两个表的内部联接[更新]

Question

tbl老师

lrn
fname
lname
email
image
schedule

tbl用户

lrn
fname
lname
email
image
account-type

$ id = $ _ SESSION ['id'];

$ get_record_sched = mysqli_query（$ db，“从教师INNER JOIN用户上的选择时间表，在Teacher.lrn = users.lrn WHERE Teacher.lrn ='$ id'“中）；

我想根据登录帐户的ID回显确切的计划。但是有了这个查询，就没有输出...

Answer 1

您说您有$id个会话。假设lrn列是ID，则

Select t.schedule
From teacher t
Where t.lrn = $id

无需将表与用户表联接。

请注意，建议进行参数绑定。

$statement = $connection->prepare("Select t.schedule From teacher t Where t.lrn = ?");
$statement->bind_param("s", $id);

Answer 2

在您的条件不正确的地方。他们没有任何参数可以验证条件。所以要这样更改

"SELECT schedule FROM teacher INNER JOIN users  ON teacher.id = users.id WHERE 1  "

OR

在连接条件相同的地方使用

"SELECT schedule FROM teacher INNER JOIN users  ON teacher.id = users.id AND teacher.schedule = users.id   WHERE 1  "

OR

如果您已在会话中记录了已使用的ID，例如

$id=$_SESSION['user_id'];

"SELECT schedule FROM teacher INNER JOIN users ON teacher.id = users.id WHERE teacher.schedule='$id' "