Question

我试图让用户创建一个给定信息的数据库，所以为此，我有以下内容：

<?php
    //ini_set('display_errors', 'Off');
    if(isset($_POST['siteName'])) {
        $DBS = htmlspecialchars($_POST['databaseServer']);
        $DBN = htmlspecialchars($_POST['databaseName']);
        $DBU = htmlspecialchars($_POST['databaseUser']);
        $DBP = htmlspecialchars($_POST['databasePass']);
        $SN = htmlspecialchars($_POST['siteName']);
        $AU = htmlspecialchars($_POST['adminUser']);
        $AP = htmlspecialchars($_POST['adminPass']);
        $con = new Mysqli($DBS,$DBU,$DBP,$DBN);
        if($con->connect_error) {
            echo "Error : ".$con->connect_error;
        }
        else {
            $sql = "CREAT DATABASE '$DBN'";
            $sql .= "CREATE TABLE options (sitename VARCHAR(30),adminUser VARCHAR(30),adminPass VARCHAR(30),reg_date TIMESTAMP);";
            $sql .= "CREATE TABLE users (ID int(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, userName VARCHAR(30),password VARCHAR(30),reg_date TIMESTAMP);";
            $sql .= "INSERT INTO '$DBN' (sitename,adminUser,adminPass) VALUES ($SN,$AU,$AP);";
            if($con->multi_query($sql) === true) {
                echo " Enjoy .. ! ";
            }
            else {
                echo "Error : ".$sql." : ".$con->error;
            }
        }
    }
?>

但是，当我为数据库键入示例名称时，出现此错误：

Error : Unknown database 'name'

当手动创建数据库并在我的表单中键入它的名称时，请收到以下错误：

Error : CREAT DATABASE 'ex'CREATE TABLE options (sitename
VARCHAR(30),adminUser VARCHAR(30),adminPass VARCHAR(30),reg_date 
TIMESTAMP);CREATE TABLE users (ID int(6) UNSIGNED AUTO_INCREMENT PRIMARY
KEY, userName VARCHAR(30),password VARCHAR(30),reg_date     
TIMESTAMP);INSERT 
INTO 'ex' (sitename,adminUser,adminPass) VALUES (My Phone 
Book,admin,102030); : You have an error in your SQL syntax; check the    
manual that corresponds to your MySQL server version for the right
syntax to use near 'CREAT DATABASE 'ex'CREATE TABLE options (sitename 
VARCHAR(30),adminUser VARCHAR(' at line 1

我该如何解决这个问题？

Answer 1

将行更改为

$sql="CREATE DATABASE ". $DBN.";";
$sql.="CREATE TABLE options (sitename VARCHAR(30),adminUser VARCHAR(30),adminPass VARCHAR(30),reg_date TIMESTAMP);";
    $sql.="CREATE TABLE users (ID int(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, userName VARCHAR(30),password VARCHAR(30),reg_date TIMESTAMP);";
    $sql.="INSERT INTO  options  (sitename,adminUser,adminPass) VALUES ($SN,$AU,$AP);";

Answer 2

如果要使用PHP变量创建数据库，则需要使用USE语句。之后，您可以在MySQL中创建表。

请尝试以下代码：

<?php
//ini_set('display_errors', 'Off');
if (isset($_POST['siteName'])) {
    $DBS = htmlspecialchars($_POST['databaseServer']);
    $DBN = htmlspecialchars($_POST['databaseName']);
    $DBU = htmlspecialchars($_POST['databaseUser']);
    $DBP = htmlspecialchars($_POST['databasePass']);
    $SN = htmlspecialchars($_POST['siteName']);
    $AU = htmlspecialchars($_POST['adminUser']);
    $AP = htmlspecialchars($_POST['adminPass']);
    if ($con->connect_error) {
        echo "Error : " . $con->connect_error;
    } else {
        $sql = "CREATE DATABASE $DBN";
        $con->query($sql);
        $con->query("use `$DBN`");
        $sql = "CREATE TABLE options (sitename VARCHAR(30),adminUser VARCHAR(30),
           adminPass VARCHAR(30),reg_date TIMESTAMP);";
        $con->query($sql);
        $sql = "CREATE TABLE users (ID int(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
            userName VARCHAR(30),password VARCHAR(30),reg_date TIMESTAMP);"
            $con->query($sql);
    }
    if ($con->query($sql) === true) {
        echo " Enjoy .. ! ";
    } else {
        echo "Error : " . $sql . " : " . $con->error;
    }

Answer 3

最后，我借助亲爱的@MagnusEriksson解决了我的问题： 1.我应该从连接到数据库的代码中删除@DBN 2. sql="CREAT DATABASE '$DBN'";应更改为sql="CREAT DATABASE $DBN";

用PHP创建数据库

3 个答案: