如何在asp.net中保存数据库中的弹出屏幕记录 [英] how to save popup screen record in the database in asp.net

问题描述

本课程资料是否有效Dropdownlist1（差）

本课程实用有效Dropdownlist2（差）




当用户从dropdownlist1中选择穷人时，弹出屏幕将会打开。

弹出屏幕如下。





Negativ_Feedback



textbox

提交（按钮）


$ b上面文本框中的$ b用户输入原因并单击提交按钮。我想在数据库中保存如下记录。



表格结构如下

类型原因


数据库中的
我想保存下拉列表名称和评论。





我想在表格中输出如下。



类型原因

Dropdownlist1课程资料不是有效的。

解决方案

带下拉列表将value和Textbox值合并为两个不同的字符串变量然后合并两个变量并将此值传递给数据库...

like ...



string str = ddl.SelectedItem.Text;

string str1 = txt.Text.ToString（）;



并传递给数据库



propertylayer.property = str + str1;

你可以合计得到这两个值

Was this course material is effective Dropdownlist1 (Poor)
was this course practical is effective Dropdownlist2 (poor)


When user select the poor from the dropdownlist1 the pop screen will open.
popup screen as follows.


Negativ_Feedback

textbox
Submit (Button)

in the above textbox user types the reason and click the submit button.i want to save the record as follows in database.

Table structure as follows
Type reason

in database i want to save the dropdownlist name and comments.


I want the output as follows in table.

Type reason
Dropdownlist1 the course material is not effective one.

解决方案

Bring Dropdownlist value and Textbox value into two different string variables then concatinate both the variables and pass this value to database...
like...

string str= ddl.SelectedItem.Text;
string str1=txt.Text.ToString();

and pass to database as

propertylayer.property=str+str1;
you can get the both the values combindly

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