我正在编写一个程序,在数据库(Mysql-Workbench)中比较 2 个不同表的 2 列(difficulty,difficulty_student)
(EXERCISES,ANSWERS) 表 ANSWERS 的 difficulty_choice
列中。
这就是我的意思:
我正在使用 VARCHAR(是或否)比较两个表。如果用户更改了练习的难度,则单元格将为"is",如果没有更改,则单元格将为“否”。
这些是我的表格:
CREATE TABLE exercises (
exercise_id INT,
difficulty VARCHAR(30),
PRIMARY KEY(exercise_id)
);
CREATE TABLE answers(
exercise_id_fk INT,
student_id INT,
difficulty_change VARCHAR(3),
difficulty_student VARCHAR(30),
FOREIGN KEY(exercise_id_fk) REFERENCES exercises(exercise_id)
);
我的问题是,在用户按下程序中的SUBMIT按钮之前,ANSWERS 表的行并不存在。因此,我只能在 Mysql-Workbench 中使用以下命令来比较表中的列。
我需要的是当用户按提交时比较difficulty_change
中的列。你能帮我做吗?我无法得到它。
我已经成功地使用以下代码比较了列,但我希望从程序中调用它们,这样我就不必每次都去 Mysql-Workbench 执行它们。
SELECT e.difficulty, a.difficulty_student,
case when e.difficulty = a.difficulty_student then 'NO' else 'YES'
END as difficulty_change
FROM exercises e
INNER JOIN answers a on e.exercise_id=a.exercise_id_fk;
UPDATE answers a
INNER JOIN exercises e on e.exercise_id=a.exercise_id_fk
set a.difficulty_change = case
when e.difficulty = a.difficulty_student then 'NO' else 'YES' END
where e.exercise_id=a.exercise_id_fk;
这是我的 PHP,它可能会有所帮助:
<?php
if (isset($_POST['submit'])) {
$user_id = $_SESSION['user_id'];
$user_check_query = "SELECT * FROM users WHERE id='$user_id'";
if(isset($_POST['choice'], $_POST['choose'])){
$choice_answer=$_POST['choice'];
$difficulty=$_POST['choose'];
// */$user_id = $_SESSION['user_id'];*/
$query = "INSERT INTO answers (exercise_id_fk, student_id, difficulty_student, choice_answer) VALUES ('$id','$user_id', '$difficulty', '$choice_answer')";
$sql=mysqli_query($conn,$query);
}
}
?>
最佳答案
尝试:
$query = "INSERT INTO answers (exercise_id_fk, student_id, difficulty_change, difficulty_student, choice_answer) VALUES ('$id','$user_id', (SELECT IF(difficulty='$difficulty','NO','YES') FROM exercises WHERE exercise_id=$id), '$difficulty', '$choice_answer')";
关于php - 通过 PHP 比较数据库中的列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51874744/