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 564095
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 13, 20262026-05-13T12:43:49+00:00 2026-05-13T12:43:49+00:00

2 tables: items(item_id, …) tags(tag_id, item_id) How do I select the items (from the

  • 0

2 tables:
items(item_id, …)
tags(tag_id, item_id)

How do I select the items (from the “items” table) that have a particular tag associated with them (in the “tags” table) along with all the tags associated with these items?

  • 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
    2026-05-13T12:43:49+00:00Added an answer on May 13, 2026 at 12:43 pm

    Use:

    SELECT i.*, 
           t.tag_id
      FROM ITEMS i
      JOIN TAGS t ON t.item_id = i.item_id
     WHERE i.item_id IN (SELECT x.item_id
                           FROM TAGS x 
                          WHERE x.tag_id = ?)
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I have items and itemgroup tables in my database: CREATE TABLE [items]( [item_id] [int]
I have 2 tables in my database: CREATE TABLE [items]( [item_id] [int] IDENTITY(1,1) NOT
3 tables: items(item_id, ...) tags(item_id, tag_name) downloads(item_id, ...) How do I select a single
I have the following tables: create temporary table Items (item_id int, item_name varchar(10)); create
In a postgresql database I have 2 tables like these ones: Table items: item_id
I have 3 tables that look like this: Table 1 (main table): item_id (item_id,
I have two tables that looks like this: Table: items id | itemId ---|------
I have three mysql tables items =========== id title items_in_categories ============================ id item_id category_id
I have three tables filters (id, name) items(item_id, name) items_filters(item_id, filter_id, value_id) values(id, filter_id,
Given m2m relation: items-categories I have three tables: items , categories and items_categories that

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.