisense/Autocomplete like the one you find in Visual Studio. It's all fine up until when the list contains probably 2000+ items. I'm using a simple LINQ statement for doing the filtering: var filterCollection = from s in listCollection where s.FilterValue.IndexOf(currentWord, StringComparison.OrdinalIgnoreCase) >= 0 orderby s.FilterValue select s; I then assign this collection to a WPF Listbox's ItemSource, and that's the end of it, works fine. Noting that, the Listbox is also virtualised as well, so there will only be at most 7-8 visual elements in memory and in the visual tree. However the caveat right now is that, when the user types extremely fast in the richtextbox, and on every key up I execute the filtering + binding, there's this semi-race condition, or out of sync filtering, like the first key stroke's filtering could still be doing it's filtering or binding work, while the fourth key stroke is also doing the same. I know I could put in a delay before applying the filter, but I'm trying to achieve a seamless filtering much like the one in Visual Studio. I'm not sure where my problem exactly lies, so I'm also attributing it to IndexOf's string operation, or perhaps my list of string's could be optimised in some kind of index, that could speed up searching. Any suggestions of code samples are much welcomed. Thanks.

  Read the article

• Double Linked List Insertion Sorting Bug

  - by house

  Hello, I have implemented an insertion sort in a double link list (highest to lowest) from a file of 10,000 ints, and output to file in reverse order. To my knowledge I have implemented such a program, however I noticed in the ouput file, a single number is out of place. Every other number is in correct order. The number out of place is a repeated number, but the other repeats of this number are in correct order. Its just strange how this number is incorrectly placed. Also the unsorted number is only 6 places out of sync. I have looked through my program for days now with no idea where the problem lies, so I turn to you for help. Below is the code in question, (side note: can my question be deleted by myself? rather my colleges dont thieve my code, if not how can it be deleted?) void DLLIntStorage::insertBefore(int inValue, node *nodeB) { node *newNode; newNode = new node(); newNode->prev = nodeB->prev; newNode->next = nodeB; newNode->value = inValue; if(nodeB->prev==NULL) { this->front = newNode; } else { nodeB->prev->next = newNode; } nodeB->prev = newNode; } void DLLIntStorage::insertAfter(int inValue, node *nodeB) { node *newNode; newNode = new node(); newNode->next = nodeB->next; newNode->prev = nodeB; newNode->value = inValue; if(nodeB->next == NULL) { this->back = newNode; } else { nodeB->next->prev = newNode; } nodeB->next = newNode; } void DLLIntStorage::insertFront(int inValue) { node *newNode; if(this->front == NULL) { newNode = new node(); this->front = newNode; this->back = newNode; newNode->prev = NULL; newNode->next = NULL; newNode->value = inValue; } else { insertBefore(inValue, this->front); } } void DLLIntStorage::insertBack(int inValue) { if(this->back == NULL) { insertFront(inValue); } else { insertAfter(inValue, this->back); } } ifstream& operator>> (ifstream &in, DLLIntStorage &obj) { int readInt, counter = 0; while(!in.eof()) { if(counter==dataLength) //stops at 10,000 { break; } in >> readInt; if(obj.front != NULL ) { obj.insertion(readInt); } else { obj.insertBack(readInt); } counter++; } return in; } void DLLIntStorage::insertion(int inValue) { node* temp; temp = this->front; if(temp->value >= inValue) { insertFront(inValue); return; } else { while(temp->next!=NULL && temp!=this->back) { if(temp->value >= inValue) { insertBefore(inValue, temp); return; } temp = temp->next; } } if(temp == this->back) { insertBack(inValue); } } Thankyou for your time.

