需要用到
slice() 方法
参考
http://www.w3school.com.cn/jsref/jsref_slice_array.asp
前端HTML使用标签是
type=file的input标签，可以直接选择文件上传
JS：
function () { ???????var file = $("#SelectImageFile")[0].files[0]; ???????var fileName = file.name; ???????var fileSplitSize = 1 * 1024 * 1024; //文件分段大小5M ???????var size = file.size; ???????var posturl = postWebUrl;//上传文件的后台方法地址 ???????var formData = new FormData(); ???????var index1 = fileName.lastIndexOf("."); ???????var index2 = fileName.length; ???????var suffix = fileName.substring(index1, index2);//后缀名 ???????//如果需要限制后缀名，可以这么写 ???????if (suffix != ".rar" && suffix != ".zip" && suffix != ".7z") { ???????????msg("上传的影像信息必须为.rar,.zip,.7z格式"); ???????} else { ???????????formData.append("file", file.slice(start, start +fileSplitSize)); ???????????formData.append("name", fileName); ???????????$.ajax({ ???????????????url: posturl, ???????????????type: ‘POST‘, ???????????????data: formData, ?// 告诉jQuery不要去处理发送的数据 ????????????????processData: false, ?// 告诉jQuery不要去设置Content-Type请求头 ????????????????contentType: false, ???????????????beforeSend: function () { ???????????????????//console.log("正在进行，请稍候"); ???????????????}, ???????????????//分段上传，1-5上传成功后，再上传6-11，以此类推，成功后，给出提示，在成功之前，会一直显示上传进度 ???????????????success: function (data) { ???????????????????$("#process").show();//进度条显示 ???????????????????if (start + fileSplitSize >= size) { ???????????????????????$("#spnProcessShow").css("width", "100%"); ???????????????????????$("#spnProcessText").html("上传完成，已上传100%"); ???????????????????} else { ???????????????????????start += fileSplitSize; ???????????????????????Percentage = Math.round(start / size * 100); ???????????????????????$("#spnProcessShow").css("width", Percentage + "%"); ???????????????????????$("#spnProcessText").html("正在上传，已上传" + Percentage + "%"); ???????????????????} ???????????????}, ???????????????error: function (data) { ???????????????????start = 0; ???????????????????$("#process").hide();//隐藏进度条 ???????????????????$("#spnProcessShow").css("width", "0%"); ???????????????????$("#lblMsg").html("影像上传失败"); ???????????????????????????????????} ???????????}); ???????} ???} ???前端接受文件调用后台上传文件的方法
原文地址：http://www.cnblogs.com/llcdbk/p/7903423.html