我对 SQL 世界很陌生,很抱歉我对此一无所知。
我在管理页面上有一个表格,可以将球员添加到球队数据库。提交表单后,我需要发生的是:
玩家被插入到玩家表中(player_id 是主键并在下一步中使用)。
运行一个 select 语句来获取 player_id。
- 然后将其插入到另外两个表中:
- team_players 和卡片。
以下是我尝试过的最佳表现:
if(isset($_POST['submit'])){
$first_name = mysqli_real_escape_string($con2, $_POST['first_name']);
$last_name = mysqli_real_escape_string($con2, $_POST['last_name']);
$email = mysqli_real_escape_string($con2, $_POST['email']);
$validation_code = md5($email + microtime());
$sql0 ="INSERT INTO players
(first_name, last_name, email, validation_code)
VALUES ('$first_name', '$last_name','$email', '$validation_code')";
$sql01 = "SELECT player_id FROM players WHERE email='$email'";
$result01 = $con2->query($sql01);
if ($result01->num_rows > 0) {
$row01 = $result01->fetch_assoc();
$playerID = $row01['player_id'];
echo $playerID; //In for debugging. Sometimes it works sometimes it doesn't
$sql02 = "INSERT INTO team_players, cards (player_id, team_id)
VALUES('$playerID', '$id')";
感谢您对此的任何帮助。