  Read the article

• Linked list in C

  - by ScReYm0

  I am still new at lists in C and i got a big problem... First i wanna show you my code for inserting item to the list: void input_books_info(int number_of_books, BOOK *current) { int i; for(i = 0; i < number_of_books; i++) { while(current->next != NULL) current = current->next; current->next = (BOOK *)malloc(sizeof(BOOK)); printf_s("%d book catalog number: ", i + 1); scanf_s("%s", &current->next->catalog_number , 20); printf_s("%d book title: ", i + 1); scanf_s("%s", current->next->title ,80); printf_s("%d book author: ", i + 1); scanf_s("%s", current->next->author ,40); printf_s("%d book publisher: ", i+1); scanf_s("%s", current->next->publisher,80); printf_s("%d book price: ", i + 1); scanf_s("%f", &current->next->price, 5); printf_s("%d book year published: ", i + 1); scanf_s("%d", &current->next->year_published, 5); current->next->next = NULL; printf_s("\n\n"); } } And this is my main function: void main (void) { int number_of_books, t = 1; char book_catalog_number[STRMAX]; char book_title[STRMAX]; char book_author[STRMAX]; char reading_file[STRMAX]; char saving_file[STRMAX]; first = malloc(sizeof(BOOK)); first->next = NULL; /* printf_s("Enter file name: "); gets(saving_file); first->next = book_open(first, saving_file); */ while(t) { char m; printf_s("1. Input \n0. Exit \n\n"); printf_s("Choose operation: "); m = getch(); switch(m) { case '1': printf_s("\ninput number of books: "); scanf_s("%d", &number_of_books); input_books_info(number_of_books, first); printf_s("\n"); break; default: printf_s("\nNo entry found!\n\n\n\n\n"); break; } } } and last maybe here is the problem the printing function: void print_books_info(BOOK *current) { while(current->next != NULL && current != NULL) { printf_s("%s, ", current->next->catalog_number); printf_s("%s, ", current->author); printf_s("%s, ", current->next->title); printf_s("%s, ", current->next->author); printf_s("%s, ", current->next->publisher); printf_s("%.2f, ", current->next->price); printf_s("%d", current->next->year_published); printf_s("\n\n"); current = current->next; } } And my problem is that, when i run the app, program is moving good. But when I start the app, the program is storing only first input of data second and third are lost ... Can you help me to figure out it... ???

  Read the article

• Unable to add separator in list view

  - by Suru

  This is my code @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.email_list_main); emailResults = new ArrayList<GetEmailFromDatabase>(); //int[] colors = {0,0xFFFF0000,0}; //getListView().setDivider(new GradientDrawable(Orientation.RIGHT_LEFT, colors)); //getListView().setDividerHeight(2); emailListFeedAdapter = new EmailListFeedAdapter(this, R.layout.email_listview_row, emailResults); setListAdapter(this.emailListFeedAdapter); getResults(); if(emailResults != null && emailResults.size() > -1){ emailListFeedAdapter.notifyDataSetChanged(); for(int i=0;i< emailResults.size();i++){ try { Here I getting email Sent date emailListFeedAdapter.add( emailResults.get(i)); datetime_text1 = emailResults.get(i).getDate(); formatter1 = new SimpleDateFormat(); formatter1 = DateFormat.getDateInstance((DateFormat.MEDIUM)); Calendar currentDate1 = Calendar.getInstance(); Item_Date1 = formatter1.parse(datetime_text1); current_Date1 = formatter1.format(currentDate1.getTime()); current_System_Date1 = formatter1.parse(current_Date1); currentDate1.add(Calendar.DATE, -1); yesterdaydate = formatter1.format(currentDate1.getTime()); yeaterday_Date = formatter1.parse(yesterdaydate); currentDate1.add(Calendar.DATE, -2); threeDaysback = formatter1.format(currentDate1.getTime()); Three_Days_Back = formatter1.parse(threeDaysback); Here I am comparing current date with list view item date, and here is my problem, dates are matching but it is not entering in if condition I tried in so many ways but nothing worked the code for separator is bellow. if(Item_Date.compareTo(current_System_Date)==0){ if(index1){ emailListFeedAdapter.addSeparatorItem("SEPARATOR"); //i--; index1=false; } } else if(yeaterday_Date.compareTo(Item_Date1)==0){ if(index2){ emailListFeedAdapter.addSeparatorItem("SEPARATOR"); //i--; index2 = false; } } else if(Item_Date1.compareTo(Three_Days_Back)==0){ if(index3){ emailListFeedAdapter.addSeparatorItem("SEPARATOR"); //i--; index3 = false; } } } catch (ParseException e) { // TODO Auto-generated catch block e.printStackTrace(); } } } } In EmailListFeedAdapter private TreeSet<Integer> mSeparatorsSet = new TreeSet<Integer>(); public void addSeparatorItem(final String item) { //itemss.add(emailResults.get(0)); // save separator position mSeparatorsSet.add(itemss.size() - 1); notifyDataSetChanged(); } @Override public int getItemViewType(int position) { return mSeparatorsSet.contains(position) ? TYPE_SEPARATOR : TYPE_ITEM; } holder = new ViewHolder(); switch (type) { case TYPE_ITEM: emailView= inflater.inflate(R.layout.email_listview_row, null); break; case TYPE_SEPARATOR: emailView= inflater.inflate(R.layout.item2, null); holder.textView = (TextView)emailView.findViewById(R.id.textSeparator); emailView.setTag(holder); holder.textView.setText("SEPARATOR"); break; } Here is ViewHolder class public static class ViewHolder { public TextView textView; } if anybody knows then please tell me where I am doing wrong. Thanx

