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Case "upload.htm"

wb.AddForm("", "form1", "receive.htm")

With wb.AddInputGroup("form1", "ipg2", "文件上传")

With .AddUploader("up128", "照片", True)

.AllowDelete = True '允许删除

.Incremental = True '允许重复选择文件或连续拍照

End With

With wb.AddInputGroup("form1", "ipg1", "增加单位")

.AddInput("责任单位", "责任单位", "text")

End With

With wb.AddButtonGroup("form1", "btg2", True)

.Add("btn1", "确定上传", "button").Attribute = "" '调用js函数上传

End With

wb.AddToast("", "tst1", "正在上传", 1)

wb.AddToast("", "tst2", "上传成功", 0)

wb.AddToast("", "tst3", "上传失败", 0).Icon = "warn"

wb.AppendHTML("<script src='./lib/ajaxform.js'></script>") '引入脚本文件

e.WriteString(wb.Build) '生成网页

Case "receive.htm"

Dim dr As DataRow = DataTables("飞机引进工作单").AddNew()

Dim nms() As String = {"责任单位"}

For Each nm As String In nms

dr(nm) = e.PostValues(nm)

For Each key As String In e.Files.Keys

If key = "up128" Then

For Each fln As String In e.Files(key)

e.SaveFile(key, fln, "d:\web\uploadfiles\" & fln)

dr.Lines("照片") = e.Files(key)

End If

dr.Save()

e.WriteString("OK")

End Select

[此贴子已经被作者于2022/7/10 20:38:36编辑过]

服务器也是使用的d:\web\目录吗，js 文件有放到服务器了吗

我照帮助测试没有问题

Select Case e.Path

Case "test.htm"

Dim wb As New WeUI

wb.AddTopTips("", "toptip1", "请输入姓名和密码!")

wb.AddForm("", "form1", "t2.htm").Attribute = "onsubmit='return valid()'" '调用函数

With wb.AddInputGroup("form1", "ipg1", "用户登录")

.AddInput("xm", "户名", "text")

.AddInput("pw", "密码", "password")

End With

With wb.AddButtonGroup("form1", "btg1", True)

.Add("btn1", "确定", "submit")

End With

wb.AppendHTML("<script src='./lib/valid.js'></script>") '引入脚本文件

e.WriteString(wb.Build)

Case "t2.htm"

Dim sb As New StringBuilder

sb.AppendLine("<meta name='viewport' content='width=device-width,initial-scale=1,user-scalable=0'>")

For Each key As String In e.PostValues.Keys

sb.AppendLine(key & ":" & e.PostValues(key) & "</br>")

e.WriteString(sb.ToString)

End Select

请上传实例说明

submit上传肯定是不行的，原因仔细看：http://www.foxtable.com/mobilehelp/topics/0106.htm

submit上传，测试没有问题

此主题相关图片如下：1.png

submitform上传，ajax是不会触发办法的验证的，需要自己在函数里调用一下submitForm函数改为

function submitForm(){

if (valid()){

show("tst1",2000);

var result = submitAjaxForm('form1','afterSubmit');

}

主题：集成照片上传和录入功能