所以这些是2个查询:
SELECT * FROM CourseGrade;
SELECT * FROM LetterGrades;
当我执行此UNION以查看哪个LetterGradeId具有哪个LetterGrade:
SELECT *
FROM CourseGrade
INNER JOIN LetterGrades
ON CourseGrade.LetterGradeId=LetterGrades.LetterGradeId;
我收到错误说
Invalid column name 'LetterGradeId'.
谁能告诉我哪里出错?
答案 0 :(得分:2)
你拼错了名字cg.LetterGradedId
SELECT *
FROM CourseGrade cg
INNER JOIN LetterGrades lg
ON cg.LetterGradedId=lg.LetterGradeId;
答案 1 :(得分:1)
尝试这样:
SELECT *
FROM CourseGrade cg
INNER JOIN LetterGrades lg
ON cg.LetterGradedId=lg.LetterGradeId;
答案 2 :(得分:0)
您的查询是正确的,但您错过了拼写第一个表格中的列名称。
这应该有效:
SELECT *
FROM CourseGrade
INNER JOIN LetterGrades
ON CourseGrade.LetterGradedId=LetterGrades.LetterGradeId;
请注意,您将colum命名为“LetterGrade d Id”而不是LetterGradeId:)