Can SELECT, SELECT COUNT and cross reference tables be handled by just one query?

问题

I have a page that displays a list of projects. With each project is displayed the following data retrieved from a mysqli database:

Title
Subtitle
Description
Part number (1 of x)
The total number of photos associated with that project
A randomly selected photo from the project
A list of tags

Projects are displayed 6 per page using a pagination system

As this is based on an old project of mine, it was originally done with sloppy code (I was just learning and did not know any better) using many queries. Three, in fact, just for items 5-7, and those were contained within a while loop that worked with the pagination system. I'm now quite aware that this is not even close to being the right way to do business.

I am familiar with INNER JOIN and the use of subqueries, but I'm concerned that I may not be able to get all of this data using just one select query for the following reasons:

Items 1-4 are easy enough with a basic SELECT query, BUT...
Item 5 needs a SELECT COUNT AND...
Item 6 needs a basic SELECT query with an ORDER by RAND LIMIT 1 to select one random photo out of all those associated with each project (using FilesystemIterator is out of the question, because the photos table has a column indicating 0 if a photo is inactive and 1 if it is active)
Item 7 is selected from a cross reference table for the tags and projects and a table containing the tag ID and names

Given that, I'm not certain if all this can (r even should for that matter) be done with just one query or if it will need more than one query. I have read repeatedly how it is worth a swat on the nose with a newspaper to nest one or more queries inside a while loop. I've even read that multiple queries is, in general, a bad idea.

So I'm stuck. I realize this is likely to sound too general, but I don't have any code that works, just the old code that uses 4 queries to do the job, 3 of which are nested in a while loop.

Database structure below.

Projects table:

+-------------+---------+----------+---------------+------+
| project_id  | title   | subtitle | description   | part |
|---------------------------------------------------------|
|       1     |   Chevy | Engine   | Modify        |  1   |
|       2     |   Ford  | Trans    | Rebuild       |  1   |
|       3     |   Mopar | Diff     | Swap          |  1   |
+-------------+---------+----------+---------------+------+

Photos table:

+----------+------------+--------+
| photo_id | project_id | active |
|--------------------------------|
|     1    |     1      |    1   |
|     2    |     1      |    1   | 
|     3    |     1      |    1   |
|     4    |     2      |    1   |
|     5    |     2      |    1   |
|     6    |     2      |    1   |
|     7    |     3      |    1   |
|     8    |     3      |    1   |
|     9    |     3      |    1   |
+----------+------------+--------+

Tags table:

+--------+------------------+
| tag_id |        tag       |
|---------------------------|
|    1   | classic          |
|    2   | new car          |
|    3   | truck            |
|    4   | performance      |
|    5   | easy             |
|    6   | difficult        |
|    7   | hard             |
|    8   | oem              |
|    9   | aftermarket      |
+--------+------------------+

Tag/Project cross-reference table:

+------------+-----------+
| project_id | tag_id    |
|------------------------|
|      1     |     1     |
|      1     |     3     |
|      1     |     4     |
|      2     |     2     |
|      2     |     5     |
|      3     |     6     |
|      3     |     9     |
+------------+-----------+

I'm not asking for the code to be written for me, but if what I'm asking makes sense, I'd sincerely appreciate a shove in the right direction. Often times I struggle with both the PHP and MySQLi manuals online, so if there's any way to break this down, then fantastic.

Thank you all so much.

回答1:

You're able to do subqueries inside your SELECT clause, like this:

SELECT 
    p.title, p.subtitle, p.description, p.part,
    (SELECT COUNT(photo_id) FROM Photos where project_id = p.project_id) as total_photos, 
    (SELECT photo_id FROM Photos where project_id = p.project_id ORDER BY RAND LIMIT 1) as random_photo
FROM projects as p

Now, for the list of tags, as it returns more than one row, you can't do a subquery and you should do one query for every project. Well, in fact you can if you return all the tags in some kind of concatenation, like a comma separated list: tag1,tag2,tag3... but I don't recommend this one time that you will need to explode the column value. Do it only if you have many many projects and the performance to retrieve the list of tags for each individual project is fairly low. If you really want, you can:

SELECT 
    p.title, p.subtitle, p.description, p.part,
    (SELECT COUNT(photo_id) FROM Photos where project_id = p.project_id) as total_photos, 
    (SELECT photo_id FROM Photos where project_id = p.project_id ORDER BY RAND LIMIT 1) as random_photo,
    (SELECT GROUP_CONCAT(tag SEPARATOR ', ') FROM tags WHERE tag_id in (SELECT tag_id FROM tagproject WHERE project_id = p.project_id)) as tags
FROM projects as p

回答2:

As you said from item 1 to 4 you already have the solution.

Add to the same query a SQL_CALC_FOUND_ROWS instead of a SELECT COUNT to solve the item 5.

For the item 6 you can use a subquery or maybe a LEFT JOIN limiting to one result.

For the latest item you can also use a subquery joining all the tags in a single result (separated by comma for instance).

来源：https://stackoverflow.com/questions/41646175/can-select-select-count-and-cross-reference-tables-be-handled-by-just-one-query

标签

php

mysqli

subquery

inner-join

nested-query