Create a table in PHP for certain variable (MySQLi)

Question

This code should check if a user is in a table and if not create a table for the user. Right now it just names the table $MyServer. How do I make it so the name is the variable $MyServer.

<?php
    require "conn.php";
    echo "debug 1";
    $stmt = $conn->prepare("SELECT * FROM UserData WHERE username = ?");
    $stmt->bind_param('s',$username);
    $username = $_POST["username"];
    $stmt->execute();
    $stmt->store_result();

    echo "debug 2";
    if ($stmt->num_rows == 0){ // username not taken
        echo "debug 3";
        $stmt2 = $conn->prepare("INSERT INTO UserData (username, password) VALUES (?, ?)");
        $password =($_POST["password"]);
        $username =($_POST["username"]);
        $stmt2->bind_param('ss', $username, $password);
        $stmt2->execute();
        $MyServer =($_POST["username"]);
        $sql = ('CREATE TABLE $MyServer (
            id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
            username VARCHAR(30) NOT NULL
            )');
        if($conn->query($sql) === TRUE){
            echo "Table created successfully";
        } else {
            echo "Table is not created successfully ";
        }

        if ($stmt2->affected_rows == 1){
            echo 'Insert was successful.';

        }else{ echo 'Insert failed.';
            var_dump($stmt2);
        }
    }else{ echo 'That username exists already.';}

    ?>

If you're creating a table for each user, you need to evaluate your data structure, that's generally a monumentally bad idea. — Difster
– Difster, Commented Aug 5, 2017 at 9:52

B. Desai · Accepted Answer · 2017-08-05 09:57:54Z

1

creating table for each user is not good idea. You need to create single table in which you can store all users records

But if you really want to do this. You need to change ' to " do your variable can take value

$sql = "CREATE TABLE $MyServer (
            id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
            username VARCHAR(30) NOT NULL
            )";

See the difference:

http://docs.php.net/manual/en/language.types.string.php

answered Aug 5, 2017 at 9:57

B. Desai

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Comments

Difster · Accepted Answer · 2017-08-05 09:53:29Z

0

Despite my comment above, here is what you're looking for: Notice that now, "$MyServer" is no longer seen as a string, but an actual variable.

$sql = ('CREATE TABLE ' . $MyServer . '(
        id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, 
        username VARCHAR(30) NOT NULL
        )');

answered Aug 5, 2017 at 9:53

Difster

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