如何在mysql中使用存储过程获取问题及其相关标签？

Question

I have a questions table with

1. id
2. title
3. body
   

And

A tags table with

1. id
2. name
3. description

And finally a question_tags relation table with

1. id
2. tag_id
3. question_id

I have the below code to get all the questions using mysql store procedures .

I need to get the question details and its related tag details using a store procedures (like this <question details,tag1 details,tag2 details... tag5 details> ).
I thought about storing the tag's id along with the question table ,but for retrieving the tag details I had to write another query which store procedure doesn't allow me to do.Same issue when trying to get tag details after getting the question id.
I tried my level best ,but yet don't know how to solve it.
any ideas or suggestions? thanks
I tried my best to explain the situation .

Answer 1

Here's the basic query for finding out the output structure like below:

 question_id    title   FirstTagName    FirstTag    SecondTagName   SecondTag   ThirdTagName    ThirdTag    FourthTagName   FourthTag   FifthTagName    FifthTag

Query:

SELECT 
t.question_id,
t.title,
MAX(CASE WHEN t.constantTagNumber = 1 THEN t.name END) AS FirstTagName,
MAX(CASE WHEN t.constantTagNumber = 1 THEN t.description END) AS FirstTag,
MAX(CASE WHEN t.constantTagNumber = 2 THEN t.name END) AS SecondTagName,
MAX(CASE WHEN t.constantTagNumber = 2 THEN t.description END) AS SecondTag,
MAX(CASE WHEN t.constantTagNumber = 3 THEN t.name END) AS ThirdTagName,
MAX(CASE WHEN t.constantTagNumber = 3 THEN t.description END) AS ThirdTag,
MAX(CASE WHEN t.constantTagNumber = 4 THEN t.name END) AS FourthTagName,
MAX(CASE WHEN t.constantTagNumber = 4 THEN t.description END) AS FourthTag,
MAX(CASE WHEN t.constantTagNumber = 5 THEN t.name END) AS FifthTagName,
MAX(CASE WHEN t.constantTagNumber = 5 THEN t.description END) AS FifthTag
FROM
(
    SELECT
        question_id,
        questions.title,
        tag_id,
        tags. NAME,
        tags.description,
        IF (@prev = question_id ,@c := @c + 1,@c := 1) constantTagNumber,
        @prev := question_id
    FROM    (   SELECT @prev := 0 ,@c := 1) var,    questions_tags
    INNER JOIN tags ON tags.id = questions_tags.tag_id
    INNER JOIN questions ON questions_tags.question_id = questions.ID
    ORDER BY    question_id,    tag_id
) t
GROUP BY t.question_id;

======>DEMO HERE<======

Explanation:

First look at the inner query:

SELECT
            question_id,
            questions.title,
            tag_id,
            tags. NAME,
            tags.description,
            IF (@prev = question_id ,@c := @c + 1,@c := 1) constantTagNumber,
            @prev := question_id
        FROM    (   SELECT @prev := 0 ,@c := 1) var,    questions_tags
        INNER JOIN tags ON tags.id = questions_tags.tag_id
        INNER JOIN questions ON questions_tags.question_id = questions.ID
        ORDER BY    question_id,    tag_id;

DEMO OF IT

Look there's no group by clause in this query. What I tried to achieve through this query is :

• To get all the questions and tags so that all the records selected from questions_tags table stick together having increasing number of tag ids (done by ORDER BY question_id,tag_id clause)
• I used two mysql user variables ( eg @prev and @c ). The question ID is stored in @prev variable. While traversing the rows in questions_tags table if I see a new question_id then I reset the variable @c to 1 .
• The @c variable assigns a tag number for the tags.id field which is irrespective of the original tag_number . Since you told that there can be infinite number of tags but each question can have at most five tags so the range of variable @c will be 1 to 5 ie [1,5] .
• The inner query generates this result set
• Now the outer query comes into play. There's a group by clause in the outer most query. So you will get a single row for each question.
• MAX(CASE WHEN t.constantTagNumber = 1 THEN t.name END) the role of this line is to ensure that the first tag description will be put in the corresponding column in the final result set. It's called pivoting .

Please spend some time looking on the inner query. Then think yourself "How can I generate the final result set from this table?".

If you need any help then feel free to ask.