当我试图在其中包含图片时,我在创建新对象时遇到了问题。
我的模特:
public Dog()
{
public int Id { get; set; }
public string Name { get; set; }
public string DogImageUrl { get; set; }
}
我的HTML代码:
@if (User.Identity.IsAuthenticated)
{
<h7>New Dog</h7>
<input id="dogName" type="text" placeholder="dog name" />
<input type="file" name="file" style="width: 100%;" />
<input id="createNewDog" type="button" value="Go" />
</div>
</div>
}
我的Ajax邮政编码: $(&#39; #addNewDog&#39;)。点击(function(e){
var imgToUpload = $("input[name=file]").get(0).files[0];
var data = {
"Name": $('#dogName').val(),
"DogImageUrl ": "nullForNow",
};
var url = "@Url.Action("NewDog", "Home")";
$.ajax({
url: url,
data: { model: data, file: Request.File[imgToUpload] },
cache: false,
type: "POST",
contentType: "multipart/form-data",
processData: false,
success: function (respone) {
alert(respone);
location.reload();
},
error: function (reponse) {
alert(reponse);
}
});
e.stopPropagation();
});
我的控制员:
public ActionResult NewDog(Dog model, HttpPostedFileBase file)
{
using (var contex = new DogContext())
{
if (User.Identity.IsAuthenticated)
{
if (file != null)
{
try
{
if (file.ContentType == "image/jpeg")
{
if (file.ContentLength < 500000)
{
string pic = System.IO.Path.GetFileName(file.FileName);
string path = System.IO.Path.Combine(
Server.MapPath("~/Content/GroupImages"), pic);
// file is uploaded
file.SaveAs(path);
model.DogImageUrl = path;
}
}
}
catch (Exception ex)
{
return Content(ex.ToString());
}
}
contex.Dogs.Add(model);
contex.SaveChanges();
}
}
我的img总是空的..我在互联网上搜索并没有弄明白。 目标是将PATH保存到我的数据库中的图像。
答案 0 :(得分:0)
首先定义模型
public class DogUpload
{
public int Id { get; set; }
public string Name { get; set; }
public HttpPostedFileBase Image { get; set; }
}
然后你的行动
[HttpPost]
public ActionResult Upload(DogUpload model)
{
var image = model.Image;
var name = model.Name;
var fileName = image.FileName;
var path = Server.MapPath("~/Content/GroupImages" + fileName);
image.SaveAs(path);
// create valid url to dog image
// create Dog and save to database
var dog = dbContext.Dogs.FirstOrDefault(d => d.Id == model.Id);
if (dog == null)
{
dog = new Dog
{
Name = model.Name,
DogImageUrl = url
};
...
}
return Json(new { message = "Created", id = dog.Id, name = dog.Name, url = dog.DogImageUrl });
}
您的观点
@using(Html.BeginForm("Upload", "Dog", FormMethod.Post,
new { id="dogForm", enctype="multipart/form-data" })
{
<input type="file" name="Image" />
<input type="text" name="Name" />
<button type="submit">Submit</button>
}
AJAX
$("#dogForm").on("submit", function(e) {
e.preventDefault();
var form = $(this);
var formData = new FormData(form.get(0));
$.ajax({
url: form.attr("action"),
method: form.attr("method"),
data: formData,
processData: false,
contentType: false
})
.then(function(result) {
console.log(result);
});
});