asp.net-mvc – 如何为ASP.NET 4.5 Web API创建MultipartFormFormatter

这些链接对我没有帮助：

例：

//Model:
public class Group
{
    public int Id { get; set; }        
    public File File { get; set; }
}
 
//Controller:
[HttpPost]
public void SaveGroup([FromBody]Group group) {}
 
//Formatter:
public class MultipartFormFormatter : MediaTypeFormatter
{
    private const string StringMultipartMediaType = "multipart/form-data";
 
    public MultipartFormFormatter()
    {
        this.SupportedMediaTypes.Add(new MediaTypeHeaderValue(StringMultipartMediaType));
 
    }
 
    public override bool CanReadType(Type type)
    {
        return true;
    }
 
    public override bool CanWriteType(Type type)
    {
        return false;
    }
 
    public async override Task<object> ReadFromStreamAsync(Type type,Stream readStream,HttpContent content,IFormatterLogger formatterLogger)
    {
        //Implementation? What here should be?
    }       
}

ReadFromStreamAsync方法应该返回什么？

如何使它能够正确地将参数传递给动作？

解决方法

public class MultipartFormFormatter : FormUrlEncodedMediaTypeFormatter
{
    private const string StringMultipartMediaType = "multipart/form-data";
    private const string StringApplicationMediaType = "application/octet-stream";
 
    public MultipartFormFormatter()
    {
        this.SupportedMediaTypes.Add(new MediaTypeHeaderValue(StringMultipartMediaType));
        this.SupportedMediaTypes.Add(new MediaTypeHeaderValue(StringApplicationMediaType));
    }
 
    public override bool CanReadType(Type type)
    {
        return true;
    }
 
    public override bool CanWriteType(Type type)
    {
        return false;
    }
 
    public override async Task<object> ReadFromStreamAsync(Type type,IFormatterLogger formatterLogger)
    {
        var parts = await content.ReadAsMultipartAsync();
        var obj = Activator.CreateInstance(type);
        var propertiesFromObj = obj.GetType().GetRuntimeProperties().ToList();
 
        foreach (var property in propertiesFromObj.Where(x => x.PropertyType == typeof(FileModel)))
        {
            var file = parts.Contents.FirstOrDefault(x => x.Headers.ContentDisposition.Name.Contains(property.Name));
 
            if (file == null || file.Headers.ContentLength <= 0) continue;
 
            try
            {
                var fileModel = new FileModel(file.Headers.ContentDisposition.FileName,Convert.ToInt32(file.Headers.ContentLength),ReadFully(file.ReadAsStreamAsync().Result));
                property.SetValue(obj,fileModel);
            }
            catch (Exception e)
            {
            }
        }
 
        foreach (var property in propertiesFromObj.Where(x => x.PropertyType != typeof(FileModel)))
        {
            var formData = parts.Contents.FirstOrDefault(x => x.Headers.ContentDisposition.Name.Contains(property.Name));
 
            if (formData == null) continue;
 
            try
            {
                var strValue = formData.ReadAsStringAsync().Result;
                var valueType = Nullable.GetUnderlyingType(property.PropertyType) ?? property.PropertyType;
                var value = Convert.ChangeType(strValue,valueType);
                property.SetValue(obj,value);
            }
            catch (Exception e)
            {
            }
        }
 
        return obj;
    }
 
    private byte[] ReadFully(Stream input)
    {
        var buffer = new byte[16 * 1024];
        using (var ms = new MemoryStream())
        {
            int read;
            while ((read = input.Read(buffer,buffer.Length)) > 0)
            {
                ms.Write(buffer,read);
            }
            return ms.ToArray();
        }
    }
}
 
public class FileModel
{
    public FileModel(string filename,int contentLength,byte[] content)
    {
        Filename = filename;
        ContentLength = contentLength;
        Content = content;
    }
 
    public string Filename { get; set; }
 
    public int ContentLength { get; set; }
 
    public byte[] Content { get; set; }
 
}

asp.net-mvc – 如何为ASP.NET 4.5 Web API创建MultipartFormFormatter

解决方法

猜你在找的asp.Net相关文章