运行SELECT * FROM用户后,该表没有任何区别。
$firstName = $_POST["firstName"];
$lastName = $_POST["lastName"];
$dateOfBirth = $_POST["dateOfBirth"];
$gender = $_POST["gender"];
$fitnessLevel = $_POST["fitnessLevel"];
$number = $_POST["number"];
$address = $_POST["address"];
$password = $_POST["password"];
$user = 'root';
$pass = 'root';
$db = 'gymmembers';
$db = new mysqli('localhost',$user,$pass,$db) or die("Error, try again");
mysqli_query($db, "INSERT INTO users(`firstName`,`lastName`,`dateOfBirth`,`gender`,`fitnessLevel`,`number`,`address`,`password`)
VALUES('$firstName','$lastName','$dateOfBirth','$gender','$fitnessLevel','$number','$address','$password')" or die(mysqli_error()));
我可以回显任何变量并显示它们,因此表单中的数据将传递到此处。
谢谢 :-)
最佳答案
实际上,您将die(mysqli_error())
放入了不正确的mysqli_query()
内,请执行以下操作:-
mysqli_query($db, "INSERT INTO users(`firstName`,`lastName`,`dateOfBirth`,`gender`,`fitnessLevel`,`number`,`address`,`password`)
VALUES('$firstName','$lastName','$dateOfBirth','$gender','$fitnessLevel','$number','$address','$password')") or die(mysqli_error($db));
注意:-在
$db
中添加mysqli_error()
,以便在发生任何错误时都可以知道。关于php - PHP INSERT INTO本地主机不起作用,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/36742484/