  Read the article

• Restart list numbering in word for each new list created

  - by Feena

  Hi, I am exporting content from a jsp page into MS Word using javascript. When the user is in Word there is a table with 10 rows and 2 columns, A & B. The user creates an ordered list in row 1, column A like this: 1 dog 2 cat 3 mouse if the user then creates a second list in row 1 column B is turns out like this: 4 car 5 truck 6 bike instead of: 1 car 2 truck 3 bike Word is set up to continue the numbering in lists from prior lists automatically. I know this can be reset easily but the users dont want to have to do this. They want the numbering of any potential lists created to restarted at 1. when the document is exported into Word and opened in front of them. So this must be set up in the javasctipt code or using a style or something prior to getting into Word. This is what I dont know how to do. Any help is much appreciated. Thanks, Feena.

  Read the article

• Django: getting the list of related records for a list of objects

  - by Silver Light

  Hello! I have two models related one-to many: class Person(models.Model): name = models.CharField(max_length=255); surname = models.CharField(max_length=255); age = models.IntegerField(); class Dog(models.Model): name = models.CharField(max_length=255); owner = models.ForeignKey('Person'); I want to output a list of persons below each person a list of dogs he has. Here's how I can do it: in view: persons = Person.objects.all()[0:100]; in template: {% for p in persons %} {{ p.name }} has dogs:<br /> {% for d in persons.dog_set.all %} - {{ d.name }}<br /> {% endfor %} {% endfor %} But if I do it like that, Django will execute 101 SQL queries which is very inefficient. I tried to make a custom manager, which will get all the persons, then all the dogs and links them in python, but then I can't use paginator (my another question: http://stackoverflow.com/questions/2532475/django-paginator-raw-sql-query ) and it looks quite ugly. Is there a more graceful way doing this?

  Read the article

• Trying to get these list items to display inline

  - by Joel

  I have several unordered lists that I want to display like this: <ul> <li><img></li> <li><p></li> //inline </ul> //linebreak <ul> <li><img></li> <li><p></li> //inline </ul> ...etc I'm not able to get the li items to be inline with eachother. They are stacking vertically. I have stripped away most styling but still can't figure out what I'm doing wrong: html: <ul class="instrument"> <li class="imagebox"><img src="/images/matepe.jpg" width="247" height="228" alt="Matepe" /></li> <li class="textbox"><p>The matepe is a 24 key instrument that is played by the Kore-Kore people in North-Eastern Zimbabwe and Mozambique. It utilizes four fingers-each playing an individual melody. These melodies also interwieve to create resultant melodies that can be manipulated thru accenting different fingers. The matepe is used in Rattletree as the bridge from the physical world to the spirit world. The matepe is used in the Kore-Kore culture to summon the Mhondoro spirits which are thought to be able to communicate directly with Mwari (God) without the need of an intermediary.</p></li> </ul> <ul class="instrument"> <li class="imagebox"><img src="/images/soprano_little.jpg" border="0" width="247" height="170" alt="Soprano" /></li> <li class="textbox"><p>The highest voice of the Rattletree Marimba orchestra is the Soprano marimba. The soprano is used to whip up the energy on the dancefloor and help people reach ecstatic state with it's high and clear singing voice. The range of these sopranos goes much lower than 'typical' Zimbabwean style sopranos. The sopranos play the range of the right hand of the matepe and go two notes higher and five notes lower. Rattletree uses two sopranos.</p></li> </ul> <ul class="instrument"> <li class="imagebox"><img src="/images/bari_little.jpg" border="0" width="247" height="170" alt="Baritone" /></li> <li class="textbox"><p>The Baritone is the next lower voice in the orchestra. The bari is where the funk is. Generally bubbling over the Bass line, the baritone creates the syncopations and polyrhythms that messes with the 'minds' of the dancers and helps seperate the listener from the physical realm of thought. The range of the baritone covers the full range of the left hand side of the matepe.</p></li> </ul> <ul class="instrument"> <li class="imagebox"><img src="/images/darren_littlebass.jpg" border="0" width="247" height="195" alt="Bass"/><strong>Bass Marimba</strong></li> <li class="textbox"><p>The towering Bass Marimba is the foundation of the Rattletree Marimba sound. Putting out frequencies as low as 22hZ, the bass creates the drive that gets the dancefloor moving. It is 5.5' tall, 9' long, and 4' deep. It is played by standing on a platform and struck with mallets that have lacross-ball size heads (they are actually made with rubber dog balls). The Bass marimba's range covers the lowest five notes of the matepe and goes another five notes lower.</p></li> </ul> <ul class="instrument"> <li class="imagebox"><img src="/images/wayne_little.jpg" border="0" width="247" height="177" alt="Drums"/><strong>Drumset</strong></li> <li class="textbox"><p>All the intricate polyrhythms are held together tastefully with the drumset. The drums provides the consistancy and grounding that the dancers need to keep going all night. While the steady kick and high-hat provide that grounding function, the toms and snare and allowed to be another voice in the poylrhythmic texture-helping the dancers abandon the concept of a "one" within this cyclical music.</p></li> </ul> css: ul.instrument { text-align:left; display:inline; } ul.instrument li { list-style-type: none; } li.imagebox { } li.textbox { } li.textbox p{ width: 247px; }

