如何重構這個 if-else 條件，使其更簡潔、更高效？

Question

我將這段代碼作為 Azure function 應用程序，我想知道if else部分，我該如何最好地處理它。

有大約 100 個具有不同客戶名稱的項目。

最好的方法是什么？

如果有人能給我舉個例子。

string customerName = string.Empty;
foreach( var doc in result )
{
    var data =(JObject)JsonConvert.DeserializeObject( doc.ToString() );
    if( (string)data["Project"] == "HPD_Oid" )
    {
        customerName = "OPPO";
    }
    else if( (string)data["Project"] == "HPD_Oreal" )
    {
        customerName = "RealMe";
    }
    else
    {
        customerName = "OnePlus";
    }
    string partitionkeyValue = string.Concat( (string)data["class"], "|", (string)data["Project"], "|", customerName );
    data.Add( new JProperty( "PartitionKey", partitionkeyValue ) );

閱讀客戶價值：

CustomerSection customer = GetConfiguration( context.FunctionAppDirectory, "CustomerSection.json" );

獲取配置值：

private static CustomerSection GetConfiguration( string basePath, string fileName )
        {
            var config = new ConfigurationBuilder()
                   .SetBasePath( basePath )
                   .AddJsonFile( fileName, optional: false )
                   .Build();
            var customerNameOutput = new CustomerSection();
            config.GetSection( "ProjectCustomerMapping" ).Bind( customerNameOutput );
            return customerNameOutput;
        }

public class CustomerSection
    {
        public Dictionary<string, string> CustomerName { get; set; }
    }

Answer 1

很簡單，使用字典：

Dictionary<string, string> projectCustomerNameMapping = new Dictionary<string, string>()
{
    { "HPD_Oid", "OPPO" },
    { "HPD_Oreal", "RealMe" }
};

然后使用查找：

if (!projectCustomerNameMapping.TryGetValue((string)data["Project"], out customerName))
{
    // if the value wasn't found in the dictionary, use the default
    customerName = "OnePlus";
}

TryGetValue 文檔

Answer 2

我有一堆IDictionary<K, V>的擴展方法，如下所示：

public static class IDictionaryExt
{
    public static Func<K, V> Map<K, V>(this IDictionary<K, V> source, Func<V> @default)
    {
        return key => (key == null || !source.ContainsKey(key)) ? @default() : source[key];
    }
}

我可以這樣使用它：

Func<string, string> projectCustomerNameMapping = new Dictionary<string, string>()
{
    { "HPD_Oid", "OPPO" },
    { "HPD_Oreal", "RealMe" }
}.Map(() => "OnePlus");

然后你的代碼變成：

string customerName = projectCustomerNameMapping((string)data["Project"]);