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 652171
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 13, 20262026-05-13T22:14:36+00:00 2026-05-13T22:14:36+00:00

I want to get the data every day within one year backward but I

  • 0

I want to get the data every day within one year backward but I have to use 365 queries for each day like:

for ($i = 0; $i<365; $i++){
    $end_day = ...; // the end time of each day
    $start_day = ...; // the start time of each day
    $query = select count(*)....where created < $end_day AND created > $start_day
}

I think that my current solution makes the system very slow. Is there any way to just use one query only?

  • 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
    2026-05-13T22:14:36+00:00Added an answer on May 13, 2026 at 10:14 pm

    Assuming that your created column is a DATETIME or TIMESTAMP, and the data stored is more granular than a day:

    SELECT COUNT(*) … GROUP BY YEAR(created), MONTH(created), DAY(created)

    COUNT is an aggregate function and will apply to each group, i.e. you will have one row per group and that row will have the number of records in that group. As we’ve made a group a day, this is the data you want.

    You cannot just group by DAY(created) as DAY returns the day of the month, so we also need to put month and year in to make sure days are discrete.

    You probably want to have a WHERE created > $start AND created < $finish to limit it to some time frame (or a WHERE YEAR(created) == 2010).

    References:

    MySQL Query GROUP BY day / month / year

    http://dev.mysql.com/doc/refman/5.1/en/group-by-functions.html

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I want to get data using CURL but I have a problem. When I
I want to get data from two tables. I have patient first name and
Whenever I want to get data from a plist file I use the following
I want to get data from a table in my MySQL database. $linkID =
I want to get data from two tables with single Entity class. How?? public
Here's the scenario, I want to get Data from the Service to Activity Whenever
i am new to asp.net, ı want to get data from url on asp.net.
I want to get only data until the <img tag. my codes are below:
I am using http://lite.facebook.com And i want to get some data from my account.
i have a table with snow data which i get delivered per hour. so

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.