  Read the article

• Link List Implementation Help - Visual C++

  - by Greenhouse Gases

  Hi there I'm trying to implement a link list which stores the city name (though you will see this commented out as I need to resolve the issue of not being able to use string and needing to use a primitive data type instead during the declaration), longitude, latitude and of course a pointer to the next node in the chain. I am new to the Visual C++ environment and my brain is somewhat scrambled after coding for several straight hours today so I wondered if anyone could help resolve the 2 errors I am getting (ignore the #include syntax as I had to change them to avoid the browser interpreting html!): 1U08221.obj : error LNK2028: unresolved token (0A000298) "public: __thiscall Locations::Locations(void)" (??0Locations@@$$FQAE@XZ) referenced in function "int __clrcall main(cli::array^)" (?main@@$$HYMHP$01AP$AAVString@System@@@Z) 1U08221.obj : error LNK2019: unresolved external symbol "public: __thiscall Locations::Locations(void)" (??0Locations@@$$FQAE@XZ) referenced in function "int __clrcall main(cli::array^)" (?main@@$$HYMHP$01AP$AAVString@System@@@Z) The code for my header file is here: include string struct locationNode { //char[10] nodeCityName; double nodeLati; double nodeLongi; locationNode* Next; }; class Locations { private: int size; public: Locations(); // constructor for the class locationNode* Head; int Add(locationNode* Item); }; and here is the code for the file containing the main method: // U08221.cpp : main project file. include "stdafx.h" include "Locations.h" include iostream include string using namespace std; int n = 0; int x; string cityNameInput; bool acceptedInput = false; int Locations::Add(locationNode *NewItem) { locationNode *Sample = new locationNode; Sample = NewItem; Sample-Next = Head; Head = Sample; return size++; } void CorrectCase(string name) // Correct upper and lower case letters of input { x = name.size(); int firstLetVal = name[0], letVal; n = 1; // variable for name index from second letter onwards if((name[0] 90) && (name[0] < 123)) // First letter is lower case { firstLetVal = firstLetVal - 32; // Capitalise first letter name[0] = firstLetVal; } while(n <= x - 1) { if((name[n] = 65) && (name[n] <= 90)) { letVal = name[n] + 32; name[n] = letVal; } n++; } cityNameInput = name; } void nameValidation(string name) { n = 0; // start from first letter x = name.size(); while(!acceptedInput) { if((name[n] = 65) && (name[n] <= 122)) // is in the range of letters { while(n <= x - 1) { while((name[n] =91) && (name[n] <=97)) // ERROR!! { cout << "Please enter a valid city name" << endl; cin name; } n++; } } else { cout << "Please enter a valid city name" << endl; cin name; } if(n <= x - 1) { acceptedInput = true; } } cityNameInput = name; } int main(array ^args) { cout << "Enter a city name" << endl; cin cityNameInput; nameValidation(cityNameInput); // check is made up of valid characters CorrectCase(cityNameInput); // corrects name to standard format of capitalised first letter, and lower case subsequent letters cout << cityNameInput; cin cityNameInput; Locations::Locations(); Locations *Parts = new Locations(); locationNode *Part; Part = new locationNode; //Part-nodeCityName = "London"; Part-nodeLati = 87; Part-nodeLongi = 80; Parts-Add(Part); } I am familiar with the concepts but somewhat inexperienced with OOP so am making some silly errors that you can never find when you've stared at something too long. Any help you can offer will be appreciated! Thanks

