如果不存在，php mysql 创建数据库
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Php mysql create database if not exists
提问by erdomester
I want to create a database. Why is not the db created with this code?
我想创建一个数据库。为什么不使用此代码创建数据库？
$dbname = 'regulations_db';
$con = mysql_connect("localhost","root","pass");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
if (mysql_num_rows(mysql_query("SELECT SCHEMA_NAME FROM INFORMATION_SCHEMA.SCHEMATA WHERE SCHEMA_NAME = '". $dbname ."'"))) {
echo "Database $dbname already exists.";
}
else {
mysql_query("CREATE DATABASE '". $dbname ."'",$con);
echo "Database $dbname created.";
}
This is working, but I think the first one is the best practice:
这是有效的，但我认为第一个是最佳实践：
if (mysql_query("CREATE DATABASE IF NOT EXISTS regulations_db",$con))
{
echo "Database created";
}
else
{
echo "Error creating database: " . mysql_error();
}
回答by buley
Just do a simple mysql_select_db()and if the result is false then proceed with the creation.
只需做一个简单的mysql_select_db()，如果结果为假，则继续创建。
As an example, check out the first answer hereby another very smart StackOverflower.
例如，请查看另一个非常聪明的 StackOverflower的第一个答案。
<?php
// Connect to MySQL
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
// Make my_db the current database
$db_selected = mysql_select_db('my_db', $link);
if (!$db_selected) {
// If we couldn't, then it either doesn't exist, or we can't see it.
$sql = 'CREATE DATABASE my_db';
if (mysql_query($sql, $link)) {
echo "Database my_db created successfully\n";
} else {
echo 'Error creating database: ' . mysql_error() . "\n";
}
}
mysql_close($link);
?>
回答by Anyone
Three steps to fix this:
解决这个问题的三个步骤：
- Don't specify the database name when connecting.
- Your SQL statement should be CREATE DATABASE IF NOT EXISTS php1.
- Call mysqli_select_db($link, 'php1') to make that the default database for your connection.
- 连接时不要指定数据库名称。
- 你的 SQL 语句应该是 CREATE DATABASE IF NOT EXISTS php1。
- 调用 mysqli_select_db($link, 'php1') 使其成为连接的默认数据库。
回答by 1uffyD9
If you're using MySQLi Object-oriented method, you can use following code, this code is similar to previous answer and only the method is different, I just put this because if anyone using MySQLi Object-oriented method, you can use this code directly.
如果您使用的是 MySQLi 面向对象的方法，则可以使用以下代码，此代码与之前的答案类似，只是方法不同，我之所以这么说是因为如果有人使用 MySQLi 面向对象的方法，则可以使用此代码直接地。
$servername = "localhost";
$username = "mysql_user";
$password = "user_password";
$dbName = "databaseName";
// Connect to MySQL
$conn = new mysqli($servername, $username, $password);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// If database is not exist create one
if (!mysqli_select_db($conn,$dbName)){
$sql = "CREATE DATABASE ".$dbName;
if ($conn->query($sql) === TRUE) {
echo "Database created successfully";
}else {
echo "Error creating database: " . $conn->error;
}
}
Furthermore you can refer W3school site here.
Good Luck! :D
祝你好运！:D
