jquery
<script type="text/javascript">
$(document).ready(function () {
$("#upload").click(function (evt) {
var fileUpload = $("#files").get(0);
var files = fileUpload.files;
var data = new FormData();
for (var i = 0; i < files.length ; i++) {
data.append(files[i].name, files[i]);
}
$.ajax({
type: "POST",
url: "/Picture/UploadFilesAjax",
contentType: false,
processData: false,
data: data,
success: function (message) {
alert(message);
},
error: function () {
alert("There was error uploading files!");
}
});
});
});
</script>
欢迎大家交流~ 以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持易采站长站。








