Question

我已编写此代码，但仍无法在数据库中创建表。只输出是：

“数据库已创建....对不起表未创建”。

<?php                                   
    include 'connection.php';
    //Creating database
    $creatingdatabase = "CREATE DATABASE IF NOT EXISTS quest";
    $query1 = mysqli_query($con, $creatingdatabase);
    if($query1){ 

        echo "database created....";
        //creating tables
        $creatingtable = "CREATE TABLE quiz (
                        id INT(6) AUTO_INCREMENT PRIMARY KEY,
                        questions VARCHAR(255) NOT NULL,
                        optiona VARCHAR(255) NOT NULL,
                        optionb VARCHAR(255) NOT NULL,
                        optionc VARCHAR(255) NOT NULL,
                        optiond VARCHAR(255) NOT NULL, 
                        answer VARCHAR(50) NOT NULL
                                                )";
        $query2 = mysqli_query($con, $creatingtable);
        if($query2){echo "table created...";}
        else { echo "sorry... table not created"; }

    }
    else { echo "database can not be created"; }
?>

Answer 1

<?php
$servername = 'localhost';
$username = 'root';
$password = 'xxxxx';

$conn = new mysqli($servername, $username, $password);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
// Create database
$sql = "CREATE DATABASE guest"; //Set name database here
if ($conn->query($sql) === TRUE) {
    $conn =  mysqli_connect($servername, $username, $password,'guest');//Set name database here
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 
    $query = "CREATE TABLE quiz (
        id INT(6) AUTO_INCREMENT PRIMARY KEY,
        questions VARCHAR(255) NOT NULL,
        optiona VARCHAR(255) NOT NULL,
        optionb VARCHAR(255) NOT NULL,
        optionc VARCHAR(255) NOT NULL,
        optiond VARCHAR(255) NOT NULL, 
        answer VARCHAR(50) NOT NULL
    )";
    if ($conn->query($query) === TRUE) {
        echo "Table users created successfully";
    } else {
        echo "Error creating table: " . $conn->error;
    }
} else {
    echo "Error creating database: " . $conn->error;
}

$conn->close();
?>

但我不建议用户在sql server中创建新数据库。

就安全性而言，最好是已经创建了一个数据库，然后只需添加用户数据，即使它是第一条记录。用户无法将数据输入Web表单，然后间接运行SQL代码，从而无法获得任何结果。

Answer 2

CREATE TABLE语句需要指定要使用的数据库，或者需要使用mysqli_select_db()设置默认数据库。

$creatingdatabase = "CREATE DATABASE IF NOT EXISTS quest";
$query1 = mysqli_query($con, $creatingdatabase);
mysqli_select_db($con, "quest");

无法创建表

2 个答案: