到MYSQL DB然后我想生成一个弹出菜单显示ID没有提交数据请指导我先生... [英] TO MYSQL DB THEN I WANT TO GENERATE A POPUP MENU SHOW ID NO OF SUBMIT DATA PLEASE GUIDE ME SIR...

问题描述

当我向MYSQL数据库提交数据时，我想要生成一个弹出菜单显示ID提交数据请不要指导我先生

WHEN I SUBMIT A DATA TO MYSQL DB THEN I WANT TO GENERATE A POPUP MENU SHOW ID NO OF SUBMIT DATA PLEASE GUIDE ME SIR

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "TB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
     die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT id FROM COMP";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
     echo "<table><tr><th>ID</th></tr>";
     // output data of each row
     while($row = $result->fetch_assoc()) {
         echo "<tr><td>" . $row["id"]. "</td></tr>";
     }
     echo "</table>";
} else {
     echo "0 results";
}

$conn->close();
?>

推荐答案

servername = localhost;

servername = "localhost";

用户名 = 根;

username = "root";

密码 = ;

password = "";

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