在javascript / jQuery中,在调用文件上传插件之前,先设置了一个变量:
doc_type = "q";
然后,插件被初始化。插件选项中包含:
onSelect:
,在选择文件时调用。代码如下:var doc_type = "q";
$(function(){
var project_num = $("#pnum").val();
var uploadObj = $("#fileuploader").uploadFile({
url: "upload_files_processor.php",
method: "POST",
onSelect: function(){
doc_type = "W";
//Or:
//doc_type = $('#hidden_input').val(); <-- What I really need to do
return true;
},
allowedTypes:"pdf,doc,docx,xls,xlsx,ppt,pptx,bmp,jpg,png,zip",
fileName: "myfile",
formData: {"project_num":project_num,"doc_type":doc_type},
multiple: true,
autoSubmit: true,
showStatusAfterSuccess:false,
onSuccess:function(files,data,xhr) {
//Refresh documents table
},
});
}); //END document.ready()
问题:
在上载处理器
upload_files_processor.php
中,收到的doc_type
值为:$doc_type = $_POST["doc_type"]; // q
如何接收值
W
?参考:heyageek jquery upload file plugin website-单击
API & Options
选项卡 最佳答案
在您的代码中,而不是formData: {"project_num":project_num,"doc_type":doc_type},
用这个:
dynamicFormData: function()
{
return {
project_num: $("#pnum").val(),
doc_type: $('#hidden_input').val()
};
},
并删除行
var doc_type = "q";
和var project_num = $("#pnum").val()
关于javascript - 此特定实例中的javascript变量范围,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/23071410/