Ваш оператор select
получает 8 полей из таблицы ITEMS
:
SELECT
...
ITEMS.ItemName,
ITEMS.PhysDef,
ITEMS.MDef,
ITEMS.Dodge,
ITEMS.Damage,
ITEMS.CritMultiplier,
ITEMS.Range,
ITEMS.AttackSpeed,
...
Однако таблица ITEMS
является , а не , на которую ссылается ваше предложение from
:
FROM
(
(
(
(
(
(
(
CHARACTER INNER JOIN RACE_CHARACTER ON
CHARACTER.CharacterID=RACE_CHARACTER.CharacterID
)
LEFT JOIN JOB_CHARACTER ON
CHARACTER.CharacterID=JOB_CHARACTER.CharacterID
)
LEFT JOIN INVENTORY ON
CHARACTER.CharacterID=INVENTORY.CharacterID
)
INNER JOIN INVENTORY ON ----------< INVENTORY table referenced twice
ITEMS.ItemID =INVENTORY.ItemID
)
INNER JOIN JOB ON
JOB.JobID=JOB_CHARACTER.JobID
)
LEFT JOIN SKILLS ON
JOB.JobID=SKILLS.JobID
)
INNER JOIN RACE ON
RACE.RaceID=RACE_CHARACTER.RaceID
)
Я должен представить, что код SQL должен быть изменен на что-то вроде:
SELECT
CHARACTER.CharacterName,
CHARACTER.Alignment,
INVENTORY.Equipped,
ITEMS.ItemName,
ITEMS.PhysDef,
ITEMS.MDef,
ITEMS.Dodge,
ITEMS.Damage,
ITEMS.CritMultiplier,
ITEMS.Range,
ITEMS.AttackSpeed,
JOB_CHARACTER.JobLevel,
RACE_CHARACTER.RacialLevel,
RACE.RaceName,
RACE.Strength,
RACE.Skill,
RACE.Vitality,
RACE.Arcane,
RACE.Spirit,
RACE.Charisma,
RACE.Luck,
JOB.JobName,
JOB.HP,
JOB.AttackBonus,
JOB.Agility,
JOB.Might,
JOB.SpellPower,
JOB.Vital,
JOB.Nimble,
JOB.Mental,
JOB.Curese,
JOB.SpellCasting,
JOB.ManaBase,
JOB.ManaType,
JOB.Ki,
SKILLS.Alchemy,
SKILLS.Awareness,
SKILLS.Climb,
SKILLS.Coach,
SKILLS.Construction,
SKILLS.Decieve,
SKILLS.DisarmMechanism,
SKILLS.DiscernTruth,
SKILLS.Dishearten,
SKILLS.Fly,
SKILLS.Forge,
SKILLS.Gymnastics,
SKILLS.Identify,
SKILLS.Leadership,
SKILLS.Lore_9Realms,
SKILLS.Lore_Alternative,
SKILLS.Lore_Arcane,
SKILLS.Lore_Arithmancy,
SKILLS.Lore_Divine,
SKILLS.Lore_Geography,
SKILLS.Lore_Nature,
SKILLS.Lore_Nobility,
SKILLS.Lore_Religion,
SKILLS.Lore_Spiritual,
SKILLS.Medical,
SKILLS.Performance,
SKILLS.Ride,
SKILLS.Steal,
SKILLS.Stealth,
SKILLS.Subterfuge,
SKILLS.Swim,
SKILLS.Tailor,
SKILLS.UseContraption,
SKILLS.Wilderness
FROM
(
(
(
(
(
(
(
CHARACTER INNER JOIN RACE_CHARACTER ON
CHARACTER.CharacterID=RACE_CHARACTER.CharacterID
)
LEFT JOIN JOB_CHARACTER ON
CHARACTER.CharacterID=JOB_CHARACTER.CharacterID
)
LEFT JOIN INVENTORY ON
CHARACTER.CharacterID=INVENTORY.CharacterID
)
LEFT JOIN ITEMS ON
ITEMS.ItemID =INVENTORY.ItemID
)
LEFT JOIN JOB ON
JOB.JobID=JOB_CHARACTER.JobID
)
LEFT JOIN SKILLS ON
JOB.JobID=SKILLS.JobID
)
INNER JOIN RACE ON
RACE.RaceID=RACE_CHARACTER.RaceID
)
WHERE
CHARACTER.CharacterID = 1 AND
JOB.JobID = 3 AND
RACE.RaceID = 6
Обратите внимание, что я изменил пару inner joins
на left joins
, потому что если вы используете inner join
на столе, который является правым от left join
(или слева от right join
) , тогда вы получите неоднозначные внешние объединения ошибка.