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  • Flex List ItemRenderer with image looses BitmapData when scrolling

    - by Dominik
    Hi i have a mx:List with a DataProvider. This data Provider is a ArrayCollection if FotoItems public class FotoItem extends EventDispatcher { [Bindable] public var data:Bitmap; [Bindable] public var id:int; [Bindable] public var duration:Number; public function FotoItem(data:Bitmap, id:int, duration:Number, target:IEventDispatcher=null) { super(target); this.data = data; this.id = id; this.duration = duration; } } my itemRenderer looks like this: <?xml version="1.0" encoding="utf-8"?> <mx:VBox xmlns:fx="http://ns.adobe.com/mxml/2009" xmlns:s="library://ns.adobe.com/flex/spark" xmlns:mx="library://ns.adobe.com/flex/mx" > <fx:Script> <![CDATA[ import mx.collections.ArrayCollection; ]]> </fx:Script> <s:Label text="index"/> <mx:Image source="{data.data}" maxHeight="100" maxWidth="100"/> <s:Label text="Duration: {data.duration}ms"/> <s:Label text="ID: {data.id}"/> </mx:VBox> Now when i am scrolling then all images that leave the screen disappear :( When i take a look at the arrayCollection every item's BitmapData is null. Why is this the case?

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  • CSS Hover on parent list Item only

    - by Daniel O'Connor
    Hey Everyone, So I have some nested lists (only one level deep) and I'm running into trouble with the CSS :hover feature. I only want the hover to apply to the parent class, but I can't figure that one out. Here's my CSS <style type="text/css" media="screen"> .listblock li img { visibility: hidden; } .listblock li:hover img { visibility: visible; } </style> And here is a sample of one of the lists. <ul> <li>One <a href="#"><img src="img/basket.png" height="16" width="16" alt="Buy" class="buy" onClick="pageTracker._trackEvent('Outbound Links', 'Amazon');"/></a></li> <li>Two <a href="#"><img src="img/basket.png" height="16" width="16" class="buy" /></a> <ul> <li>Uno<a href="#"><img src="img/basket.png" height="16" width="16" class="buy" /></a></li> <li>Dos <a href="#"><img src="img/basket.png" height="16" width="16" class="buy" /></a></li> </ul> </li> <li>Three <a href="#"><img src="img/basket.png" height="16" width="16" alt="Buy" class="buy" onClick="pageTracker._trackEvent('Outbound Links', 'Amazon');"/></a></li> </ul> The problem is that the image in the Uno and Dos list items also hovers. :( Help please! Thanks a lot

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  • Excel CSV into Nested Dictionary; List Comprehensions

    - by victorhooi
    heya, I have a Excel CSV files with employee records in them. Something like this: mail,first_name,surname,employee_id,manager_id,telephone_number [email protected],john,smith,503422,503423,+65(2)3423-2433 [email protected],george,brown,503097,503098,+65(2)3423-9782 .... I'm using DictReader to put this into a nested dictionary: import csv gd_extract = csv.DictReader(open('filename 20100331 original.csv'), dialect='excel') employees = dict([(row['employee_id'], row) for row in gp_extract]) Is the above the proper way to do it - it does work, but is it the Right Way? Something more efficient? Also, the funny thing is, in IDLE, if I try to print out "employees" at the shell, it seems to cause IDLE to crash (there's approximately 1051 rows). 2. Remove employee_id from inner dict The second issue issue, I'm putting it into a dictionary indexed by employee_id, with the value as a nested dictionary of all the values - however, employee_id is also a key:value inside the nested dictionary, which is a bit redundant? Is there any way to exclude it from the inner dictionary? 3. Manipulate data in comprehension Thirdly, we need do some manipulations to the imported data - for example, all the phone numbers are in the wrong format, so we need to do some regex there. Also, we need to convert manager_id to an actual manager's name, and their email address. Most managers are in the same file, while others are in an external_contractors CSV, which is similar but not quite the same format - I can import that to a separate dict though. Are these two items things that can be done within the single list comprehension, or should I use a for loop? Or does multiple comprehensions work? (sample code would be really awesome here). Or is there a smarter way in Python do it? Cheers, Victor