  Read the article

• list images from directory by function

  - by osc2nuke

  i'm using this function: function getmyimages($qid){ $imgdir = 'modules/Projects/uploaded_project_images/'. $qid .''; // the directory, where your images are stored $allowed_types = array('png','jpg','jpeg','gif'); // list of filetypes you want to show $dimg = opendir($imgdir); while($imgfile = readdir($dimg)) { if(in_array(strtolower(substr($imgfile,-3)),$allowed_types)) { $a_img[] = $imgfile; sort($a_img); reset ($a_img); } } $totimg = count($a_img); // total image number for($x=0; $x < $totimg; $x++) { $size = getimagesize($imgdir.'/'.$a_img[$x]); // do whatever $halfwidth = ceil($size[0]/2); $halfheight = ceil($size[1]/2); $mytest = 'name: '.$a_img[$x].' width: '.$size[0].' height: '.$size[1].'<br /><a href="'. $imgdir .'/'.$a_img[$x].'">'. $a_img[$x]. '</a>'; } return $mytest; } And i call this function between a while row as: $sql_select = $db->sql_query('SELECT * from '.$prefix.'_projects WHERE topic=\''.$cid.'\''); OpenTable(); while ($row2 = $db->sql_fetchrow($sql_select)){ $qid = $row2['qid']; $project_query = $db->sql_query('SELECT p.uid, p.uname, p.subject, p.story, p.storyext, p.date, p.topic, p.pdate, p.materials, p.bidoptions, p.projectduration, pd.id_duration, pm.material_id, pbo.bidid, pc.cid FROM ' . $prefix . '_projects p, ' . $prefix . '_projects_duration pd, ' . $prefix . '_project_materials pm, ' . $prefix . '_project_bid_options pbo, ' . $prefix . '_project_categories pc WHERE p.topic=\''.$cid.'\' and p.qid=\''.$qid.'\' and p.bidoptions=pbo.bidid and p.materials=pm.material_id and p.projectduration=pd.id_duration'); while ($project_row = $db->sql_fetchrow($project_query)) { //$qid = $project_row['qid']; $uid = $project_row['uid']; $uname = $project_row['uname']; $subject = $project_row['subject']; $story = $project_row['story']; $storyext = $project_row['storyext']; $date = $project_row['date']; $topic = $project_row['topic']; $pdate = $project_row['pdate']; $materials = $project_row['materials']; $bidoptions = $project_row['bidoptions']; $projectduration = $project_row['projectduration']; //Get the topic name $topic_query = $db->sql_query('SELECT cid,title from '.$prefix.'_project_categories WHERE cid =\''.$cid.'\''); while ($topic_row = $db->sql_fetchrow($topic_query)) { $topic_id = $topic_row['cid']; $topic_title = $topic_row['title']; } //Get the material text $material_query = $db->sql_query('SELECT material_id,material_name from '.$prefix.'_project_materials WHERE material_id =\''.$materials.'\''); while ($material_row = $db->sql_fetchrow($material_query)) { $material_id = $material_row['material_id']; $material_name = $material_row['material_name']; } //Get the bid methode $bid_query = $db->sql_query('SELECT bidid,bidname from '.$prefix.'_project_bid_options WHERE bidid =\''.$bidoptions.'\''); while ($bid_row = $db->sql_fetchrow($bid_query)) { $bidid = $bid_row['bidid']; $bidname = $bid_row['bidname']; } //Get the project duration $duration_query = $db->sql_query('SELECT id_duration,duration_value,duration_alias from '.$prefix.'_projects_duration WHERE id_duration =\''.$projectduration.'\''); while ($duration_row = $db->sql_fetchrow($duration_query)) { $id_duration = $duration_row['id_duration']; $duration_value = $duration_row['duration_value']; $duration_alias = $duration_row['duration_alias']; } } echo '<br/><b>id</b>--->' .$qid. '<br/><b>uid</b>--->' .$uid. '<br/><b>username</b>--->' .$uname. '<br/><b>subject</b>--->'.$subject. '<br/><b>story1</b>--->'.$story. '<br/><b>story2</b>--->'.$storyext. '<br/><b>postdate</b>--->'.$date. '<br/><b>categorie</b>--->'.$topic_title . '<br/><b>project start</b>--->'.$pdate. '<br/><b>materials</b>--->'.$material_name. '<br/><b>bid methode</b>--->'.$bidname. '<br/><b>project duration</b>--->'.$duration_alias.'<br /><br /><br/><b>image url</b>--->'.getmyimages($qid).'<br /><br />'; } CloseTable(); the result outputs only the "last" file from the directories. if i do a echo instead of a return $mytest; it read the whole directory but ruïns the output.

