Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 8762131
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 13, 20262026-06-13T15:27:14+00:00 2026-06-13T15:27:14+00:00

I have a table with 2 columns, user and userMessage with table name as

  • 0

I have a table with 2 columns, user and userMessage with table name as Messages
I have another table with 2 columns as user and userMessageCount with table name as MessageCount.
I am very new to programming. I need to write an update query or procedure something that will count the number of messages per user from Messages and store the count in MessageCount.

I am not able to understand the proper way of doing this. I understand this is a very basic question and I should study to solve this. i am running very late for a deadline. Please help me on this query.

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-06-13T15:27:15+00:00Added an answer on June 13, 2026 at 3:27 pm
    Insert Into MessageCount 
    Select user, count(*)
    From Messages
    Group by user
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I have a user profile table with columns User Name , Manager and many
i have a winform datagridview: DataTable table = new DataTable(); table.Columns.Add(Check, typeof(bool)); table.Columns.Add(User, typeof(string));
I have a database test with table name table with two columns user &
I have a MySQL table containing columns for user IP (IP) and their name
I have a table with columns id, user I want to group by column
I have a table called LOGS with the following columns: id - user -
In my table accounts I have four columns like user, pass, column1, column2 I
I have a table in HTML which has three columns i.e Error - User
I have a table of users with three columns. Username Accepted Rejected User 1
Consider i have a user table and I have three columns mobilePhone , homePhone

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.