Advertisement

04.26.2008 at 10:26AM PDT, ID: 23356022
[x]
Attachment Details
[x]
The Solution Rating System

With so many solutions, how can you tell which solutions are most likely to help you and which ones are not? To provide you with a tool to use, we rate our solutions based on various elements that most accurately determine if a solution is a quality solution. To explain what factors affect the solution rating, here are the elements we take into consideration when formulating our solution rating.

  • The Grade of the Solution
  • The Zone Rank of the Expert Providing the Solution
  • The Number of Author and Expert Comments
  • The Number of Experts Contributing
  • The Feedback of the Community

Your Input Matters
Because of the way the system is set up, the most important variable in this equation is you. As a member of Experts Exchange, you are able to cast your vote on the quality of the solutions in regard to how complete, accurate, helpful and easy to understand each solution is. When you provide your feedback, each rating is adjusted accordingly. So, if you see a solution that has a poor rating that you think is a good solution, let us know by rating it. As you do, the rating will be adjusted and will become more accurate for other members of our site.

If you have any suggestions that you would like to make for our rating system, please ask a question in the Suggestions Zone of Community Support.

Thank you!

Adding a new record to an sql database and retrieving data

Tags: VB.net
I have a form with plenty of textboxes and a button. When the button is clicked the information from the textboxes is added to a table within my sql database. I can manage to insert the data fine but I'm having trouble trying to achieve one thing...

I have 4 columns in my table, one called IDNumber, Name, Age, Address.
The information for name, age and address is from the textboxes in my form and is inserted when the user clicks the button. IDNumber is an int set to identity specification = yes, so when a record is added the number is incremented by 1.

I wish to show the user the IDNumber for the record they have just inserted, how could I achieve that? I would like to be able to achieve this with in the same button press event as the information is uploaded to the database the IDNumber is then shown on the form.

Here's the code I have so far...

Dim con As New SqlClient.SqlConnection(Connection)
        Dim cmdstr As String = "INSERT INTO testTBL(name, age, address) VALUES (@name, @age, @address)"
        Dim cmd As New SqlClient.SqlCommand(cmdstr, con)
        With cmd.Parameters
            .Add(New SqlClient.SqlParameter("@Name", name.Text))
            .Add(New SqlClient.SqlParameter("@age", age.Text))
            .Add(New SqlClient.SqlParameter("@address", address.Text))
        End With
        con.Open()
        cmd.ExecuteNonQuery()
        con.Close()
Start your free trial to view this solution
Question Stats
Zone: Programming
Question Asked By: Cannibal_Jack
Solution Provided By: FernandoSoto
Participating Experts: 2
Solution Grade: A
Views: 9
Translate:
Loading Advertisement...
04.26.2008 at 11:01AM PDT, ID: 21446314

Rank: Guru

All comments and solutions are available to Premium Service Members only.

Start your 7 day free trial and see for yourself why Experts Exchange is the easiest and most proven technology resource in the world. Get Started

Already a member? Login to view this solution.

 
04.26.2008 at 11:39AM PDT, ID: 21446457

Rank: Genius

All comments and solutions are available to Premium Service Members only.

Start your 7 day free trial and see for yourself why Experts Exchange is the easiest and most proven technology resource in the world. Get Started

Already a member? Login to view this solution.

 
04.29.2008 at 07:09AM PDT, ID: 21462398

All comments and solutions are available to Premium Service Members only.

Start your 7 day free trial and see for yourself why Experts Exchange is the easiest and most proven technology resource in the world. Get Started

Already a member? Login to view this solution.

 
04.29.2008 at 09:26AM PDT, ID: 21463743

Rank: Genius

All comments and solutions are available to Premium Service Members only.

Start your 7 day free trial and see for yourself why Experts Exchange is the easiest and most proven technology resource in the world. Get Started

Already a member? Login to view this solution.

 
04.29.2008 at 02:02PM PDT, ID: 21466171

Rank: Guru

All comments and solutions are available to Premium Service Members only.

Start your 7 day free trial and see for yourself why Experts Exchange is the easiest and most proven technology resource in the world. Get Started

Already a member? Login to view this solution.

