form.ajaxSubmit获取上传的文件名

    var form = $("form[name=fileForm]");
    $("#uploadTip").html("正在上传...");
    var options = {
        action: "/optimizationJob/uploadFile.action",
        type: "post",
        data: {fileName:fileName },//传递文件名到服务器
        success: function (data) {
            var success = data.success;
            var errMsg = data.errMsg;
            if (success == "Y") {
                console.log("上传成功,返回success=Y,errMsg:" + errMsg);
                $("#uploadTip").html("上传成功");
                // uploadFileToJSS(fileName);
            } else {
                console.log("上传失败,返回success=N,errMsg:" + errMsg);
                $("#uploadTip").html("上传失败");
            }
            $("#submitBtn").attr("disabled", false);
        },
        error: function (data) {
            console.log("上传失败,返回error");
            $("#uploadTip").html("上传失败");
            $("#submitBtn").attr("disabled", false);
        }
    };
    form.ajaxSubmit(options);

java中的Action写好fileName的set,get方法，就能使用了