将嵌套SQL SELECT语句中的SELECT结果分配给变量-PHP

Question

In PHP-MySQL, I am using a nested select statement as follows:

SELECT object_id FROM form_entry WHERE id=(SELECT entry_id FROM form_entry_values WHERE value='32711178-888' AND field_id = (SELECT id FROM form_field WHERE name='dealer'))

Each SELECT returns only one cell as a result. However, I need each result of these SELECTs for example I need (depth 3):

SELECT id FROM oform_field WHERE name='dealer'

also I need (depth 2):

SELECT entry_id FROM form_entry_values WHERE value='32711178-888' AND field_id = (SELECT id FROM form_field WHERE name='dealer')

and I need (depth 1):

SELECT object_id FROM form_entry WHERE id=(SELECT entry_id FROM form_entry_values WHERE value='32711178-888' AND field_id = (SELECT id FROM form_field WHERE name='dealer'))

Is there any way to assign the results of these statements during only one execution of the largest SQL query (indicated as "depth 1")? In other words, without executing three sql queries, I am looking for a way assigning the results of the SELECT statements in each depth to a variable like $result_1, $result_2, etc.

It sounds there is no way to deal with it, but thanks in advance...

Answer 1

Your query is equivalent with:

SELECT e.object_id
FROM form_entry e
  INNER JOIN form_entry_values v ON e.id=v.entry_id
  INNER JOIN form_field f ON v.field_id = f.id
WHERE v.value='32711178-888' 
  AND f.name='dealer'

and in this form the answer is straight forward: just add v.entry_id and f.id to the list of expressions in the SELECT clause:

SELECT e.object_id, v.entry_id, f.id
FROM form_entry e
  INNER JOIN form_entry_values v ON e.id=v.entry_id
  INNER JOIN form_field f ON v.field_id = f.id
WHERE v.value='32711178-888' 
  AND f.name='dealer'

Answer 2

try something like:

SELECT depth1.object_id, depth2.entry_id, depth3.id
FROM form_entry depth1
JOIN form_entry_values depth2 ON depth1.id=depth2.entry_id
JOIN form_field depth3 ON depth3.id=depth2.entry_id
WHERE
depth2.value='32711178-888'
AND depth3.name='dealer';