<?php
$servername = "localhost:3307";
$username = "root";
$password = "";
$dbname = "pageant"; //database
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname); //open ng connection
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['submit'])){
$category=$_POST['category'];
$sql = "ALTER TABLE men
ADD " . $category . " INT BEFORE TOTAL";
if ($conn->query($sql) === TRUE) {
header('Location:welcome_admin.php');
echo "New record created successfully";
exit();
}else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
$conn->close();
?>
<?php
$servername = "localhost:3307";
$username = "root";
$password = "";
$dbname = "pageant"; //database
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname); //open ng connection
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST['submit'])){
$category=$_POST['category'];
$sql = "ALTER TABLE women
ADD " . $category . " INT BEFORE TOTAL";
if ($conn->query($sql) === TRUE) {
header('Location:welcome_admin.php');
echo "New record created successfully";
exit();
}else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
$conn->close();
?>
This is my code.
2番目のクエリでエラーがありましたか? – Muthu17
それはちょうど私の2番目のテーブルに追加されません。 – callme123
$ sqlクエリを実行するコードを入力してください。 – Muthu17