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Скачать или смотреть How to Find Records Where All Values in a Column Are Contained in Another Table Using SQL

  • vlogize
  • 2025-09-21
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How to Find Records Where All Values in a Column Are Contained in Another Table Using SQL
How to find records which ALL values in column are contained in another table?sqlsql server
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Описание к видео How to Find Records Where All Values in a Column Are Contained in Another Table Using SQL

Learn how to efficiently retrieve all records in one table that have corresponding values in another table utilizing SQL. Perfect for database queries and data analysis!
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This video is based on the question https://stackoverflow.com/q/62778030/ asked by the user 'AtlasPromotion' ( https://stackoverflow.com/u/8891499/ ) and on the answer https://stackoverflow.com/a/62778117/ provided by the user 'Gordon Linoff' ( https://stackoverflow.com/u/1144035/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.

Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: How to find records which ALL values in column are contained in another table?

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The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license.

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Understanding the Problem

In database management, it’s often necessary to ensure that a certain set of values in one table exists in another. A common question arises: How can we retrieve all records from one table where all values from a specific column are present in another table?

Let’s imagine a scenario with three tables:

Table1 which holds IDs and names.

Table2 that contains a mapping of FindIDs to Table1 IDs.

Table3 listing specific FindIDs we’re interested in.

The Objective

Our goal is to find all IDs from Table1 such that every value present in Table3 is also found in the FindID column of Table2 for the corresponding Table1 ID.

For example, if Table3 lists values 1 and 2, we want to return:

ID123This is because IDs 1 and 2 have both values in Table2, and ID 3 has all 1, 2, and 3.

If Table3 instead contains values 1, 2, and 3, then only ID 3 would be returned, as it is the only ID that encompasses all listed FindIDs.

The SQL Query Solution

To achieve this result, we will need to leverage SQL's aggregation and grouping functions. Below, we break down the SQL query that will help us find the required IDs.

Step-by-Step Breakdown

Select relevant fields: We want to select unique Table1 IDs from Table2.

Join the tables: We will join Table2 with Table1 based on the mapping from FindID to the corresponding Table1 ID.

Group by: We group by the Tabe1 ID in order to apply an aggregate function on the results.

Count values: Use the HAVING clause to ensure that the count of corresponding FindIDs matches the total number of rows in Table3.

SQL Code

Here is the SQL code that reflects the above strategy:

[[See Video to Reveal this Text or Code Snippet]]

Explanation of the SQL Code

SELECT t2.Table1ID: This specifies that we want the Table1 IDs from Table2.

FROM Table2 t2 JOIN Table1 t1 ON t2.FindID = t1.ID: Here we are joining Table2 and Table1 on the condition that the FindID from Table2 corresponds to an ID in Table1.

GROUP BY t2.Table1ID: This groups our results by the Table1 ID so that we can perform aggregate calculations.

HAVING COUNT() = (SELECT COUNT() FROM Table3): This condition ensures that only those Table1 IDs that cover all FindIDs in Table3 are selected.

Conclusion

Using SQL effectively can significantly simplify data validation and retrieval processes. This approach not only helps find the desired records but also enhances data integrity by ensuring that necessary relationships between tables are maintained.

If you ever find yourself needing to check for complete containment of values across tables, this SQL method will be your go-to solution!

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