asp.net MVC中的jQuery Ajax上传文件

小编典典

asp.net MVC中的jQuery Ajax上传文件

ajax

我认为有一个文件

<form id="upload" enctype="multipart/form-data">
   <input type="file" name="fileUpload" id="fileUpload" size="23" />
</form>

和一个ajax请求

$.ajax({
    url: '<%=Url.Action("JsonSave","Survey")  %>',
    dataType: 'json',
    processData: false,
    contentType: "multipart/mixed",
    data: {
        Id: selectedRow.Id,
        Value: 'some date was added by the user here :))'
    },
    cache: false,
    success: function (data) {}
});

但是 Request.Files中 没有文件。Ajax请求出了什么问题？

阅读 719

2020-07-26

共1个答案

小编典典

在ASP.Net MVC中使用AJAX上传文件

自HTML5以来情况发生了变化

的JavaScript

document.getElementById('uploader').onsubmit = function () {
    var formdata = new FormData(); //FormData object
    var fileInput = document.getElementById('fileInput');
    //Iterating through each files selected in fileInput
    for (i = 0; i < fileInput.files.length; i++) {
        //Appending each file to FormData object
        formdata.append(fileInput.files[i].name, fileInput.files[i]);
    }
    //Creating an XMLHttpRequest and sending
    var xhr = new XMLHttpRequest();
    xhr.open('POST', '/Home/Upload');
    xhr.send(formdata);
    xhr.onreadystatechange = function () {
        if (xhr.readyState == 4 && xhr.status == 200) {
            alert(xhr.responseText);
        }
    }
    return false;
}

控制者

public JsonResult Upload()
{
    for (int i = 0; i < Request.Files.Count; i++)
    {
        HttpPostedFileBase file = Request.Files[i]; //Uploaded file
        //Use the following properties to get file's name, size and MIMEType
        int fileSize = file.ContentLength;
        string fileName = file.FileName;
        string mimeType = file.ContentType;
        System.IO.Stream fileContent = file.InputStream;
        //To save file, use SaveAs method
        file.SaveAs(Server.MapPath("~/")+ fileName ); //File will be saved in application root
    }
    return Json("Uploaded " + Request.Files.Count + " files");
}

编辑：HTML

<form id="uploader">
    <input id="fileInput" type="file" multiple>
    <input type="submit" value="Upload file" />
</form>

2020-07-26