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var uniqueId

var files = ""

var db = 0

upload.render({

elem: '#ID-upload-demo-btn-2',

url: '../pnxpsc', //

data: {

bj: function() {

return $('#bjxz').val();

db: function() {

$('input[type="checkbox"][name="BBB"]:checked').each(function() {

var number = parseFloat($(this).attr('value'))

db += number

});

return db;

filename:function(){

return uniqueId;

headers: {

token: 'sa'

multiple: true,

auto: false,

bindAction: "#ID-upload-demo-btn-3",

choose: function(obj) {

// $('#upload-demo-preview').empty;

files = obj.pushFile();

console.log(files)

obj.preview(function(index, file, result) {

$('#upload-demo-preview').append('<img src="' + result + '">').find('img').css({

widhth: 92,

height: 92,

margin: '0 10px 10px 0'

})

}

});

Dim e As RequestEventArgs = args(0)

Dim bj As String = e.values("bj")

Dim db As Double=e.values("db")

Dim xpid=e.values("filename")

'Dim i As Integer=1

For Each key As String In e.files.keys

For Each fl As String In e.files(key)

e.savefile(key,fl,"/web/xp/wjxp/" & xpid & "-" & i &".jpg"

i=i+1

Dim obj As new Jobject

obj("status")="ok"

e.Writestring(obj.Tostring)

e.Handled=True ''

请问老师,你好，这个是前端牟取预览有多个文件，想上传后在后台将每个文件用 xpid连个i的形式来保存文件，现在是几个图片传过来文件名只有第一个，请问问题在哪里，该怎样改？谢谢

估计是前端添加文件时没有添加文件名。自行使用浏览器开发者工具跟踪一下前端的执行过程

有文件，现在的问题是传后台想把上传的几个图片文件分别存在为xpid-"X"的形式，如AA-1，AA-2
图片点击可在新窗口打开查看

此主题相关图片如下：微信图片_20241008215642.png

调试看看接收到几个

For Each key As String In e.files.keys

msgbox(key )

For Each fl As String In e.files(key)

msgbox(fl )

e.savefile(key,fl,"/web/xp/wjxp/" & xpid & "-" & i &".jpg"

i=i+1

Dim e As RequestEventArgs = args(0)

Dim bj As String = e.values("bj")

Dim db As Double=e.values("db")

Dim xpid=e.values("filename")

Dim i As Integer=1

For Each key As String In e.files.keys

For Each fl As String In e.files(key)

msgbox(fl )

e.savefile(key,fl,"/web/xp/wjxp/" & xpid & "-" & i &".jpg")

i=i+1

前端测试了弹出的了三个文件，但这样最终后只有存在xpid-1一个，也就是每次都存为1号文件了去了，所以不前端一次多个文件，最后保存结果都只有xpid-1,这怎么办

Dim i As Integer=1

For Each key As String In e.files.keys

For Each fl As String In e.files(key)

msgbox(fl )

e.savefile(key,fl,"/web/xp/wjxp/" & xpid & "-" & i &".jpg")

i=i+1

Dim e As RequestEventArgs = args(0)

Dim bj As String = e.values("bj")

Dim db As Double=e.values("db")

Dim xpid=e.values("filename")

Dim i As Integer = rand.Next(1000000)

For Each key As String In e.files.keys

For Each fl As String In e.files(key)

msgbox(fl )

e.savefile(key,fl,"/web/xp/wjxp/" & xpid & "-" & i &".jpg")

i=i+1

老师，终于懂起了layui前端多文件上传时是有多少个就调多少次接口，所以每一次i都为1，所以就成这样了，我用了随机数，这样就能解决了，虽解决了上传不履盖，但这不是我想要的最好结果

老师：接着再提点问，现在多文件传成功了。现有表中图片文件名为AAAA,现在对应文件夹中存有AAAA-ahdjhk.jpg,AAAA-djlskl.jpg等个数不定的文件，相在fox中点到表记录时用图片控件来显示各对应的多个文件，怎么做，没有思路，请指点，谢谢

[此贴子已经被作者于2024/10/10 20:37:16编辑过]

获取文件夹中所有AAAA-开头的文件名参考：

Dim files() As String = System.IO.Directory.GetFiles("d:\xxx文件夹" , "AAAA-*.jpg")

然后遍历逐个添加到图片控件

主题：多文件上传