[求助] submitAjaxForm未选择文件时提交报错如何处理 - 专家坐堂 - Foxtable(狐表) - 新一代数据库软件，完美融合Access、Foxpro、Excel、vb.net之优势，人人都能掌握的快速软件开发工具!

下面帮助中例子，如果未选择图片点击【提交】提示下面错误，如何处理？

http://www.foxtable.com/mobilehelp/scr/0139.htm

此主题相关图片如下：上传err.png

贴出后台相关代码

提示的地址中有2个“web”是不是重复了

[此贴子已经被作者于2018/2/24 9:44:18编辑过]

服务器先创建所提示的目录。

如果未选择图片点击【提交】提示错误，选择了图片上传正常，如何处理？

原因很明确，保存文件时没有文件名，只是一个目录，想做个友好的提示“‘请选择文件后提交’”，不知如何修改！

Case "upload.htm"

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

With wb.AddInputGroup("form1","ipg1","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.AddButtonGroup("form1","btg1",True)

.Add("btn1", "确定", "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='./web/lib/ajaxform.js'></script>") '引入脚本文件

e.WriteString(wb.Build) '生成网页

Case "receive.htm"

For Each key As String In e.Files.Keys

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

e.SaveFile(key,fln,"D:\web\web\images\" & fln) '这是我指定额保存接收到的文件的文件夹

e.WriteString("OK")

[此贴子已经被作者于2018/2/24 10:12:24编辑过]

Case "receive.htm"

For Each key As String In e.Files.Keys

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

If fln > "" Then

e.SaveFile(key,fln,"D:\web\web\images\" & fln) '这是我指定额保存接收到的文件的文件夹

End If

e.WriteString("OK")

==htm

‘不能显示上传成功，如何写呀！

Case "receive.htm"

For Each key As String In e.Files.Keys

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

If fln > ""

e.SaveFile(key,fln,"D:\web\web\images\" & fln) '保存接收到的文件

Else

e.WriteString("No")

'Return

End If

e.WriteString("OK")

==ajaxform.js

function submitForm(){

show("tst1",2000);

var result = submitAjaxForm('form1','afterSubmit',true,'tst1');

}

function afterSubmit(result){

hide("tst1");

if (result=='OK') {

show("tst2",2000);

location="upload.htm"; ‘不能显示上传成功

}

if (result=='No') {

location="msg.htm";

}

else

show("tst3",2000);

}

[此贴子已经被作者于2018/2/24 10:27:00编辑过]

Function afterSubmit(result){

hide("tst1");

If (result=='OK') {

show("tst2",2000);

}

Else

{

show("tst3",2000);

}

实现了，哈哈！

==ajaxform.js

function submitForm(){

show("tst1",2000);

var result = submitAjaxForm('form1','afterSubmit',true,'tst1');

}

function afterSubmit(result){

hide("tst1");

if (result=='OK') {

show("tst2",2000);

location="upload.htm";

}

else

if (result=='No') {

location="msg.htm";

}

else

show("tst3",2000);

}

主题：[求助] submitAjaxForm未选择文件时提交报错如何处理