Question

我在字典中有两个值，但是当我尝试在循环外获取两个值时，我只得到一个值。 locationdesc变量值被覆盖。有没有更好的方法来创建唯一的变量来处理这个问题

有两个键location-1和location-2。我试图找出如何在循环外得到两个值。我做错了吗？

string locationDesc = "";
string locationAddress = "";

int count = dictionary.Count(D => D.Key.StartsWith("location-"));

for (int i = 1; i <= count; i++)
{
    if (dictionary.ContainsKey("location-"+i))
    {
        string locationData = dictionary["location-"+i];
        string[] locationDataRow = locationData.Split(':');
        locationDesc = locationDataRow[0];
        locationAddress = locationDataRow[1];
    }
}
// Only getting location-2 value outside this loop since locationDesc is not unique.
Debug.WriteLine("Location Desc from dictionary is : " + locationDesc);
Debug.WriteLine("Location Add from dictionary is : " + locationAddress);

What I would like to get here is get both the values like locationDesc1 and locationDesc2 instead of locationDesc

我正在寻找的是创建locationDesc和locationAddress唯一，这样我就可以访问for循环之外的两个值。

更多解释因为我不太清楚：

我有一个将在前端创建的动态表。每次创建位置时，我都会创建一个cookie。对于例如location-1，location-2 ... location-n，其中位置描述和位置值为cookie中的值。我试图通过创建一个字典来在后端访问这些值，这样我就可以将所有值分配给唯一变量，这将使我更容易将这些值传递给api调用。我认为我过于复杂化一个简单的问题，可能做错了。

我的api电话会是这样的：

<field="" path="" value=locationDesc1>
<field="" path="" value=locationDesc2>

Answer 1

循环问题在于您依赖于字典中与循环中的索引匹配的条目的位置。你的第一行代码几乎有它：

int count = dictionary.Count(D => D.Key.StartsWith("location-"));

这告诉我的是，您正在寻找字典中所有以“location-”开头的条目。那么为什么不直接这样做呢？

var values = dictionary.Where(d => d.Key.StartsWith("location-"));

同时进行提取/字符串拆分：

var values = dictionary
    .Where(d => d.Key.StartsWith("location-"))
    .Select(d => d.Item.Split(':')
    .Select(s => new 
    { 
        LocationDesc = s[0], 
        LocationAddress = s[1]
    });

这将为您提供一个可以循环的IEnumerable LocationDesc / LocationAddress对：

foreach(var pair in values)
{
    Debug.WriteLine(pair.LocationDesc);
    Debug.WriteLine(pair.LocationAddress);
}

Answer 2

试试这个：

int count = dictionary.Count(D => D.Key.StartsWith("location-"));
Dictionary<string, string> values = new Dictionary<string, string>();
for (int i = 1; i <= count; i++)
{
    if (dictionary.ContainsKey("location-"+i))
    {
        string locationData = dictionary["location-"+i];
        string[] locationDataRow = locationData.Split(':');
        values.Add(locationDataRow[0],locationDataRow[1]);
    }
}
foreach (var item in values)
{
    Debug.WriteLine(item.Key + " : " + item.Value);
}

Answer 3

当您处理多个值时，您应该使用一个容器来存储所有值。

如果您只处理两个唯一值，请使用以下代码。

int count = dictionary.Count(D => D.Key.StartsWith("location-"));
string[] locationDesc = new string[2];
string[] locationAddress = new string[2];
    for (int i = 1; i <= count; i++)
    {
        if (dictionary.ContainsKey("location-"+i))
        {
            string locationData = dictionary["location-"+i];
            string[] locationDataRow = locationData.Split(':');
            locationDesc[i-1] = locationDataRow[0];
            locationAddress[i-1] = locationDataRow[1];
        }
    }

for (int i = 0; i <= locationDesc.Length-1; i++)
{
Debug.WriteLine("Location Desc from dictionary is : " + locationDesc[i]);
Debug.WriteLine("Location Add from dictionary is : " + locationAddress[i]);
}

如果未修复唯一值的数量，则使用ArrayList

int count = dictionary.Count(D => D.Key.StartsWith("location-"));
ArrayList locationDesc = new ArrayList();
ArrayList locationAddress = new ArrayList();
    for (int i = 1; i <= count; i++)
    {
        if (dictionary.ContainsKey("location-"+i))
        {
            string locationData = dictionary["location-"+i];
            string[] locationDataRow = locationData.Split(':');
            locationDesc.Add(locationDataRow[0]);
            locationAddress.Add(locationDataRow[1]);
        }
    }

for (int i = 0; i < locationDesc.Count; i++)
{
    Debug.WriteLine("Location Desc from dictionary is : " + locationDesc[i]);
    Debug.WriteLine("Location Add from dictionary is : " + locationAddress[i]);
}

简单的一个。如果您只想使用Debug.WriteLine显示结果，请使用以下代码

int count = dictionary.Count(D => D.Key.StartsWith("location-"));

    for (int i = 1; i <= count; i++)
    {
        if (dictionary.ContainsKey("location-"+i))
        {
            string locationData = dictionary["location-"+i];
            string[] locationDataRow = locationData.Split(':');
            Debug.WriteLine("Location Desc from dictionary is : " + locationDataRow[0]);
            Debug.WriteLine("Location Add from dictionary is : " + locationDataRow[1]);
        }
    }

目前无法在Visual Studio中准备代码，因此可能存在一些语法错误。

Answer 4

很难判断你正在尝试做什么。我不会只是将你已经拥有的对象转移到其他对象中以获得乐趣。如果您只是尝试在循环中公开值以与其他函数一起使用，则可以使用LINQ迭代字典。如果你想要一个特定的值，只需添加一个LINQ表达式。在我相信3.5之后，LINQ应该在任何.NET框架中。

public static void ApiMock(string s)
{
  Console.WriteLine($"I worked on {s}!");
}

static void Main(string[] args)
{
  var d =  new Dictionary<int, string> {
      {  1, "location-1" },
      {  2, "location-2" },
      {  3, "location-3" }
  };

  d.ToList().ForEach(x => ApiMock(x.Value));

  //I just want the second one
  d.Where(x => x.Value.Contains("-2")).ToList().ForEach(x => ApiMock(x.Value));

  //Do you want a concatenated string
  var holder = string.Empty;
  d.ToList().ForEach(x => holder += x.Value + ", ");
  holder = holder.Substring(0, holder.Length - 2);

  Console.WriteLine(holder);
}

C＃变量没有得到循环外的所有值

4 个答案: