MYSQL/SQLWave - 从 Join 生成不同的数据

Question

目前我有两个表，students 和 schools。它们通过 school_id(在 students 中)和 id(在 schools 中)连接。如果它有什么意义的话，这是来自带有 MYSQL 数据库的 cakePHP 应用程序，我使用 SQLwave 来运行查询。

我想要做的是生成一份特定学校 graduation_year 列为 2016 年的学生列表。我使用了以下查询，目的是生成学生的姓名、学校名称和毕业年份:

SELECT DISTINCT students.first_name, students.last_name, schools.name, students.graduation_year
FROM students JOIN schools
ON schools.id = students.school_id
WHERE schools.name = "Central City High School"
OR schools.name = "Gotham High School"
OR schools.name = "Metropolis High School"
OR schools.name = "Ivy Town High School"
AND students.graduation_year = 2016;

但是，当我运行查询时，我得到了这些学校的所有学生的列表，而不仅仅是 2016 年毕业的学生(即，我得到了那些没有列出毕业年份的学生，2014 年、2016 年等) 。我如何改进此代码以仅显示 2016 年毕业的学生？预先感谢您。

Answer 1

由于 AND 位于 OR 之前，因此您必须将 school.name 条件放在括号内。

SELECT DISTINCT students.first_name, students.last_name, schools.name, students.graduation_year
FROM students JOIN schools
ON schools.id = students.school_id
WHERE (schools.name = "Central City High School"
       OR schools.name = "Gotham High School"
       OR schools.name = "Metropolis High School"
       OR schools.name = "Ivy Town High School")
  AND students.graduation_year = 2016;

但是，使用 IN 会使代码更漂亮:

SELECT DISTINCT students.first_name, students.last_name, schools.name, students.graduation_year
FROM students JOIN schools
ON schools.id = students.school_id
WHERE schools.name IN ("Central City High School", "Gotham High School",
                       "Metropolis High School", "Ivy Town High School")
  AND students.graduation_year = 2016;