 
 
Loading Advertisement...
Microsoft
  • Internet Protocols
  • Applications
  • Development
  • OS
  • Hardware
  • Windows Security
Apple
  • Operating Systems
  • Hardware
  • Programming
  • Networking
  • Software
Internet
  • Search Engines
  • File Sharing
  • WebTrends / Stats
  • Spy / Ad Blockers
  • Web Browsers
  • New Net Users
  • Web Development
  • Chat / IM
  • Anti Spam
  • Web Servers
  • Anti-Virus
  • Email Clients
Gamers
  • Tips
  • Online / MMORPG
  • Puzzle
  • Emulators
  • Action / Adventure
  • Role Playing
  • Consoles
  • Game Programming
  • Strategy
  • Sports
  • Misc
  • Computer Games
Digital Living
  • Hardware
  • New Net Users
  • New Users
  • Software
  • Digital Music
  • Gaming World
  • Home Security
  • Apple
  • Networking Hardware
Virus & Spyware
  • Vulnerabilities
  • IDS
  • Encryption
  • Anti-Virus
  • Operating Systems Security
  • Software Firewalls
  • WebApplications
  • Cell Phones
  • Operating Systems
  • Internet
  • Hardware Firewalls
Hardware
  • Handhelds / PDAs
  • Displays / Monitors
  • Components
  • Networking Hardware
  • Peripherals
  • Laptops/Notebooks
  • Storage
  • Servers
  • Desktops
  • New Users
  • Misc
  • Apple
Software
  • System Utilities
  • Industry Specific
  • Network Management
  • Photos / Graphics
  • Page Layout
  • VMWare
  • Misc
  • Web Development
  • OS
  • CYGWIN
  • Voice Recognition
  • Message Queue
  • Quality Assurance
  • Security
  • Firewalls
  • MultiMedia Applications
  • Development
  • Database
  • Office / Productivity
  • Business Management
  • OS/2 Apps
  • Server Software
  • Internet / Email
ITPro
  • OS
  • Storage
  • Encryption
  • Operating Systems Security
  • Apple Hardware
  • Laptops & Notebooks
  • Servers
  • Networking Hardware
  • Peripherals
  • Devices
  • Displays / Monitors
  • WebTrends / Stats
  • Search Engines
  • Firewalls
  • WebApplications
  • IDS
  • Vulnerabilities
  • Email Clients
  • File Sharing
  • Spy / Ad Blockers
  • Web Browsers
  • Web Servers
  • Networking
  • Anti-Virus
  • Chat / IM
  • Anti Spam
Developer
  • Web Servers
  • Web Browsers
  • Game Programming
  • Dev Tools
  • Industry Specific
  • Office / Productivity
  • Database
  • CYGWIN
  • Web Development
  • Search Engines
  • File Sharing
  • WebTrends / Stats
  • Programming
  • Content Management
  • Application Servers
  • Protocols
Storage
  • Removable Backup Media
  • Storage Technology
  • Servers
  • Grid
  • Remote Access
  • Backup / Restore
  • Misc
  • Hard Drives
OS
  • Miscellaneous
  • Security
  • Development
  • Linux
  • VMWare
  • MainFrame OS
  • Unix
  • Apple
  • OS / 2
  • AS / 400
  • BeOS
  • Microsoft
  • VMS / OpenVMS
Database
  • Oracle
  • Miscellaneous
  • MySQL
  • Software
  • Sybase
  • Contact Management
  • PostgreSQL
  • Data Manipulation
  • Clarion
  • InterSystems Cache
  • Siebel
  • MUMPS
  • OLAP
  • SQLBase
  • SAS
  • GIS & GPS
  • 4GL
  • Berkeley DB
  • DB2
  • Informix
  • Interbase / Firebird
  • FoxPro
  • Reporting
  • LDAP
  • Filemaker Pro
  • MS SQL Server
  • dBase
  • MS Access
Security
  • Misc
  • Web Browsers
  • Software Firewalls
  • Operating Systems Security
  • File Sharing
  • Spy / Ad Blockers
  • Vulnerabilities
  • WebApplications
  • IDS
  • Anti-Virus
  • Encryption
  • Anti Spam
  • Email Clients
  • VPN
  • Chat / IM
Programming
  • Editors IDEs
  • Installation
  • Handhelds / PDAs
  • Multimedia Programming
  • System / Kernel
  • Algorithms
  • Game
  • Signal Processing
  • Project Management
  • Open Source
  • Database
  • Misc
  • Languages
  • Processor Platforms
  • Theory
Web Development
  • Scripting
  • Blogs
  • Web Servers
  • Software
  • Search Engines
  • Web Graphics
  • Images
  • Internet Marketing
  • Images and Photos
  • Components
  • Document Imaging
  • Web Languages/Standards
  • Illustration
  • WebApplications
  • Fonts
  • WebTrends / Stats
  • Authoring
  • Digital Camera Software
  • Miscellaneous
Networking
  • Protocols
  • Apple Networking
  • Network Management
  • Message Queue
  • Application Servers
  • Content Management
  • File Servers
  • Email Servers
  • Misc
  • Java Editors & IDEs
  • Wireless
  • Networking Hardware
  • Backup / Restore
  • System Utilities
  • ISPs & Hosting
  • Web Servers
  • Storage Technology
  • Removable Backup Media
  • Servers
  • Broadband
  • Grid
  • OS / 2
  • Novell Netware
  • Unix Networking
  • Windows Networking
  • Security
  • Telecommunications
  • Operating Systems
  • Linux Networking
Other
  • Community Advisor
  • Lounge
  • Community Support
  • New Net Users
  • Philosophy / Religion
  • Math / Science
  • Miscellaneous
  • URLs
  • Expert Lounge
  • Politics
  • Puzzles / Riddles
Community Support
  • Suggestions
  • New to EE
  • New Topics
  • Community Advisor
  • CleanUp
  • Announcements
  • General
  • Feedback
  • Input
  • EE Bugs
 
