为什么这个php代码不允许我连接到数据库？
[英]Why isn't this php code allowing me to connect to my database?
问题描述
I'm using a WAMP server and on it I created a database and a table. 
我正在使用WAMP服务器，并在其上创建了一个数据库和一个表。 All the names are correct and the user has full access to everything. 所有名称都是正确的，并且用户可以完全访问所有内容。 When I run the code, it prints out "Unable to select database". 当我运行代码时，它会打印出“无法选择数据库”。 Thanks. 谢谢。
<?php
if(isset($_POST["Submit"])){
print_r ($_POST["nutrient"]);
}
session_start();
//establish connection
$server = "localhost";
$db_username = "root";
$db_password = "";
$database = "gainlife_cavin";
$table = "cavintable";
//connect PHP script to database
$connection = mysqli_connect($server, $db_username, $db_password, $database);
//select database to use
@mysql_select_db($database) or die( "Unable to select database");
//$query = "INSERT INTO $table VALUES("")"
//mysql_query($query)
mysql_close();
?>
<body>
</form>
4 个解决方案
解决方案1
1 2014-06-23 04:22:48
Try something like the following. 尝试类似以下的方法。
<?php
//establish connection
$server = "localhost";
$db_username = "root";
$db_password = "";
$database = "gainlife_cavin";
$table = "cavintable";
//connect PHP script to database
$connection =mysqli_connect("$server","$db_username","$db_password","$database");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
//Your query here
mysqli_close($connection);
?>
解决方案2
0 2014-06-23 04:21:15
I use simple code 1 line. 我使用简单的代码1行。 here is my code that i'm using. 这是我正在使用的代码。
$conn = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
// Evaluate the connection
if (mysqli_connect_errno()) {
echo mysqli_connect_error();
exit();
}
解决方案3
-1 2014-06-23 04:03:43
您从MySQLI转变为在选择数据库以及其余代码时不使用MySQLI。
解决方案4
-2 2014-06-23 04:15:22
Try this. 尝试这个。 I have change mysqli_connect to mysql_connect and mysql_select_db variable. 我将mysqli_connect更改为mysql_connect和mysql_select_db变量。
//connect PHP script to database
$connection = mysql_connect($server, $db_username, $db_password, $database);
//select database to use
$select = mysql_select_db($connection) or die( "Unable to select database");
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