SQL Multiple Left joins - mysql

Question

I have a table called documents with a row called Nigel Harding
DOCUMENTS
id | label
24 | Nigel Harding
He has been tagged with two other documents one with an ID of 1 & 12 that table is called document tags
DOCUMENT_TAGS
id | label | Document_id
1 | TAG A | 24
12 | TAG B | 24
I am trying to create a query where I can find one result where Nigel Harding will appear once if searching for the tags 1 AND 12 but i'm having no luck.
I figured out the query for searching one tag id but i'm trying to do the query for both tags.
SELECT documents.id
FROM documents
LEFT JOIN documents_tags
ON documents.id=documents_tags.document_id
WHERE documents_tags.tag_id = 1 ORDER BY documents.label
I understand why adding...
AND documents_tags.tag_id = 12
...to the end of that will not work but i'm not sure what i need to do get the correct query display my one result as my understanding of SQL is very basic.

Answer 1

If I understand your question, you're just lacking IN and DISTINCT:
SELECT DISTINCT documents.id
FROM documents
LEFT JOIN documents_tags
ON documents.id=documents_tags.document_id
WHERE documents_tags.tag_id IN (1, 12) ORDER BY documents.label
Using IN you'll be getting documents for any of the tag ids specified, and using DISTINCT you'll make sure to get each document id only once.
EDIT: Since you're ordering by documents.label, I'm guessing that's what you want displayed. Don't forget you can change your SELECT statement to:
SELECT DISTINCT documents.label
or even to:
SELECT DISTINCT documents.id, documents.label

Answer 2

I think the most flexible way to approach this type of problem is to use aggregation with a having clause. Here is one example:
SELECT dt.document_id
FROM documents_tags dt
GROUP BY dt.document_id
HAVING sum(dt.tag_id = 1) > 0 and
sum(dt.tag_id = 12) > 0;
Each condition in the having clause counts the number of document tags that are 1 (or 12) and the filter passes only when both are found. You could also write this as:
SELECT dt.document_id
FROM documents_tags dt
WHERE dt.tag_id in (1, 12)
GROUP BY dt.document_id
HAVING count(distinct dt.tag_id) = 2;

Answer 3

Another way to accomplish this (adjusted to search by tag label which I think is your goal, if not just replace dt.Label = 'etc' with dt.ID = etc):
SELECT documents.id
FROM documents d
WHERE
exists (select dt.id from documents_tags dt
where dt.document_id = d.document_id and dt.label = 'TAG A')
AND
exists (select dt.id from documents_tags dt
where dt.document_id = d.document_id and dt.label = 'TAG B')

Answer 4

one WHERE word is enough:
SELECT documents.id
FROM documents
LEFT JOIN documents_tags
ON documents.id=documents_tags.document_id
WHERE (documents_tags.tag_id = 1 OR documents_tags.tag_id = 12) ORDER BY documents.label
alternatively, if you are going to work with n different tags and don't want to add OR documents_tags.tag_id = x ever time you can use IN operator
SELECT documents.id
FROM documents
LEFT JOIN documents_tags
ON documents.id=documents_tags.document_id
WHERE documents_tags.tag_id IN(1,12) ORDER BY documents.label

Answer 5

It sounds like you're trying to aggregate the results when you search for multiple document tags, so you should use GROUP BY, which is for grouping multiple rows into one row:
SELECT documents.id
FROM documents
LEFT JOIN documents_tags ON documents.id = documents_tags.document_id
WHERE documents_tags.tag_id IN (1, 12)
GROUP BY documents.id
This will give you one row for each document that matches the given tag ids. And you can generalize to any list of tag ids by adding to the list without too much pain. You could even add a COUNT(*) to the select statement to find out how many tag ids where matched for the given document.