问题列表 - 第28461页

考虑代码：

ObservableCollection<string> cities = new ObservableCollection<string>();\nObservableCollection<string> states = new ObservableCollection<string>();\n\nListBox list;\n\ncities.Add("Frederick");\ncities.Add("Germantown");\ncities.Add("Arlington");\ncities.Add("Burbank");\ncities.Add("Newton");\ncities.Add("Watertown");\ncities.Add("Pasadena");\n\nstates.Add("Maryland");\nstates.Add("Virginia");\nstates.Add("California");\nstates.Add("Nevada");\nstates.Add("Ohio");\n\nCompositeCollection cmpc = new CompositeCollection();\nCollectionContainer cc1 = new CollectionContainer();\nCollectionContainer cc2 = new CollectionContainer();\n\ncc1.Collection = cities;\ncc2.Collection = states;\n\ncmpc.Add(cc1);\ncmpc.Add(cc2);\n\nlist.ItemsSource = cmpc;\n\nforeach(var itm in cmpc)\n{\n    // itm is CollectionContainer and there are only two itm\xe2\x80\x99s\n    // I need the strings\n}\n

虽然列表在 GUI 上显示了正确的数据

我需要这些数据（不引用列表框）但我没有得到它

使Beyond Compare 3使用Eclipse/Subclipse冲突解决方案的正确参数是什么？

在Preferences > Team > SVN > Diff/Merge那里可以选择指定外部程序来解决冲突.

默认参数是:

"${yours}" "${theirs}" "${base}" "${merged}"

它建议TortoiseMerge设置:

/theirs:"${theirs}" /base:"${base}" /mine:"${yours}" /merged:"${merged}"

但Beyond Compare的适当配置是什么？

我有一个文本框并使用数据绑定到一个对象.这很好,直到我尝试选择一个新产品:

product = new Product(id);
textbox.DataBindings.Add("Text", product, "ProductName");

// After user action:
product = new Product(newId); // <- the textbox isn't updated

产品更新后,是否必须清除数据绑定并再次设置？

我有一个表单发布到MVC的一个动作.我想从操作中的FormCollection中提取所选的下拉列表项.我该怎么做？

我的Html表格:

<% using (Html.BeginForm())
    {%>
    <select name="Content List">
    <% foreach (String name in (ViewData["names"] as IQueryable<String>)) { %>
          <option value="<%= name %>"><%= name%></option>
    <% } %>
    </select>
    <p><input type="submit" value="Save" /></p>
<% } %>

我的行动:

[HttpPost]
public ActionResult Index(FormCollection collection)
{
    //how do I get the selected drop down list value?
    String name = collection.AllKeys.Single();
    return RedirectToAction("Details", name);
}

我正在使用cPickle将我的数据库保存到文件中.代码看起来像这样:

def Save_DataBase():
import cPickle
from scipy import *
from numpy import *
a=Results.VersionName
#filename='D:/results/'+a[a.find('/')+1:-a.find('/')-2]+Results.AssType[:3]+str(random.randint(0,100))+Results.Distribution+".lft"
filename='D:/results/pppp.lft'
plik=open(filename,'w')


DataOutput=[[[DataBase.Arrays.Nodes,DataBase.Arrays.Links,DataBase.Arrays.Turns,DataBase.Arrays.Connectors,DataBase.Arrays.Zones],
             [DataBase.Nodes.Data,DataBase.Links.Data,DataBase.Turns.Data,DataBase.OrigConnectors.Data,DataBase.DestConnectors.Data,DataBase.Zones.Data],
             [DataBase.Nodes.DictionaryPy2Vis,DataBase.Links.DictionaryPy2Vis,DataBase.Turns.DictionaryPy2Vis,DataBase.OrigConnectors.DictionaryPy2Vis,DataBase.DestConnectors.DictionaryPy2Vis,DataBase.Zones.DictionaryPy2Vis],
             [DataBase.Nodes.DictionaryVis2Py,DataBase.Links.DictionaryVis2Py,DataBase.Turns.DictionaryVis2Py,DataBase.OrigConnectors.DictionaryVis2Py,DataBase.DestConnectors.DictionaryVis2Py,DataBase.Zones.DictionaryVis2Py],
             [DataBase.Paths.List]],[Results.VersionName,Results.noZones,Results.noNodes,Results.noLinks,Results.noTurns,Results.noTrips,
                                     Results.Times.VersionLoad,Results.Times.GetData,Results.Times.GetCoords,Results.Times.CrossTheTime,Results.Times.Plot_Cylinder,
                                     Results.AssType,Results.AssParam,Results.tStart,Results.tEnd,Results.Distribution,Results.tVector]]



