[英]How to insert data into multiple php sql tables with foreign keys?
我正在寻找运行三个sql查询以将3组数据插入3个不同的sql表中的登录/注册系统。 我需要能够一次更新所有三个表,同时保持所有键的准确性和最新性。 我能够使两个表正常工作,但我的登录表不允许插入任何信息。
任何帮助将是有用的:)
PHP代码:
if ($password == $password2){
$password = md5($password);
$sql = "INSERT INTO customer(FirstName, SecondName, Address, EmailAddress, TelephoneNumber) VALUES ('$firstname', '$surname', '$address', '$email', '$telephonenumber')";
mysqli_query($db, $sql);
$CustomerID = mysqli_insert_id($db);
$sql2 = "INSERT INTO login(CustomerID, Username, PassW) VALUES ('$CustomerID', '$username', '$password')";
mysqli_query($db, $sql2);
$sql3 = "INSERT INTO inkstyle(CustomerID, Colour, Blackwork, Realism, Traditional, Japanese, Watercolour) VALUES ('$CustomerID', '$colour', '$blackwork', '$realism', '$traditional', '$japanese', '$watercolour')";
mysqli_query($db, $sql3);
我的html代码在按钮和文本框方面很好。
以下是在if语句之前声明的变量ive:
$username= mysqli_real_escape_string($db, $_POST['username']);
$password= mysqli_real_escape_string($db, $_POST['password']);
$password2= mysqli_real_escape_string($db, $_POST['password2']);
$firstname= mysqli_real_escape_string($db, $_POST['firstname']);
$surname= mysqli_real_escape_string($db, $_POST['surname']);
$address= mysqli_real_escape_string($db, $_POST['address']);
$email= mysqli_real_escape_string($db, $_POST['email']);
$telephonenumber= mysqli_real_escape_string($db, $_POST['telephonenumber']);
$colour= mysqli_real_escape_string($db, $_POST['cb1']);
$blackwork= mysqli_real_escape_string($db, $_POST['cb2']);
$realism= mysqli_real_escape_string($db, $_POST['cb3']);
$traditional= mysqli_real_escape_string($db, $_POST['cb4']);
$japanese= mysqli_real_escape_string($db, $_POST['cb5']);
$watercolour= mysqli_real_escape_string($db, $_POST['cb6']);
我已经解决了这个问题。 这是因为我正在散列密码,但是我没有在sql表中设置足够的字符:)谢谢您的帮助!
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