  Read the article

• SPSiteDataQuery Returns Only One List Type At A Time

  - by Brian Jackett

  The SPSiteDataQuery class in SharePoint 2007 is very powerful, but it has a few limitations.  One of these limitations that I ran into this morning (and caused hours of frustration) is that you can only return results from one list type at a time.  For example, if you are trying to query items from an out of the box custom list (list type = 100) and document library (list type = 101) you will only get items from the custom list (SPSiteDataQuery defaults to list type = 100.)  In my situation I was attempting to query multiple lists (created from custom list templates 10001 and 10002) each with their own content types. Solution     Since I am only able to return results from one list type at a time, I was forced to run my query twice with each time setting the ServerTemplate (translates to ListTemplateId if you are defining custom list templates) before executing the query.  Below is a snippet of the code to accomplish this. SPSiteDataQuery spDataQuery = new SPSiteDataQuery(); spDataQuery.Lists = "<Lists ServerTemplate='10001' />"; // ... set rest of properties for spDataQuery   var results = SPContext.Current.Web.GetSiteData(spDataQuery).AsEnumerable();   // only change to SPSiteDataQuery is Lists property for ServerTemplate attribute spDataQuery.Lists = "<Lists ServerTemplate='10002' />";   // re-execute query and concatenate results to existing entity results = results.Concat(SPContext.Current.Web.GetSiteData(spDataQuery).AsEnumerable());   Conclusion     Overall this isn’t an elegant solution, but it’s a workaround for a limitation with the SPSiteDataQuery.  I am now able to return data from multiple lists spread across various list templates.  I’d like to thank those who commented on this MSDN page that finally pointed out the limitation to me.  Also a thanks out to Mark Rackley for “name dropping” me in his latest article (which I humbly insist I don’t belong in such company)  as well as encouraging me to write up a quick post on this issue above despite my busy schedule.  Hopefully this post saves some of you from the frustrations I experienced this morning using the SPSiteDataQuery.  Until next time, Happy SharePoint’ing all.         -Frog Out   Links MSDN Article for SPSiteDataQuery http://msdn.microsoft.com/en-us/library/microsoft.sharepoint.spsitedataquery.lists.aspx

  Read the article

• Convert multi-dimensional list to a 1D list in Python

  - by Sayan

  Hello, A multidimensional list like l=[[1,2],[3,4]] could be converted to a 1D one by doing sum(l,[]). Can anybody please explain how that happens? Thanks, Sayan

  Read the article

• Why can I not convert from List<IScreen> to List<IRenderable> even when IScreen implements IRenderab

  - by Blub

  Pretty self explanatory question, glad for any answer. For me it doesn't make sense. I don't care about losing information. I just want to concatenate all my lists that inherit IRenderable somehow into a single list and then call IRenderable methods on it. I have a feeling that this is some basic rule of programming I forgot...

