Question

我通过AjaxUpload选项上传文件。我需要将一些动态参数值传递给action。我的代码如下：

$("#btnImportUtility_Upload").ajaxSubmit(function () {
new AjaxUpload('#btnImportUtility_Upload', {
    action: 'FileUploader.ashx?path=ExcelImportUtility&companyId=' + $("#ddlCompany option:selected").val() + '',
    name: 'upload',
    onSubmit: function (file, ext) {
        if (!(ext && /^(txt|csv|xls|xml|xlsx)$/i.test(ext))) {
            alert('Invalid File Format.');
            return false;
        }
    },
    onComplete: function (file, response) {            
    }
});

}）;

我想获取当前选定的下拉值，但它始终获得其初始值。

Answer 1

尝试这个

$("#btnImportUtility_Upload").ajaxSubmit(function () {
new AjaxUpload('#btnImportUtility_Upload', {
action: 'FileUploader.ashx?path=ExcelImportUtility&companyId=' + $("#ddlCompany").val() + '',
name: 'upload',
onSubmit: function (file, ext) {
    if (!(ext && /^(txt|csv|xls|xml|xlsx)$/i.test(ext))) {
        alert('Invalid File Format.');
        return false;
    }
},
onComplete: function (file, response) {            
}

}）;

Answer 2

试试这个，

  $("#btnImportUtility_Upload").ajaxSubmit(function () {
new AjaxUpload('#btnImportUtility_Upload', {
action: 'FileUploader.ashx?path=ExcelImportUtility&companyId=' + $('#ddlCompany').val() + '',
name: 'upload',
onSubmit: function (file, ext) {
    if (!(ext && /^(txt|csv|xls|xml|xlsx)$/i.test(ext))) {
        alert('Invalid File Format.');
        return false;
    }
},
onComplete: function (file, response) {            
}

你的下拉菜单＆lt;选择＆GT;对？请发帖也是

如何在ajax提交时获取当前选定的AjaxUpload OnSubmit事件下拉值

2 个答案: