如何解决使用onclick函数和ajax上传图像文件?
我尝试使用onclick
函数上传图像文件,并且可以使用ajax吗?我搜索有关如何上传图像的信息,但只有我可以看到是通过单击功能。但是我有一个原因,这就是为什么我使用onclick函数。控制器始终为空值
<div class="col-md-2">
<div class="form-group">
@Html.LabelFor(x => x.ImagePath,new { @class = "form-label" })
<input type="file" name="file" id="files" />
</div>
</div>
<div class="col-md-2">
<div class="form-group">
<button type="submit" id="btnUpload" class="btn btn-primary btn-sm" onclick="saveImage()"><i class="fas fa-save"></i> Submit Image</button>
</div>
</div>
这是我的jquery和ajax
function saveImage() {
var formData = new FormData();
formData.append('file',$('#files')[0].files[0]);
$.ajax({
url: '@Url.Action("index","payable")',type: 'POST',data: formData,Traditional: true,success: function (data) {
alert("success");
}
});
}
我的控制器
public ActionResult Index()
{
HttpPostedFileBase file = Request.Files[0];
string fileName = Path.GetFileNameWithoutExtension(file.FileName);
string extension = Path.GetExtension(file.FileName);
fileName = fileName + extension;
var ImagePath = "~/Image/" + fileName;
fileName = Path.Combine(Server.MapPath("~/Image/"),fileName);
file.SaveAs(fileName);
return View();
}
解决方法
尝试此代码:
var ufiles = $("#fileInput")[0].files;
if (ufiles.length > 0) {
if (window.FormData !== undefined) {
var fileData = new FormData();
//handling multiple files or single file
for (var x = 0; x < ufiles.length; x++) {
fileData.append(ufiles[x].name,ufiles[x]);
}
$.ajax({
type: "POST",url: "yourUrl",contentType: false,processData: false,data: fileData,success: function (result) {
//do something with result
},});
} else {
alert("This browser doesn't support HTML5 file uploads!"); //bootbox dependency
}
}
在服务器端:
[HttpPost]
public JsonResult Index()
{
HttpPostedFileBase file = HttpContext.Request.Files[0];//this can be put under foreach for multiple files
string fileName = Path.GetFileNameWithoutExtension(file.FileName);
string extension = Path.GetExtension(file.FileName);
fileName = fileName + extension;
var ImagePath = "~/Image/" + fileName;
fileName = Path.Combine(Server.MapPath("~/Image/"),fileName);
file.SaveAs(fileName);
//now cause we are hadnling ajax
return Json(fileName,JsonRequestBehavior.AllowGet);
}
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