04.26.2008 at 11:01AM PDT, ID: 21446314

Rank: Guru

Try this:



you should then be able to use the variable value as necessary

AW
1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:
14:
Dim con As New SqlClient.SqlConnection(Connection)
        Dim cmdstr As String = "INSERT INTO testTBL(name, age, address) VALUES (@name, @age, @address)"
        Dim cmd As New SqlClient.SqlCommand(cmdstr, con)
        With cmd.Parameters
            .Add(New SqlClient.SqlParameter("@Name", name.Text))
            .Add(New SqlClient.SqlParameter("@age", age.Text))
            .Add(New SqlClient.SqlParameter("@address", address.Text))
        End With
        con.Open()
        cmd.ExecuteNonQuery()
        cmdstr = "Select @@IDNumber from tstTBL"
        cmd.CommandText = cmdstr
        Dim value As Integer = CType(cmd.ExecuteScalar(), Integer)
        con.Close()
Open in New Window
 
04.26.2008 at 11:39AM PDT, ID: 21446457

Rank: Genius

Hi Cannibal_Jack;

Please note the changes I made to the cmdstr variable and the removal of the ExecuteNonQuery and addition of the ExecuteScalar.

        Dim con As New SqlClient.SqlConnection(Connection)
        Dim cmdstr As String = "INSERT INTO testTBL(name, age, address) VALUES (@name, @age, @address); Select @@IDENTITY"
        Dim cmd As New SqlClient.SqlCommand(cmdstr, con)
        With cmd.Parameters
            .Add(New SqlClient.SqlParameter("@name", namex.Text))
            .Add(New SqlClient.SqlParameter("@age", age.Text))
            .Add(New SqlClient.SqlParameter("@address", address.Text))
        End With
        con.Open()
        IDNumber.Text = cmd.ExecuteScalar()

Fernando
Accepted Solution
 
04.29.2008 at 07:09AM PDT, ID: 21462398
Hi, I'm getting this error - "Must declare the scalar variable "@@IDNumber"
How would I fix this? thanks
 
04.29.2008 at 09:26AM PDT, ID: 21463743

Rank: Genius

Who's code are you using?
 
04.29.2008 at 02:02PM PDT, ID: 21466171

Rank: Guru

I was incorrect.  You should be able to use:

AW

1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:
14:
        Dim con As New SqlClient.SqlConnection(Connection)
        Dim cmdstr As String = "INSERT INTO testTBL(name, age, address) VALUES (@name, @age, @address)"
        Dim cmd As New SqlClient.SqlCommand(cmdstr, con)
        With cmd.Parameters
            .Add(New SqlClient.SqlParameter("@Name", name.Text))
            .Add(New SqlClient.SqlParameter("@age", age.Text))
            .Add(New SqlClient.SqlParameter("@address", address.Text))
        End With
        con.Open()
        cmd.ExecuteNonQuery()
        cmdstr = "Select @@Identity from tstTBL"
        cmd.CommandText = cmdstr
        Dim value As Integer = CType(cmd.ExecuteScalar(), Integer)
        con.Close()
Open in New Window
Assisted Solution
 
 
04.29.2008 at 07:28PM PDT, ID: 21467726
glad to be of assistance.

AW
 
 
 
20080236-EE-VQP-29 / EE_QW_2_20070628