Question

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？

Answer 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...... :(

外键约束失败错误

1 个答案: