尝试更新连接表时出现以下错误:
Notice: Undefined variable: id
Notice: Undefined variable: course_id
Cannot add or update a child row: a foreign key constraint fails (`school-project`.`students_courses`, CONSTRAINT `fk_courses` FOREIGN KEY (`course_id`) REFERENCES `courses` (`id`) ON DELETE CASCADE)0 Row inserted.
3个表是:
# Name Type Collation Attributes Null Default Comments Extra
1 id Primary Index int(11) No None AUTO_INCREMENT
2 student_id Index int(11) No None
3 course_id Index int(11) Yes NULL
# Name Type Collation Attributes Null Default Comments Extra
1 idPrimaryIndex int(11) No None AUTO_INCREMENT
2 name varchar(255) latin1_swedish_ci No None
3 phone varchar(20) latin1_swedish_ci No None
4 email text latin1_swedish_ci No None
5 image text latin1_swedish_ci No None
# Name Type Collation Attributes Null Default Comments Extra
1 idPrimaryIndex int(11) No None AUTO_INCREMENT
2 nameIndex varchar(20) latin1_swedish_ci No None
3 descr text latin1_swedish_ci No None
4 image text latin1_swedish_ci No None
代码:
function joinTable() {
global $connection;
$join_table_sql = "INSERT INTO students_courses (student_id, course_id)
VALUES ('$id', '$course_id')";
if ($connection->query($join_table_sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $join_table_sql . "<br>" . $connection->error;
}
}
我错过了什么?为什么我会得到那些错误?
答案 0 :(得分:1)
如何获得$id
和$course_id
?
第一个错误出现在PHP Notice
中,您也没有在sql
中插入空值
*这应该是评论,但我没有声誉...... :(