我被赋予了MySQL查询:这是在PHP中使用MySQL @variables的正确方法吗?
SET @COD_PAIS = 3, @ID_CARTERA = 8;
SELECT
C.DES_PERMADURACION AS Item
, B.POR_MADURACION AS Percentage_Distribution
FROM dficha_mes A
JOIN det_maduraciones B
ON (A.ID_FICHA_MES = B.ID_FICHA_MES)
JOIN mpermaduraciones C
ON (B.ID_PERMADURACION = C.ID_PERMADURACION)
WHERE A.ID_CARTERA = @ID_CARTERA
AND A.COD_PAIS = @COD_PAIS
AND B.F_CORTE = (
SELECT
F_ULT_CIERRE_MENSUAL
FROM mpaises
WHERE COD_PAIS = @COD_PAIS);
表在西班牙的行和列。
当我运行使用此查询PHP的mysql_query()
,我得到以下错误:
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 'SELECT
C.DES_PERMADURACION AS Item
, B.POR_MADURACION AS Percentage_Distrib' at line 2
但是这个查询从MySQL工作台,甚至SQLyog的完美运行。
任何指针?
非常感谢再次提醒我这件事。我将把代码库移动到PDO;只是遗留代码有上述问题。我很快就会迁徙。谢谢。 –