  Read the article

• TCP packets larger than 4 KB don't get a reply from Linux

  - by pts

  I'm running Linux 3.2.51 in a virtual machine (192.168.33.15). I'm sending Ethernet frames to it. I'm writing custom software trying to emulate a TCP peer, the other peer is Linux running in the virtual machine guest. I've noticed that TCP packets larger than about 4 KB are ignored (i.e. dropped without an ACK) by the Linux guest. If I decrease the packet size by 50 bytes, I get an ACK. I'm not sending new payload data until the Linux guest fully ACKs the previous one. I've increased ifconfig eth0 mtu 51000, and ping -c 1 -s 50000 goes through (from guest to my emulator) and the Linux guest gets a reply of the same size. I've also increased sysctl -w net.ipv4.tcp_rmem='70000 87380 87380 and tried with sysctl -w net.ipv4.tcp_mtu_probing=1 (and also =0). There is no IPv3 packet fragmentation, all packets have the DF flag set. It works the other way round: the Linux guest can send TCP packets of 6900 bytes of payload and my emulator understands them. This is very strange to me, because only TCP packets seem to be affected (large ICMP packets go through). Any idea what can be imposing this limit? Any idea how to do debug it in the Linux kernel? See the tcpdump -n -vv output below. tcpdump was run on the Linux guest. The last line is interesting: 4060 bytes of TCP payload is sent to the guest, and it doesn't get any reply packet from the Linux guest for half a minute. 14:59:32.000057 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [S], cksum 0x8da0 (correct), seq 10000000, win 14600, length 0 14:59:32.000086 IP (tos 0x10, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 44) 192.168.33.15.22 > 192.168.33.1.36522: Flags [S.], cksum 0xc37f (incorrect -> 0x5999), seq 1415680476, ack 10000001, win 19920, options [mss 9960], length 0 14:59:32.000218 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0xa752 (correct), ack 1, win 14600, length 0 14:59:32.000948 IP (tos 0x10, ttl 64, id 53777, offset 0, flags [DF], proto TCP (6), length 66) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], cksum 0xc395 (incorrect -> 0xfa01), seq 1:27, ack 1, win 19920, length 26 14:59:32.001575 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0xa738 (correct), ack 27, win 14600, length 0 14:59:32.001585 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 65) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], cksum 0x48d6 (correct), seq 1:26, ack 27, win 14600, length 25 14:59:32.001589 IP (tos 0x10, ttl 64, id 53778, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x9257), ack 26, win 19920, length 0 14:59:32.001680 IP (tos 0x10, ttl 64, id 53779, offset 0, flags [DF], proto TCP (6), length 496) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 27:483, ack 26, win 19920, length 456 14:59:32.001784 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0xa557 (correct), ack 483, win 14600, length 0 14:59:32.006367 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 1136) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 26:1122, ack 483, win 14600, length 1096 14:59:32.044150 IP (tos 0x10, ttl 64, id 53780, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x8c47), ack 1122, win 19920, length 0 14:59:32.045310 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 312) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 1122:1394, ack 483, win 14600, length 272 14:59:32.045322 IP (tos 0x10, ttl 64, id 53781, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x8b37), ack 1394, win 19920, length 0 14:59:32.925726 IP (tos 0x10, ttl 64, id 53782, offset 0, flags [DF], proto TCP (6), length 1112) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], seq 483:1555, ack 1394, win 19920, length 1072 14:59:32.925750 IP (tos 0x10, ttl 64, id 53784, offset 0, flags [DF], proto TCP (6), length 312) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 1555:1827, ack 1394, win 19920, length 272 14:59:32.927131 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x9bcf (correct), ack 1555, win 14600, length 0 14:59:32.927148 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x9abf (correct), ack 1827, win 14600, length 0 14:59:32.932248 IP (tos 0x10, ttl 64, id 53785, offset 0, flags [DF], proto TCP (6), length 56) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], cksum 0xc38b (incorrect -> 0xd247), seq 1827:1843, ack 1394, win 19920, length 16 14:59:32.932366 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x9aaf (correct), ack 1843, win 14600, length 0 14:59:32.964295 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 104) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 1394:1458, ack 1843, win 14600, length 64 14:59:32.964310 IP (tos 0x10, ttl 64, id 53786, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x85a7), ack 1458, win 19920, length 0 14:59:32.964561 IP (tos 0x10, ttl 64, id 53787, offset 0, flags [DF], proto TCP (6), length 88) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 1843:1891, ack 1458, win 19920, length 48 14:59:32.965185 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x9a3f (correct), ack 1891, win 14600, length 