TyF*_*ude 4 json asp.net-mvc-file-upload
我在一个名为'Upload'的控制器中有一个带有参数HttpPostedFileBase对象的方法,我从视图中发布了该文件并成功保存到文件夹中.但是当我尝试返回具有以下内容的JSON字符串对象时,它会抛出一条带有消息的异常:
"从'System.Web.HttpInputStream'''ReadTimeout'获取值时出错."
如果是'files = files',如果我将其删除,则会正确返回.但我需要这些数据
public string Upload(HttpPostedFileBase files)
{
try
{
if (files != null && files.ContentLength > 0)
{
var path = Path.Combine(Server.MapPath("~/Uploads"), files.FileName);
files.SaveAs(path);
return JsonConvert.SerializeObject(
new
{
files=files,
Passed = true,
Mesaj = "item added"
});
}
}
}
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小智 8
您可以像这样创建自定义JsonConverter:
public class HttpPostedFileConverter : JsonConverter
{
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
var stream = (Stream)value;
using (var sr = new BinaryReader(stream))
{
var buffer = sr.ReadBytes((int)stream.Length);
writer.WriteValue(Convert.ToBase64String(buffer));
}
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
throw new NotImplementedException();
}
public override bool CanRead
{
get { return false; }
}
public override bool CanConvert(Type objectType)
{
return objectType.IsSubclassOf(typeof(Stream));
}
}
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并将其传递给JsonConvert.SerializeObject方法
return JsonConvert.SerializeObject(
new
{
files=files,
Passed = true,
Mesaj = "item added"
},
new HttpPostedFileConverter());
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