Question

所以我有两张桌子，“房间”和“ scheds ”。我还有一个输入表单，其中值被插入到这两个表中。

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

步骤：
当我输入房间ID和容量时，它应该按原样在房间表中插入值，然后在 scheds 表中插入2个相同的房间ID值进入 roomid 列并自动将“1st”和“2nd”连续添加到 semester 列。

我遇到问题的唯一部分是，我正在考虑如何处理这个问题的两种方法，但遗憾的是我似乎找不到合适的代码：

1.如果房间ID在房间表中具有相同的确切值，则它不应再在两个表中插入，而是回显该房间已存在。

2.如果房间ID在 scheds 表中已经具有“1st”和“2nd”学期值，则回显房间已经存在。

这是“ rooms ”表。

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

这是“ scheds ”表格。

+----+--------+----------+
| 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

如果你想检查两者是否＆＃34; id＆＃39; s＆＃34;存在于两个表中，您可以使用以下命令使用JOIN和条件语句。您可以根据自己的意愿进行插入或不进行插入。

<?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

如果您未获得结果，那么您希望使用LEFT JOIN。

PHP MySQL - 如果两个表中不存在该值，则插入

1 个答案: