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ASP.NET MVC中的jquery ajax上载文件

ajax asp.net-mvc jquery

CoffeeCode · 技术社区 · 16 年前

我有一个文件在我的视图中

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

以及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) {}
});

但是里面没有文件 请求.files . Ajax请求有什么问题?

5 回复 | 直到 8 年前

111

Ajeesh M 11 年前

Upload files using AJAX in ASP.Net MVC

自HTML5以来,情况发生了变化

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;
}

控制器

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>

M.Babcock 10 年前

Ajax文件上载现在可以通过传递 FormData 反对 data 的属性 $.ajax 请求。

正如OP特别要求jquery实现一样,这里是:

<form id="upload" enctype="multipart/form-data" action="@Url.Action("JsonSave", "Survey")" method="POST">
    <input type="file" name="fileUpload" id="fileUpload" size="23" /><br />
    <button>Upload!</button>
</form>

$('#upload').submit(function(e) {
    e.preventDefault(); // stop the standard form submission

    $.ajax({
        url: this.action,
        type: this.method,
        data: new FormData(this),
        cache: false,
        contentType: false,
        processData: false,
        success: function (data) {
            console.log(data.UploadedFileCount + ' file(s) uploaded successfully');
        },
        error: function(xhr, error, status) {
            console.log(error, status);
        }
    });
});

public JsonResult Survey()
{
    for (int i = 0; i < Request.Files.Count; i++)
    {
        var file = Request.Files[i];
        // save file as required here...
    }
    return Json(new { UploadedFileCount = Request.Files.Count });
}

有关的详细信息 FormData at MDN

Nick Craver 16 年前

您不能通过Ajax上传文件,您需要使用iframe或其他一些技巧来进行完整的回发。这主要是出于安全考虑。

Here's a decent write-up including a sample project 使用Steve Sanderson的swfupload和asp.net MVC。这是我读到的第一件事,让它与ASP.NET MVC一起正常工作(当时我也是MVC的新手),希望它对您有帮助。

sandeep 12 年前

如果使用Ajax发布表单,则不能使用$.Ajax方法发送图像, 您必须使用经典的xmlhttpobject方法保存图像, 它的其他替代方法是使用提交类型而不是按钮

-1

ddagsan 8 年前

在VueJS版本:v2.5.2上,我有一个这样的示例

<form action="url" method="post" enctype="multipart/form-data">
    <div class="col-md-6">
        <input type="file" class="image_0" name="FilesFront" ref="FilesFront" />
    </div>
    <div class="col-md-6">
        <input type="file" class="image_1" name="FilesBack" ref="FilesBack" />
    </div>
</form>
<script>
Vue.component('v-bl-document', {
    template: '#document-item-template',
    props: ['doc'],
    data: function () {
        return {
            document: this.doc
        };
    },
    methods: {
        submit: function () {
            event.preventDefault();

            var data = new FormData();
            var _doc = this.document;
            Object.keys(_doc).forEach(function (key) {
                data.append(key, _doc[key]);
            });
            var _refs = this.$refs;
            Object.keys(_refs).forEach(function (key) {
                data.append(key, _refs[key].files[0]);
            });

            debugger;
            $.ajax({
                type: "POST",
                data: data,
                url: url,
                cache: false,
                contentType: false,
                processData: false,
                success: function (result) {
                    //do something
                },

            });
        }
    }
});
</script>