substring_index(substring_index(string,'x',номер),'x',-1)

SELECT `product_id`,`attribute_id`
, TRIM(BOTH '' FROM substring_index(substring_index(`text`,'x',2),'x',-1))

from (
select 701 product_id, 629 attribute_id, 4 language_id, '7.5 x 28.8 x 51 см' `text`
union all select 719 ,629 ,4 ,'4 x 58 x 51 см'
) t0

where 
`attribute_id` = 629

SELECT `product_id`,`attribute_id`, TRIM(BOTH '' FROM substring_index(substring_index(`text`,'x',2),'x',-1))
from 
`oc_product_attribute`
where 
`attribute_id` = 629

SELECT product_id, attribute_id, language_id, attr_pos, SUBSTRING_BY_INDEX(text, 'x', attr_pos) attr
FROM oc_product_attribute, (SELECT 1 attr_pos UNION SELECT 2 UNION SELECT 3) dummy

Вероятно, SQL-запрос содержит ошибку. При наличии таковой, ниже будет выведена ошибка MySQL-сервера, облегчающая диагностику проблемы.

ERROR: Неизвестная пунктуация @ 108
STR: //
SQL: DROP FUNCTION IF EXISTS SUBSTRING_BY_INDEX;# MySQL вернула пустой результат (т.е. ноль строк).


DELIMITER //

CREATE FUNCTION SUBSTRING_BY_INDEX(Source VARCHAR(255), Delimiter VARCHAR(255), Number INT)
RETURNS VARCHAR(255)
DETERMINISTIC
NO SQL
BEGIN
	IF ((LENGTH(Source) - LENGTH(REPLACE(Source, Delimiter, ''))) / LENGTH(Delimiter) >= Number - 1) AND (Number > 0)
		THEN
			RETURN TRIM(SUBSTRING_INDEX(SUBSTRING_INDEX(Source, Delimiter, Number), Delimiter, -1));
		ELSE
			RETURN '';
	END IF;
END;
//

DELIMITER ;

DELIMITER //

CREATE FUNCTION SUBSTRING_BY_INDEX(Source VARCHAR(255), Delimiter VARCHAR(255), Number INT)
RETURNS VARCHAR(255)
DETERMINISTIC
NO SQL
BEGIN
	IF ((LENGTH(Source) - LENGTH(REPLACE(Source, Delimiter, ''))) / LENGTH(Delimiter) >= Number - 1) AND (Number > 0)
		THEN
			RETURN TRIM(SUBSTRING_INDEX(SUBSTRING_INDEX(Source, Delimiter, Number), Delimiter, -1));
		ELSE
			RETURN '';
	END IF;
END;
//

DELIMITER ;

SQL-запрос:

DELIMITER // CREATE FUNCTION SUBSTRING_BY_INDEX(Source VARCHAR(255), Delimiter VARCHAR(255), Number INT) RETURNS VARCHAR(255) DETERMINISTIC NO SQL BEGIN IF ((LENGTH(Source) - LENGTH(REPLACE(Source, Delimiter, ''))) / LENGTH(Delimiter) >= Number - 1) AND (Number > 0) THEN RETURN TRIM(SUBSTRING_INDEX(SUBSTRING_INDEX(Source, Delimiter, Number), Delimiter, -1)); ELSE RETURN ''; END IF; END; // DELIMITER ;

Ответ MySQL: Документация

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'RETURNS VARCHAR(255)
DETERMINISTIC
NO SQL
BEGIN
    IF ((LENGTH(Source) - LENGT' at line 1