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  • Linked List pop() function

    - by JKid314159
    Consider the following list: [LinkNode * head -- LinkNode * node1 -- LinkNode * node2] I'm creating a stack of FIFO. I Call pop() which I want to pop node1. LinkNode::LinkNode(int numIn) { this->numIn = numIn; next = null; } . . . int LinkNode::pop() { Link * temp = head->next; head = temp->next; int popped = head->nodeNum; delete temp; Return numOut; Question: 1) head should be a pointer or a LinkNode *? 2) Link * temp is created on the call stack and when pop finishes doesn't temp delete automatically? 3) My major confusion is on what is the value of temp-next? Does this point to node1.next which equals node2? Appreciate your help? My reference is C++ for Java Programmers by Weiss.

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  • sorting names in a linked list

    - by sil3nt
    Hi there, I'm trying to sort names into alphabetical order inside a linked list but am getting a run time error. what have I done wrong here? #include <iostream> #include <string> using namespace std; struct node{ string name; node *next; }; node *A; void addnode(node *&listpointer,string newname){ node *temp; temp = new node; if (listpointer == NULL){ temp->name = newname; temp->next = listpointer; listpointer = temp; }else{ node *add; add = new node; while (true){ if(listpointer->name > newname){ add->name = newname; add->next = listpointer->next; break; } listpointer = listpointer->next; } } } int main(){ A = NULL; string name1 = "bob"; string name2 = "tod"; string name3 = "thomas"; string name4 = "kate"; string name5 = "alex"; string name6 = "jimmy"; addnode(A,name1); addnode(A,name2); addnode(A,name3); addnode(A,name4); addnode(A,name5); addnode(A,name6); while(true){ if(A == NULL){break;} cout<< "name is: " << A->name << endl; A = A->next; } return 0; }

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  • SharePoint: Filtering a List that has Folders

    - by Gary McGill
    I have a SharePoint document library that has a folder structure used for organizing the documents (but also for controlling access, via permissions on the folders). The documents in the library are updated every month, and we store every month's version of the document in the same folder; there's a "month" column used for filtering that will contain values like Jan 09, Feb 09, etc. It looks like this: Title Month ----- ----- SubFolder 1 SubFolder 2 [] Interesting Facts Jan 09 [] Interesting Facts Feb 09 [] Interesting Facts Mar 09 [] Fascinating Numbers Jan 09 [] Fascinating Numbers Feb 09 ... Now, because users will generally be most interested in the 'current' month, I'd like them to be able to apply a filter, and select (say) Mar 09. However, if they do this using the built-in filtering, it also filters out the folders, and they can no longer navigate the folder hierarchy. This is no good - I want them to be able to move between folders with the filter intact, so that they don't need to keep switching it off and on again. I figured I might be able to use a custom view (selecting where type=folder or month=[month]), and to an extent that does work. However, I can only get it to work for a fixed month, whereas I need the user to be able to select the month - perhaps via a drop-down control on the page (and I don't want to create 60 views for 5 years' worth of months, nor do I want to have to create a new view every month). I thought it might be possible to create a view in code (rather than via the UI), but I've not been able to figure out how to get a dynamic value (a user-specific setting) into the CAML query. Any pointers gratefully appreciated! And by the way, I am aware of the dogma that folders are bad, and that everything should just be a list. However, having considered the alternatives, I still favour using folders - if I can solve this problem. Thanks in advance.

