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  • Java - Adding a Node in a doubly-linked list.

    - by jacobnlsn
    Hi, I am working on some code were I need to add a Node into a doubly linked-list, this is the code I have so far: Node tempNext = cursor.getNext(); temp = new Node(item, null, cursor, tempNext); tempNext.setPrev(temp); cursor is the Node that is right before the new added Node should go.

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  • How do I randomly select from a list in Python?

    - by Liam Block
    Basically, I've got a homework task of programming a text based battle simulator in Python. Obviously I've gone with pokémon... I would like the enemy to be randomly selected, however I don't know how to randomly select from a list... foo = ['a', 'b', 'c', 'd', 'e'] from random import choice print choice(foo) This is what I've been told to try but I've got no modules or anything imported... How can I make this work, appreciated.

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  • Elegent way to collapse or expand sub-sequences of a list in Python?

    - by forgot
    I want to collapse or expand sub-sequences of a list e.g. ['A', 'B', 'D', 'E', 'H'] -> ['AB', 'DE', 'H'] and vice versa currently I wrote some ugly code like: while True: for i, x in enumerate(s): if x == 'A' and s[i+1] == 'B': s[i:i+2] = 'AB' break else: break For people who asking 'why do that thing': Actually I'm working on a optimizing compiler and this is the peephole part. Writing pattern matching is a little annoying.

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  • How to determine whether a linked list contains a loop?

    - by ET
    During a preparation for a job interview, I encountered the following question: How can you determine whether a linked list (of any type) contains a loop, using additional space complexity of O(1)? You cannot assume that the loop starts at the first node (and of course, the loop doesn't have to contain all nodes). I couldn't find the answer, though I have the feeling it's quite simple...

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  • How can I acquire an aggregated list of known proxy IP addresses?

    - by Howard3
    I'd like to use this to help maintain a good defence against people trying to skirt the rules of my system. I've found TOR endpoints, nothing that's readily available to be shot into a script (needs to be parsed) but they work. However I need a list which goes beyond TOR yet I cannot find anything conclusive just yet. Any suggestions would be greatly appreciated.

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  • How should I grab pairs from a list in python?

    - by tomaski
    Say I have a list that looks like this: ['item1', 'item2', 'item3', 'item4', 'item5', 'item6', 'item7', 'item8', 'item9', 'item10'] Using Python, how would I grab pairs from it, where each item is included in a pair with both the item before and after it? ['item1', 'item2'] ['item2', 'item3'] ['item3', 'item4'] ['item4', 'item5'] ['item5', 'item6'] ['item6', 'item7'] ['item7', 'item8'] ['item8', 'item9'] ['item9', 'item10'] Seems like something i could hack together, but I'm wondering if someone has an elegant solution they've used before?

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  • Office 365 - Outlook shows Global Address List clicking "Rooms" during a meeting request

    - by TheCleaner
    This appears to be a "known" issue, but apparently no fix for it. However, I've been impressed before at the tenacity of the experts here to figure out an answer/fix. ISSUE When booking a New Meeting in Outlook (2013 or 2010) and choosing the Rooms button: The default list that opens is the Offline Global Address List: Which means a user has to change from the Offline Global Address List to the All Rooms list as shown here in order to easily pick from the list of actual rooms/resources: This isn't the default however for On-Premise Exchange servers. They default "correctly" to the All Rooms list when you click the Rooms button in the meeting request. While the option of using the Room Finder is there and does work, users have to know to click the Room Finder choice and it doesn't fix the actual root issue here. MY RESEARCH A few links I've found: http://community.office365.com/en-us/forums/158/t/41013.aspx http://community.office365.com/en-us/forums/148/p/24139/113954.aspx http://community.office365.com/en-us/forums/172/t/58824.aspx It was suggested that it might be that the "msExchResourceAddressLists attribute has incorrect value set". I checked my config by running: Get-OrganizationConfig | Select-Object ResourceAddressLists and the output was what it should be: ResourceAddressLists -------------------- {\All Rooms} QUESTION Does anyone have a fix that will make the All Rooms list be the default list when clicking the Rooms button in Outlook when using Office 365 / Exchange Online?

