我正在尝试将 2 个表结构复制到新数据库中。我使用了一个函数，它自己制作了SQL命令：当我在phpmyadmin上执行代码时，代码被执行，但是当我通过PHP执行时，它不会执行。
怎么可能？
sql 命令是这样的：
CREATE TABLE `tabella_1` (
`campo1` int(11) NOT NULL AUTO_INCREMENT,
`campo2` varchar(100) COLLATE latin1_general_ci NOT NULL,
`campo_3` int(11) NOT NULL,
PRIMARY KEY (`campo1`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci;
CREATE TABLE `tabella_2` (
`campo1` int(11) NOT NULL DEFAULT '0',
`campo2` varchar(100) COLLATE latin1_general_ci NOT NULL,
`campo_3` int(11) NOT NULL,
PRIMARY KEY (`campo1`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci;
谢谢
使用数据库连接详细信息更新此代码：
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// sql to create table
$sql = "CREATE TABLE `tabella_1` ( `campo1` int(11) NOT NULL AUTO_INCREMENT, `campo2` varchar(100) COLLATE latin1_general_ci NOT NULL, `campo_3` int(11) NOT NULL, PRIMARY KEY (`campo1`) ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci";
if ($conn->query($sql) === TRUE) {
echo "Table 1 created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
// sql to create table
$sql2 = "CREATE TABLE `tabella_2` ( `campo1` int(11) NOT NULL DEFAULT '0', `campo2` varchar(100) COLLATE latin1_general_ci NOT NULL, `campo_3` int(11) NOT NULL, PRIMARY KEY (`campo1`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci";
if ($conn->query($sql2) === TRUE) {
echo "Table 2 created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
$conn->close();
?>
我明白了！
我把函数放进了一个数组（）中，它可以工作！
给大家打个哆嗦！