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  • How to take a Serialized String from the DB and output a UL list with Coldfusion

    - by nobosh
    Given a string from the database as follows: ul[0][id]=main1&ul[0][children][0][id]=child2&ul[0][children][0][class]=&ul[1][id]=main3&ul[2][id]=main4&ul[3][id]=main5 Where what's after the equal sign is an ID, how can I take this string on the backend and build out a list as following from front-end output? <UL id="container"> <LI id="main1"> Lorem ipsum dolor sit amet, consectetur <UL> <LI id="child2"> In hac habitasse platea dictumst. <UL></UL> </LI> </UL> </LI> <LI id="main3"> In hac habitasse platea dictumst. <UL></UL> </LI> <LI id="main4"> In hac habitasse platea dictumst. <UL></UL> </LI> <LI id="main5"> In hac habitasse platea dictumst. <UL></UL> </LI> </UL> Thanks

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  • Finding the index of a list item in jQuery

    - by Tim Piele
    I have an unordered list of eight items. On page load the first five <li> have default thumbnail images in them and the 6th one has a 1px by 1px placeholder image with the ID of $('#last'). When a user inserts a new image it replaces the 'src' of $('#last') with their new image. It's not the most efficient way but it works. <ul> <li><img src="img1.png" /></li> <li><img src="img2.png" /></li> <li><img src="img3.png" /></li> <li><img src="img4.png" /></li> <li><img src="img5.png" /></li> <li><img src="1px.png" id="last"/></li> <li></li> <li></li> </ul> When the user adds a new image the ID of $('#last') is removed and I use each() to find the next empty <li> and insert the 1px by 1px image in it, with an ID of $('#last') so it is ready for the next image upload. At this point I need to get the index() of the <li> that now has the 1px by 1px image in it, whose ID is $('#last'), so that I can store the index in the session, so when a user comes back to the page the $('#last') ID is still set and ready to accept another image. How do I get the index of the <li> with that image in it, since it was set after page load? Is there a way to use delegate() or on() to get it? i.e. how do I get the index of an element that was set after page load?

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  • Weird margin in a list

    - by kevin
    I'm trying to style a menu, but I keep running into this weird margin that's appearing in both FF4 and IE. This is the only affecting css: #header ul { display: inline; } #header ul li { list-style-type: none; background: #000; display: inline; margin: 0; padding: 0; } #header ul li a { color: #fff; text-decoration: none; display: inline-block; width: 100px; text-align: center; } And this is the HTML: <div id="header"> <ul id="toplinks"> <li><a href="#">Hello</a></li> <li><a href="#">Herp</a></li> <li><a href="#">Derp</a></li> </ul> </div> As you can see, there's a margin appearing on both sides, and I'd like it so it would have no margin (or maybe 1px would be okay)...

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  • pointer to a pointer in a linked list

    - by user1596497
    I'm trying to set a linked list head through pointer to a pointer. I can see inside the function that the address of the head pointer is changing but as i return to the main progran it becomes NULL again. can someone tell me what I'm doing wrong ?? #include <stdio.h> #include <stdlib.h> typedef void(*fun_t)(int); typedef struct timer_t { int time; fun_t func; struct timer_t *next; }TIMER_T; void add_timer(int sec, fun_t func, TIMER_T *head); void run_timers(TIMER_T **head); void timer_func(int); int main(void) { TIMER_T *head = NULL; int time = 1; fun_t func = timer_func; while (time < 1000) { printf("\nCalling add_timer(time=%d, func=0x%x, head=0x%x)\n", time, func, &head); add_timer(time, func, head); time *= 2; } run_timers(&head); return 0; } void add_timer(int sec, fun_t func, TIMER_T *head) { TIMER_T ** ppScan=&head; TIMER_T *new_timer = NULL; new_timer = (TIMER_T*)malloc(sizeof(TIMER_T)); new_timer->time = sec; new_timer->func = func; new_timer->next = NULL; while((*ppScan != NULL) && (((**ppScan).time)<sec)) ppScan = &(*ppScan)->next; new_timer->next = *ppScan; *ppScan = new_timer; }

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  • 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?

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  • 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)

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  • 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++; }

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  • 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.

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  • 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?

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  • 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?

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  • Showing google-task only

    - by ideotop
    Google-task is available in google calendar without using the gmail-task api. Is there a way to use the same "google calendar" way to build a standalone google-task only web page ? (It looks like the google-task api is not available)

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  • 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()); } }

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  • 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.

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  • 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.

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  • 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... ???

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  • 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

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  • 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.

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  • 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?

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  • 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; }

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