【发布时间】:2016-05-19 06:15:02
【问题描述】:
在 MVC5 中单击按钮上传时要读取的 Excel 工作表。上传的 excel 文件名使用 AJAX 方法传递给操作。这里文件变量在张贴方法中获取空值。 在这里,如何在下面的 ajax 方法中将选定的文件作为 HttpPostedFileBase 传递。 `
<input style="display:none" type="file" id="fileupload1" />
<button type="button" onclick='$("#fileupload1").click()'>UPLOAD FROM EXCEL</button>
<span style="display:none" id="spnName"></span>
$(function () {$("#fileupload1").change(function () {
$("#spnName").html($("#fileupload1").val().substring($("#fileupload1").val().lastIndexOf('\\') + 1));
var file = $("#spnName").html();
$.ajax({
url: "UploadExcelForContractStaff",
type: 'POST',
data: JSON.stringify({ file: file }),
dataType: 'json',
contentType: "application/json; charset=utf-8",
success: function (data) {
}
});
});
});`
[AcceptVerbs(HttpVerbs.Post)]
public string UploadExcelForContractStaff(HttpPostedFileBase uploadFile)
{
StringBuilder strValidations = new StringBuilder(string.Empty);
try
{
if (uploadFile.ContentLength > 0)
{
string filePath = Path.Combine(HttpContext.Server.MapPath("../Uploads"),
Path.GetFileName(uploadFile.FileName));
uploadFile.SaveAs(filePath);
DataSet ds = new DataSet();
//A 32-bit provider which enables the use of
string ConnectionString = "Provider=Microsoft.ACE.OLEDB.12.0;Data Source=" + filePath + ";Extended Properties=Excel 12.0;";
using (OleDbConnection conn = new System.Data.OleDb.OleDbConnection(ConnectionString))
{
conn.Open();
using (DataTable dtExcelSchema = conn.GetSchema("Tables"))
{
string sheetName = dtExcelSchema.Rows[0]["TABLE_NAME"].ToString();
string query = "SELECT * FROM [" + sheetName + "]";
OleDbDataAdapter adapter = new OleDbDataAdapter(query, conn);
//DataSet ds = new DataSet();
adapter.Fill(ds, "Items");
if (ds.Tables.Count > 0)
{
if (ds.Tables[0].Rows.Count > 0)
{
for (int i = 0; i < ds.Tables[0].Rows.Count; i++)
{
//Now we can insert this data to database...
}
}
}
}
}
}
}
catch (Exception ex) { }
return "";
}
【问题讨论】:
-
通过更改代码解决了这个问题
<form enctype="multipart/form-data" id="frmUplaodFileAdd"> @Html.AntiForgeryToken() <input style="display:none" type="file" id="fileupload1" /> <button type="button" onclick='$("#fileupload1").click()'>UPLOAD FROM EXCEL</button> <span style="display:none" id="spnName"></span> </form>
标签: jquery-file-upload ajaxform asp.net-mvc-ajax asp.net-mvc-file-upload