我正在尝试使用INTERSECT子句构建SQL查询。我得到了一系列电子邮件ID，作为回报，我需要在输入中由电子邮件ID所提供的所有教师之间分配普通学生。
表的结构为：
id | teacher_email | student_email | valid |
41 | Tom@gmail.com | May@gmail.com | 1 |
42 | Tom@gmail.com | jerry@gmail.com | 1 |
43 | Tom@gmail.com | Kestov@gmail.com | 1 |
44 | Toni@gmail.com | hdjh@kd.com | 1 |
58 | larry@gmail.com | jerry@gmail.com | 1 |
59 | gerrad@gmail.com | jerry@gmail.com | 1 |
60 | gerrad@gmail.com | Kestov@gmail.com | 1 |
61 | gerrad@gmail.com | Katrin@gmail.com | 1 |
62 | gerrad@gmail.com | Piniyara@gniail.com | 1 |
63 | gerrad@gmail.com | Ritz@gmail.com | 1 |
64 | gerrad@gmail.com | Taz@gmail.com | 1 |
65 | gerrad@gmail.com | Fensuk@gmail.com | 1 |
66 | gerrad@gmail.com | Joe@gmail.com | 1 |
67 | gerrad@gmail.com | Mustafa@gmail.com | 1 |
69 | Tom@gmail.com | Fensuk@gmail.com | 1 |
70 | Tom@gmail.com | Taz@gmail.com | 1 |
这是API代码
router.get("/api/commonstudents", (req, res) => {
console.log("Trying to retrieve common students of given teachers...")
console.log("Teacher's email IDs: " + req.query.teacher)
const teacherEmailIDs = req.query.teacher
var commonStudentsForTeachersQueryString = "";
const connection = getConnection()
console.log("no of teachers: " + teacherEmailIDs)
teacherEmailIDs.forEach(function(teacherEmailID) {
console.log(teacherEmailID);
commonStudentsForTeachersQueryString += "select student_email from register where teacher_email=? and valid=1";
commonStudentsForTeachersQueryString += " INTERSECT "
})
commonStudentsForTeachersQueryString = commonStudentsForTeachersQueryString.substr(0, commonStudentsForTeachersQueryString.lastIndexOf(" INTERSECT "))
console.log("Query formed is :: " + commonStudentsForTeachersQueryString);
connection.query(commonStudentsForTeachersQueryString, teacherEmailIDs, (errCommonStudentTeacher, rowsCommonStudentTeacher) => {
if (errCommonStudentTeacher) {
console.log("Failed to query for users: " + errCommonStudentTeacher)
console.error(errCommonStudentTeacher);
res.status(500).json({"message": "Some Internal Error Occured"})
return
}
if (rowsCommonStudentTeacher === undefined || rowsCommonStudentTeacher.length == 0) {
//rows empty or does not exist
console.log("No common students")
res.status(404).json({"message": "No Common Students found"})
}
else{
//common students found
console.log("Common students Found" + rowsCommonStudentTeacher.length)
const commonStudents = rowsCommonStudentTeacher.map((commonStudent)=>{
return commonStudent.student_email
})
res.status(200).json({"students": commonStudents})
}
})
})
这是控制台显示的内容：
Harshs-MacBook-Air:nodejs_api harshvardhan$ node app.js
Server is up and listening on 3003...
Trying to retrieve common students of given teachers...
Teacher's email IDs: Tom@gmail.com,gerrad@gmail.com,larry@gmail.com
no of teachers: Tom@gmail.com,gerrad@gmail.com,larry@gmail.com
Tom@gmail.com
gerrad@gmail.com
larry@gmail.com
Query formed is :: select student_email from register where teacher_email=? and valid=1 INTERSECT select student_email from register where teacher_email=? and valid=1 INTERSECT select student_email from register where teacher_email=? and valid=1
Failed to query for users: Error: ER_PARSE_ERROR: 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 'INTERSECT select student_email from register where teacher_email='gerrad@gmail.c' at line 1
{ Error: ER_PARSE_ERROR: 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 'INTERSECT select student_email from register where teacher_email='gerrad@gmail.c' at line 1
at Query.Sequence._packetToError (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/mysql/lib/protocol/sequences/Sequence.js:47:14)
at Query.ErrorPacket (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/mysql/lib/protocol/sequences/Query.js:77:18)
at Protocol._parsePacket (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/mysql/lib/protocol/Protocol.js:291:23)
at Parser._parsePacket (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/mysql/lib/protocol/Parser.js:433:10)
at Parser.write (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/mysql/lib/protocol/Parser.js:43:10)
at Protocol.write (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/mysql/lib/protocol/Protocol.js:38:16)
at Socket.<anonymous> (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/mysql/lib/Connection.js:91:28)
at Socket.<anonymous> (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/mysql/lib/Connection.js:525:10)
at emitOne (events.js:116:13)
at Socket.emit (events.js:211:7)
--------------------
at Pool.query (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/mysql/lib/Pool.js:199:23)
at router.get (/Users/harshvardhan/Documents/work/test/nodejs_api/routes/admin.js:129:14)
at Layer.handle [as handle_request] (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/express/lib/router/layer.js:95:5)
at next (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/express/lib/router/route.js:137:13)
at Route.dispatch (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/express/lib/router/route.js:112:3)
at Layer.handle [as handle_request] (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/express/lib/router/layer.js:95:5)
at /Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/express/lib/router/index.js:281:22
at Function.process_params (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/express/lib/router/index.js:335:12)
at next (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/express/lib/router/index.js:275:10)
at Function.handle (/Users/harshvardhan/Documents/work/test/nodejs_api/node_modules/express/lib/router/index.js:174:3)
code: 'ER_PARSE_ERROR',
errno: 1064,
sqlMessage: '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 \'INTERSECT select student_email from register where teacher_email=\'gerrad@gmail.c\' at line 1',
sqlState: '42000',
index: 0,
sql: 'select student_email from register where teacher_email=\'Tom@gmail.com\' and valid=1 INTERSECT select student_email from register where teacher_email=\'gerrad@gmail.com\' and valid=1 INTERSECT select student_email from register where teacher_email=\'larry@gmail.com\' and valid=1' }
::ffff:127.0.0.1 - GET /api/commonstudents?teacher=Tom@gmail.com&teacher=gerrad@gmail.com&teacher=larry@gmail.com HTTP/1.1 500 41 - 65.808 ms
此查询按原样使用时效果很好。
select student_email from register where teacher_email='Tom@gmail.com' and valid=1
但是当与INTERSECT与其他查询一起使用时，会产生上述错误。
如何解决此问题，我认为这与电子邮件格式及其字符编码有关。我已经清理了论坛，并尝试了一些直到这里的事情。感谢您为正确的方法提供指导。
Here是供参考的GitHub链接。
答案 0 :(得分：0)
在mysql中没有INTERSECT。
如果您真的需要一个相交，可以使用多个内部联接
select student_email
from register r
INNER JOIN (
select student_email from
register where teacher_email=? and valid=1
) t1 ON t1.student_email = r.student_email
INNER JOIN (
INTERSECT
INTERSECT select student_email from register
where teacher_email=? and valid=1
) t2 ON t2.student_email = t1.student_email
where r.teacher_email=? and r.valid=1