cPickle.dump(DataOutput, plik, protocol=0)
plik.close()`

它工作正常.我的大多数数据库行都是列表,类似vecor或类似数组的数据集.

但是现在当我输入数据时,会发生错误:

def Load_DataBase():
    import cPickle 
    from scipy import *
    from numpy import *  
    filename='D:/results/pppp.lft'
    plik= open(filename, 'rb')
    """ first cPickle load approach """
    A= cPickle.load(plik)
    """ fail """
    """ Another approach - data format exact as in Output step above , also fails"""
    [[[DataBase.Arrays.Nodes,DataBase.Arrays.Links,DataBase.Arrays.Turns,DataBase.Arrays.Connectors,DataBase.Arrays.Zones],
                 [DataBase.Nodes.Data,DataBase.Links.Data,DataBase.Turns.Data,DataBase.OrigConnectors.Data,DataBase.DestConnectors.Data,DataBase.Zones.Data],
                 [DataBase.Nodes.DictionaryPy2Vis,DataBase.Links.DictionaryPy2Vis,DataBase.Turns.DictionaryPy2Vis,DataBase.OrigConnectors.DictionaryPy2Vis,DataBase.DestConnectors.DictionaryPy2Vis,DataBase.Zones.DictionaryPy2Vis],
                 [DataBase.Nodes.DictionaryVis2Py,DataBase.Links.DictionaryVis2Py,DataBase.Turns.DictionaryVis2Py,DataBase.OrigConnectors.DictionaryVis2Py,DataBase.DestConnectors.DictionaryVis2Py,DataBase.Zones.DictionaryVis2Py],
                 [DataBase.Paths.List]],[Results.VersionName,Results.noZones,Results.noNodes,Results.noLinks,Results.noTurns,Results.noTrips,
                                         Results.Times.VersionLoad,Results.Times.GetData,Results.Times.GetCoords,Results.Times.CrossTheTime,Results.Times.Plot_Cylinder,
                                         Results.AssType,Results.AssParam,Results.tStart,Results.tEnd,Results.Distribution,Results.tVector]]= cPickle.load(plik)` …

您将获得一个具有Integral类型值的大数组,如何以时间效率方式将其中的所有零值移动到数组的前部？

例如0,1,72,3,0,5,9,0,6,51,0,3 ---> 0,0,0,0,1,72,3,5,9,6,51, 3

问候!

开发人员学习ruby需要多长时间.并开发像stackoverflow这样的生产网站？一般.如果开发人员具有.NET经验但没有ruby和MYSQL或PostgreSQL经验.

我在编写自己的HashTable时遇到了这个问题.这一切都有效,但当我试图模仿这件事时,它给了我错误.我重新创建了如下问题:

本代码的工作原理:

typedef double Item;

class A
{
public:
    A()
    {
        v.push_back(pair<string, Item>("hey", 5.0));
    }

    void iterate()
    {
        for(Iterator iter = v.begin(); iter != v.end(); ++iter)
            cout << iter->first << ", " << iter->second << endl;
    }

private:
    vector<pair<string, double> > v;
    typedef vector< pair<string, double> >::iterator Iterator;
};

这个代码不是:

template<typename ValueType>
class B
{
public:
    B(){}

    void iterate()
    {
        for(Iterator iter = v.begin(); iter != v.end(); ++iter)
            cout << iter->first << ", " << iter->second << endl;
    }

private: …

我正在尝试编写一个同时运行许多程序的.sh文件

我试过这个

prog1 
prog2

但是运行prog1然后等到prog1结束然后开始prog2 ...

那我怎么能并行运行呢？

是MAC用户的git的任何扩展吗？我想在我的MAC中使用与TextWrangler或任何文本编辑器集成的git.