How to compare the field of a current record with the next, in order to know if it is repeated or not?

Since this is a duplicate check, a possible solution would be a query like this:

SELECT
    p.codigo,  
    GROUP_CONCAT(p.id SEPARATOR '|') ids, 
    COUNT(*) total
FROM
    presupuestosa p 
WHERE p.obra=18 AND p.empresa=1
GROUP BY p.codigo, p.obra, p.empresa;

This query will group by and return the number of records in each group codigo, obra, empresain the column . totalThe column idsobtained with GROUP_CONCATis a plus, in case you are interested in knowing, for example, what the id of each repeated value are, they will be separated by |.

This will allow you simpler and easier to analyze results.

proof of concept

VER DEMO EN REXTESTER

I did a test based on real data (I didn't put all the records you show, just a few for brevity) and I put a case where there are three duplicates, to show that the query can have an interesting scope: know once how many duplicates are there...

    CREATE TABLE IF NOT EXISTS presupuesto_20180729 
    (
        id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
        codigo INT,
        obra INT,
        empresa INT
    )ENGINE=INNODB;

    INSERT INTO presupuesto_20180729 (codigo, obra, empresa) VALUES 
                (1,1,1),
                (2,1,1),
                (3,18,1),
                (5,18,1),
                (5,18,1),
                (5,18,1),
                (7,18,1),
                (11,18,1),
                (11,18,1)
        ;

SET sql_mode = 'ONLY_FULL_GROUP_BY';


SELECT
    p.codigo,  
    GROUP_CONCAT(p.id SEPARATOR '|') ids, 
    COUNT(*) total
FROM
    presupuesto_20180729 p 
WHERE p.obra=18 AND p.empresa=1
GROUP BY p.codigo, p.obra, p.empresa;

Result:

codigo    ids      total
-------------------------
3         3          1
5         4|5|6      3
7         7          1
11        8|9        2

Using a subquey that counts the records grouped by work and code, then a left join to translate the totals to 0 or 1.

SELECT o.id, o.obra, o.Codigo, IFNULL(n.hayDupes,0) as nro, o.cosa 
FROM Table1 o
  LEFT JOIN (
    SELECT i.Codigo, COUNT(i.Codigo) AS total, 1 AS hayDupes 
    FROM Table1 i 
    WHERE i.obra = 18
    GROUP BY i.obra, i.codigo 
    HAVING total>1
    ) AS n
  ON n.Codigo = o.Codigo 
WHERE o.obra = 18;

you have to repeat the work id so that it groups well, that can perhaps be improved

a fidel: http://www.sqlfiddle.com/#!9/ca674b4/1

explaineishon

In the subquery, the records are grouped by work and code and counted, and a column is added hayDupesthat is always equal to 1.

The magic is manifested between the fact HAVING total>1that it only returns duplicate records (2 or more) and the transformation of hayDupes = NULLa 0in the outer query.

That is: in the outer query, the column is returned nrowith the value hayDupesthat is always equal to 1, or if the subquery column hayDupesdoes not exist ( IFNULL) returns 0.

Due to LEFT JOINthe fact that the outer query only includes results from the subquery that meet the condition of having the same Codigo.

Finally, it is filtered by work id.

There may be a more efficient way using group theory to filter the intersection, but I didn't get that much =P (self join is what is generally used in these cases)

I would like to provide you with this query, it allows you to know if the record is repeated more than once, on the other hand, the word ascis redundant since it is sorted by default asc.

select tp.codigo,@rownum+1 AS nro
from(
    SELECT count(1) cantidad,pa.codigo
    FROM presupuestosa pa 
    WHERE pa.obra=18 AND pa.empresa=1 
    Group by pa.codigo
    having count(1)> 1
)tp
    ORDER BY tp.codigo

Hope this can help you.