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  • Converting a generic list into JSON string and then handling it in java script

    - by Jalpesh P. Vadgama
    We all know that JSON (JavaScript Object Notification) is very useful in case of manipulating string on client side with java script and its performance is very good over browsers so let’s create a simple example where convert a Generic List then we will convert this list into JSON string and then we will call this web service from java script and will handle in java script. To do this we need a info class(Type) and for that class we are going to create generic list. Here is code for that I have created simple class with two properties UserId and UserName public class UserInfo { public int UserId { get; set; } public string UserName { get; set; } } Now Let’s create a web service and web method will create a class and then we will convert this with in JSON string with JavaScriptSerializer class. Here is web service class. using System; using System.Collections.Generic; using System.Linq; using System.Web; using System.Web.Services; namespace Experiment.WebService { /// <summary> /// Summary description for WsApplicationUser /// </summary> [WebService(Namespace = "http://tempuri.org/")] [WebServiceBinding(ConformsTo = WsiProfiles.BasicProfile1_1)] [System.ComponentModel.ToolboxItem(false)] // To allow this Web Service to be called from script, using ASP.NET AJAX, uncomment the following line. [System.Web.Script.Services.ScriptService] public class WsApplicationUser : System.Web.Services.WebService { [WebMethod] public string GetUserList() { List<UserInfo> userList = new List<UserInfo>(); for (int i = 1; i <= 5; i++) { UserInfo userInfo = new UserInfo(); userInfo.UserId = i; userInfo.UserName = string.Format("{0}{1}", "J", i.ToString()); userList.Add(userInfo); } System.Web.Script.Serialization.JavaScriptSerializer jSearializer = new System.Web.Script.Serialization.JavaScriptSerializer(); return jSearializer.Serialize(userList); } } } Note: Here you must have this attribute here in web service class ‘[System.Web.Script.Services.ScriptService]’ as this attribute will enable web service to call from client side. Now we have created a web service class let’s create a java script function ‘GetUserList’ which will call web service from JavaScript like following function GetUserList() { Experiment.WebService.WsApplicationUser.GetUserList(ReuqestCompleteCallback, RequestFailedCallback); } After as you can see we have inserted two call back function ReuqestCompleteCallback and RequestFailedCallback which handle errors and result from web service. ReuqestCompleteCallback will handle result of web service and if and error comes then RequestFailedCallback will print the error. Following is code for both function. function ReuqestCompleteCallback(result) { result = eval(result); var divResult = document.getElementById("divUserList"); CreateUserListTable(result); } function RequestFailedCallback(error) { var stackTrace = error.get_stackTrace(); var message = error.get_message(); var statusCode = error.get_statusCode(); var exceptionType = error.get_exceptionType(); var timedout = error.get_timedOut(); // Display the error. var divResult = document.getElementById("divUserList"); divResult.innerHTML = "Stack Trace: " + stackTrace + "<br/>" + "Service Error: " + message + "<br/>" + "Status Code: " + statusCode + "<br/>" + "Exception Type: " + exceptionType + "<br/>" + "Timedout: " + timedout; } Here in above there is a function called you can see that we have use ‘eval’ function which parse string in enumerable form. Then we are calling a function call ‘CreateUserListTable’ which will create a table string and paste string in the a div. Here is code for that function. function CreateUserListTable(userList) { var tablestring = '<table ><tr><td>UsreID</td><td>UserName</td></tr>'; for (var i = 0, len = userList.length; i < len; ++i) { tablestring=tablestring + "<tr>"; tablestring=tablestring + "<td>" + userList[i].UserId + "</td>"; tablestring=tablestring + "<td>" + userList[i].UserName + "</td>"; tablestring=tablestring + "</tr>"; } tablestring = tablestring + "</table>"; var divResult = document.getElementById("divUserList"); divResult.innerHTML = tablestring; } Now let’s create div which will have all html that is generated from this function. Here is code of my web page. We also need to add a script reference to enable web service from client side. Here is all HTML code we have. <form id="form1" runat="server"> <asp:ScriptManager ID="myScirptManger" runat="Server"> <Services> <asp:ServiceReference Path="~/WebService/WsApplicationUser.asmx" /> </Services> </asp:ScriptManager> <div id="divUserList"> </div> </form> Now as we have not defined where we are going to call ‘GetUserList’ function so let’s call this function on windows onload event of javascript like following. window.onload=GetUserList(); That’s it. Now let’s run it on browser to see whether it’s work or not and here is the output in browser as expected. That’s it. This was very basic example but you can crate your own JavaScript enabled grid from this and you can see possibilities are unlimited here. Stay tuned for more.. Happy programming.. Technorati Tags: JSON,Javascript,ASP.NET,WebService

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  • How can I css for List/grid in div with border?