0 14:59:32.965196 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 104) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 1458:1522, ack 1891, win 14600, length 64 14:59:32.965233 IP (tos 0x10, ttl 64, id 53788, offset 0, flags [DF], proto TCP (6), length 88) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 1891:1939, ack 1522, win 19920, length 48 14:59:32.965970 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x99cf (correct), ack 1939, win 14600, length 0 14:59:32.965979 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 568) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 1522:2050, ack 1939, win 14600, length 528 14:59:32.966112 IP (tos 0x10, ttl 64, id 53789, offset 0, flags [DF], proto TCP (6), length 520) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 1939:2419, ack 2050, win 19920, length 480 14:59:32.970059 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x95df (correct), ack 2419, win 14600, length 0 14:59:32.970089 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 616) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 2050:2626, ack 2419, win 14600, length 576 14:59:32.981159 IP (tos 0x10, ttl 64, id 53790, offset 0, flags [DF], proto TCP (6), length 72) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], cksum 0xc39b (incorrect -> 0xa84f), seq 2419:2451, ack 2626, win 19920, length 32 14:59:32.982347 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x937f (correct), ack 2451, win 14600, length 0 14:59:32.982357 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 104) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 2626:2690, ack 2451, win 14600, length 64 14:59:32.982401 IP (tos 0x10, ttl 64, id 53791, offset 0, flags [DF], proto TCP (6), length 88) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 2451:2499, ack 2690, win 19920, length 48 14:59:32.982570 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x930f (correct), ack 2499, win 14600, length 0 14:59:32.982702 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 104) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 2690:2754, ack 2499, win 14600, length 64 14:59:33.020066 IP (tos 0x10, ttl 64, id 53792, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x7e07), ack 2754, win 19920, length 0 14:59:33.983503 IP (tos 0x10, ttl 64, id 53793, offset 0, flags [DF], proto TCP (6), length 72) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], cksum 0xc39b (incorrect -> 0x2aa7), seq 2499:2531, ack 2754, win 19920, length 32 14:59:33.983810 IP (tos 0x10, ttl 64, id 53794, offset 0, flags [DF], proto TCP (6), length 88) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 2531:2579, ack 2754, win 19920, length 48 14:59:33.984100 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x92af (correct), ack 2531, win 14600, length 0 14:59:33.984139 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x927f (correct), ack 2579, win 14600, length 0 14:59:34.022914 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 104) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 2754:2818, ack 2579, win 14600, length 64 14:59:34.022939 IP (tos 0x10, ttl 64, id 53795, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.15.22 > 192.168.33.1.36522: Flags [.], cksum 0xc37b (incorrect -> 0x7d77), ack 2818, win 19920, length 0 14:59:34.023554 IP (tos 0x10, ttl 64, id 53796, offset 0, flags [DF], proto TCP (6), length 88) 192.168.33.15.22 > 192.168.33.1.36522: Flags [P.], seq 2579:2627, ack 2818, win 19920, length 48 14:59:34.027571 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 40) 192.168.33.1.36522 > 192.168.33.15.22: Flags [.], cksum 0x920f (correct), ack 2627, win 14600, length 0 14:59:34.027603 IP (tos 0x0, ttl 64, id 0, offset 0, flags [DF], proto TCP (6), length 4100) 192.168.33.1.36522 > 192.168.33.15.22: Flags [P.], seq 2818:6878, ack 2627, win 14600, length 4060

  Read the article

• Autodiscover service seems to reply with User Principal Name instead of email address

  - by Jeff McJunkin

  After this latest round of Windows updates (on 1/11/11, in fact) my Exchange 2007 server of course rebooted. This may have had the side effect of making any changes I'd inadvertently made take effect. Since then, the Autodiscover service in Exchange 2007 from Outlook 2007 seems to reply with the User Principal Name ([email protected] instead of [email protected]). I'm specifically seeing this from within the "Test Email AutoConfiguration" tool in Outlook (the UPN appears in the first text box labeled "E-mail") and when creating a new profile in Outlook. If I disregard the UPN and instead fill in my email address, Autodiscover works as expected and I can connect without issue. I've confirmed using ADSI Edit that the SMTP email address is properly set for my users. I even went a bit crazy and set the UPN to the email address using ADSI Edit. I've re-installed the Client Access role on the server in question. Exchange server is Server 2008, 64-bit of course. Clients are mostly XP 32-bit, though the issue happens from a Windows 7 machine as well.

  Read the article

< Previous Page | 26 27 28 29 30 31 32 33 34 35 36 37  | Next Page >