获取图像的资源文件夹路径

Question

I ask because I already searched everywhere without finding results...

I already try these two methods:

//first attempt
  Var filePath = System.Reflection.Assembly.GetExecutingAssembly (). Location + @"\..\..\Resources\Image.jpg";

//Second attempt
System.Reflection.Assembly.GetExecutingAssembly (). GetManifestResourceNames ();

Already probe these methods in which they did not give results I only need to obtain the route of an image of my project. help please.

Answer 1

Make sure the Build Action of your item is set to Embedded Resource.

You might find these two methods useful, pass in the name of the resource and a list of assemblies -- it'll search the assemblies in order which provides a simple override resource capability

    private static Tuple<Assembly, string> FindEmbeddedResource(string name, params Assembly[] assemblies)
    {
        foreach (var assembly in assemblies)
        {
            var resourceName = assembly.GetManifestResourceNames().FirstOrDefault(r => r.EndsWith(name));

            if (resourceName != null)
            {
                return new Tuple<Assembly, string>(assembly, resourceName);
            }
        }

        return new Tuple<Assembly, string>(null, null);
    }

    /// <summary>
    /// Gets the embedded resource scanning assemblies in the order supplied. Allows APIs to override library assets.
    /// </summary>
    /// <param name="name">The name.</param>
    /// <param name="assemblies">The assemblies.</param>
    /// <returns></returns>
    private static string GetEmbeddedResource(string name, params Assembly[] assemblies)
    {
        var embeddedResource = FindEmbeddedResource(name, assemblies);
        var assembly = embeddedResource.Item1;
        var resource = embeddedResource.Item2;

        if (assembly != null)
        {
            var stream = assembly.GetManifestResourceStream(resource);
            using (var textStreamReader = new StreamReader(stream))
            {
                resource = textStreamReader.ReadToEnd();
            }
        }

        return resource;
    }