如果不存在,则创建表失败,表已存在

2022-08-30 23:22:50

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,是否应该取消SQL查询的执行?换句话说,如果该表存在,它不应该退出 if 语句并且根本不回显任何内容,并且不尝试执行查询吗?IF NOT EXISTS

只是试图找到“如果不存在,则创建一个表”的最佳方法,而无需向用户输出任何内容。


答案 1

试试这个

$query = "SELECT ID FROM USERS";
$result = mysqli_query($dbConnection, $query);

if(empty($result)) {
                $query = "CREATE TABLE USERS (
                          ID int(11) AUTO_INCREMENT,
                          EMAIL varchar(255) NOT NULL,
                          PASSWORD varchar(255) NOT NULL,
                          PERMISSION_LEVEL int,
                          APPLICATION_COMPLETED int,
                          APPLICATION_IN_PROGRESS int,
                          PRIMARY KEY  (ID)
                          )";
                $result = mysqli_query($dbConnection, $query);
}

这将检查表中是否有任何内容,如果它返回,则您没有表。NULL

此外,mysql中没有数据类型,您应该在插入表中时将其设置为1或0。您也不需要在所有内容周围使用单引号,只需将数据硬编码到查询中即可。BOOLEANINT

喜欢这个。。。

$query = "INSERT INTO USERS (EMAIL, PASSWORD, PERMISSION_LEVEL, APPLICATION_COMPLETED, APPLICATION_IN_PROGRESS) VALUES ('foobar@foobar.com', 'fjsdfbsjkbgs', 0, 0, 0)";

答案 2

为避免输出任何内容,请在尝试创建表之前在 php 中测试该表。例如

$querycheck='SELECT 1 FROM `USERS`';

$query_result=$dbConnection->query($querycheck);

if ($query_result !== FALSE)
{
 // table exists
} else
{
// table does not exist, create here.
}

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