表存在时如果不存在则创建表失败[重复]

Question

我有以下代码：

$db_host = 'localhost';
$db_port = '3306';
$db_username = 'root';
$db_password = 'root';
$db_primaryDatabase = 'dsl_ams';

// Connect to the database, using the predefined database variables in /assets/repository/mysql.php
$dbConnection = new mysqli($db_host, $db_username, $db_password, $db_primaryDatabase);

// If there are errors (if the no# of errors is > 1), print out the error and cancel loading the page via exit();
if (mysqli_connect_errno()) {
    printf("Could not connect to MySQL databse: %s\n", mysqli_connect_error());
    exit();
}

$queryCreateUsersTable = "CREATE TABLE IF NOT EXISTS `USERS` (
    `ID` int(11) unsigned NOT NULL auto_increment,
    `EMAIL` varchar(255) NOT NULL default '',
    `PASSWORD` varchar(255) NOT NULL default '',
    `PERMISSION_LEVEL` tinyint(1) unsigned NOT NULL default '1',
    `APPLICATION_COMPLETED` boolean NOT NULL default '0',
    `APPLICATION_IN_PROGRESS` boolean NOT NULL default '0',
    PRIMARY KEY  (`ID`)
)";

if(!$dbConnection->query($queryCreateUsersTable)){
    echo "Table creation failed: (" . $dbConnection->errno . ") " . $dbConnection->error;
}

哪些输出...

Table creation failed: (1050) Table '`dsl_ams`.`USERS`' already exists

我不明白的是：如果该表已经存在，IF NOT EXISTS 不应该取消 SQL 查询的执行吗？换句话说，如果表存在，它是否应该退出该 if 语句并且根本不回显任何内容，并且不尝试执行查询？

只是试图找到最好的方法来“创建一个不存在的表”而不向用户输出任何内容。

Answer 1

I use your script in mysql is true:
mysql> select version();
+-----------+
| version() |
+-----------+
| 5.5.30    |
+-----------+
1 row in set (0.01 sec)

$queryCreateUsersTable = "CREATE TABLE IF NOT EXISTS `sun_channel` (
    `ID` int(11) unsigned NOT NULL auto_increment,
    `EMAIL` varchar(255) NOT NULL default '',
    `PASSWORD` varchar(255) NOT NULL default '',
    `PERMISSION_LEVEL` tinyint(1) unsigned NOT NULL default '1',
    `APPLICATION_COMPLETED` boolean NOT NULL default '0',
    `APPLICATION_IN_PROGRESS` boolean NOT NULL default '0',
    PRIMARY KEY  (`ID`)
)";
$res = mysql_query($queryCreateUsersTable);
var_dump($res);
die;

//show bool(true)

Answer 2

您应该收到警告，而不是错误。你运行的是什么版本？ 无论如何，如果您想在 SQL 查询之前显示错误，请键入以下内容： 设置 sql_notes = 0; 然后输入 SET sql notes=1;查询后。