PHP MySQL - 如果值不存在于两个表中则插入

Question

所以我有 2 个 table，“rooms”和“scheds”。我还有一个输入表单，其中将值插入到这 2 tables.

这是输入的样子：
房间号：________________
Capacity:________________
-[提交按钮]-

程序：
当我输入房间 ID 和容量时，它应该按原样将值插入 rooms table，然后插入 scheds table 它将在 roomid 列中插入 2 个相同的房间 ID 值，并自动将“1st”和“2nd”连续添加到 semester列。

我遇到的唯一问题是，我正在考虑两种方法来解决这个问题，但不幸的是，我似乎无法获得正确的代码：

1. 如果 Room ID 在 rooms table 中具有完全相同的值，则不应再将其插入到两个 table 中，而是回显房间已经存在。

2. 如果 Room ID 在 scheds table 中已经有“1st”和“2nd”学期值，则回显房间已经存在。

这是“房间”table.

+----+--------+----------+
| id | roomid | capacity |
+----+--------+----------+
| 1  | NB201  | 30       |
+----+--------+----------+
| 2  | NB202  | 30       |
+----+--------+----------+

这是“scheds”table。

+----+--------+----------+
| id | roomid | semester |
+----+--------+----------+
| 1  | NB201  | 1st      |
+----+--------+----------+
| 2  | NB201  | 2nd      |
+----+--------+----------+
| 3  | NB202  | 1st      |
+----+--------+----------+
| 4  | NB202  | 2nd      |
+----+--------+----------+

以下是我目前正在处理的代码。

<?php
 if($_POST){  
    try{
        //write query
        $query = "INSERT INTO rooms SET roomid = ?, capacity = ?, roomimage = ?";

        //prepare query for excecution
        $stmt = $con->prepare($query);

        //bind the parameters
        $stmt->bindParam(1, $_POST['roomid']);

        $stmt->bindParam(2, $_POST['capacity']);

        $stmt->bindParam(3, $_POST['roomimage']);

        // Execute the query
        if($stmt->execute()){
            echo "<div class='btn-success'>Room was successfully saved.</div>";
        }else{
            echo "<div class='btn-danger'>Unable to save room.</div>";
        }

    }catch(PDOException $exception){ //to handle error
        echo "<div class='btn-danger'>Error: " . $exception->getMessage() . "</div>";
    }
}
?>

<?php
 if($_POST){ 
    try{        
            //-----2 semesters-----
            if($_POST['semester']=='2'){
            //write query
            $query_roomsched1 = "INSERT INTO sched_2014_2015 SET roomid = ?, semester = ?";

            //prepare query for excecution
            $stmt = $con->prepare($query_roomsched1);

            //bind the parameters
            $stmt->bindParam(1, $_POST['roomid']);

            $stmt->bindValue(2, '1st');

                // Execute the query
                if($stmt->execute()){
                    echo "<div class='btn-success'>Schedule table for 1st semester  was successfully created.</div>";
                }else{
                    echo "<div class='btn-danger'>Unable to create schedule table for 1st semester.</div>";
                }

            //write query
            $query_roomsched2 = "INSERT INTO sched_2014_2015 SET roomid = ?, semester = ?";

            //prepare query for excecution
            $stmt = $con->prepare($query_roomsched2);

            //bind the parameters
            $stmt->bindParam(1, $_POST['roomid']);

            $stmt->bindValue(2, '2nd');

                // Execute the query
                if($stmt->execute()){
                    echo "<div class='btn-success'>Schedule table for 2nd semester was successfully created.</div>";
                }else{
                    echo "<div class='btn-danger'>Unable to create schedule table for 2nd semester.</div>";
                }
             }catch(PDOException $exception){ //to handle error
        echo "<div class='btn-danger'>Error: " . $exception->getMessage() . "</div>";
    }
}
?>

Answer 1

如果要检查两个表中是否都存在 "id's"，可以使用 JOIN 和条件语句来使用以下内容。你可以根据自己的情况来决定是否插入。

<?php
$user = 'xxxx';
$pass = 'xxxx';

$con = new PDO("mysql:host=localhost;dbname=your_db", $user, $pass);
$con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

$stmt = $con->prepare("

    SELECT * 
    FROM rooms 
    LEFT JOIN scheds 
        ON rooms.roomid=scheds.roomid 
    WHERE rooms.roomid= 'NB201'

");

$stmt->execute();

if($stmt->rowCount() > 0){
    echo "It exists."; // do NOT INSERT
}

else{
    echo "It does not exist."; // do the INSERT
}

您也可以通过更改

来尝试 INNER JOIN

LEFT JOIN scheds

到

INNER JOIN scheds

如果使用 LEFT JOIN 没有得到想要的结果。

PHP MySQL - 如果值不存在于两个表中则插入

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