    - by Saidul Haque Nayan
    I want to create a list table in DIV with border. When more content of width of a column then line break and border increase but other column border not increase. see the example: Untitled Document /list/ .list_container{ float:left; width: 550px; margin-bottom:10px; font-family: vardana; } .list_row{ float:left; width: 548px; border-bottom:1px #9F9F9F solid; } .list_row:hover{ background-color:#CCCCCC; } .list_rowHeader{ float:left; width: 548px; border-bottom:1px #9F9F9F solid; border-top:1px #9F9F9F solid; font-weight: bold; background-color: #FF0000; color: #FFFFFF; } .list_column{ float:left; padding: 3px; border-left: 1px #9F9F9F solid; } .list_columnLast{ float:left; padding: 3px; border-left: 1px #9F9F9F solid; border-right: 1px #9F9F9F solid; } .even{ background-color:#E0E0E0!important;} .odd{ background-color:#FFFFFF!important;} </style> </head> <body> <div class="list_container" > <div class="list_rowHeader" > <div class="list_column" style="width: 250px;">Name</div> <div class="list_column" style="width: 96px;"> Bid Amount</div> <div class="list_columnLast" style="width: 180px;"> Email </div> </div> <div class="list_row even" > <div class="list_column" style="width: 250px;">Saidul Haque</div> <div class="list_column" style="width: 96px;"> 2131231</div> <div class="list_columnLast" style="width: 180px;"> [email protected]</div> </div> <div class="list_row odd" > <div class="list_column" style="width: 250px;">Saidul Haque, Sonargaon, Bangladesh Dhaka, </div> <div class="list_column" style="width: 96px;"> 2131231</div> <div class="list_columnLast" style="width: 180px;"> [email protected]</div> </div> <div class="list_row even" > <div class="list_column" style="width: 250px;">Saidul Haque</div> <div class="list_column" style="width: 96px;"> 2131231</div> <div class="list_columnLast" style="width: 180px;"> [email protected]</div> </div> </div> </body> Any body solve this problem?

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  • is it wasteful/bad design to use a vector/list where in most instances it will only have one element

    - by lucid
    is it wasteful/bad design to use a vector/list where in most instances it will only have one element? example: class dragon { ArrayList<head> = new ArrayList<head> Heads; tail Tail = new tail(); body Body = new body(); dragon() { theHead=new head(); Heads.add(theHead); } void nod() { for (int i=0;i<Heads.size();i++) { heads.get(i).GoUpAndDown(); } } } class firedragon extends dragon { } class icedragon extends dragon { } class lightningdragon extends dragon { } // 10 other one-headed dragon declarations here class hydra extends dragon { hydra() { anotherHead=new head(); for (int i=0;i<2;i++) { Heads.add(anotherHead); } } } class superhydra extends dragon { superhydra() { anotherHead=new head(); for (int i=0;i<4;i++) { Heads.add(anotherHead); } } }

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  • Is it possible to return a list of all ranges from all worksheets in an Excel 2002 workbook?

    - by generalt
    Hello all. I want to extract "special" data from an Excel 2002 (client requirement, cannot change) workbook and worksheets contained therein. I have classified ranges in this "special" data category. I would like to acquire a list of all ranges in, ideally, all worksheets in a workbook. The attributes I'm interested in are the range name, and the range address. I have been googling for a while now, and have not found anything relevant. I was assuming the Excel 2002 API would expose something like this: ApplicationClass app = new ApplicationClass(); Workbook workbook = app.Workbooks.Open(@"c:\file.xls", ...); Worksheet worksheet = workbook.Worksheets["sheet1"] as Worksheet; Range[] ranges = worksheet.GetAllRanges(); or something similar. However, I am sadly mistaken. Is this possible with Excel 2002?

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  • Convert a generic list to an array

    - by Freewind
    I have searched for this, but unfortunately, I don't get the correct answer. class Helper { public static <T> T[] toArray(List<T> list) { T[] array = (T[]) new Object[list.size()]; for (int i = 0; i < list.size(); i++) { array[i] = list.get(i); } return array; } } Test it: public static void main(String[] args) { List<String> list = new ArrayList<String>(); list.add("abc"); String[] array = toArray(list); System.out.println(array); } But there is an error thrown: Exception in thread "main" java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to [Ljava.lang.String; at test.Helper.main(Helper.java:30) How to solve this?

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  • Properly removing an Integer from a List<Integer>

    - by Yuval A
    Here's a nice pitfall I just encountered. Consider a list of integers: List<Integer> list = new ArrayList<Integer>(); list.add(5); list.add(6); list.add(7); list.add(1); Any educated guess on what happens when you execute list.remove(1)? What about list.remove(new Integer(1))? This can cause some nasty bugs. What is the proper way to differentiate between remove(int index), which removes an element from given index and remove(Object o), which removes an element by reference, when dealing with lists of integers? The main point to consider here is the one @Nikita mentioned - exact parameter matching takes precedence over auto-boxing.

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