Mysql Error: Each derived table must have its own alias

Question

I’m trying to put two tables together questionario and resposta and select from the table questionario two dates, this was the code I tried in php:

$verificar=mysqli_query($link,"SELECT questionario.*, resposta.* from questionario 
JOIN resposta on questionario.pergunta_id=resposta.pergunta_id,
(select COUNT(pergunta_id) as total, classificacao from questionario 
where data BETWEEN '".$_POST["datainicial"]."' AND '".$_POST["datafinal"]."' 
group by classificacao");
        while ($linha = mysqli_fetch_assoc($verificar)) {...}

But it returns this error:

Warning: mysqli_fetch_assoc() expects Parameter 1 to be mysqli_result, Boolean Given

I’ve reviewed the Internet, and I can’t find any solution.

So I tried the following on phpMyAdmin:

SELECT questionario.*, resposta.* from questionario 
JOIN resposta on questionario.pergunta_id=resposta.pergunta_id,
     (select COUNT(pergunta_id) AS total, classificacao from questionario where data 
BETWEEN '2018/06/07' AND '2018/06/08' GROUP by classificacao)

As expected the query did not work, it returns this error

Each derived table must have its own alias...

Here is the structure of the tables

Examples of some data inserted in tables

Table inquiry


Response table

Answer 1

Ana, use this query below:

Select COUNT(questionario.pergunta_id) total,questionario.*, resposta.* from 
questionario 
JOIN resposta on questionario.pergunta_id = resposta.pergunta_id
Where questionario.data BETWEEN '2018-06-07' AND '2018-06-08'
group by questionario.pergunta_id

Answer to PHP error: Mysql error "expects Parameter 1 to be Resource, Boolean Given in"