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Php在本地主机上工作,但不在live网站上工作_Php_Mysqli - Fatal编程技术网

Php在本地主机上工作,但不在live网站上工作

Php在本地主机上工作,但不在live网站上工作,php,mysqli,Php,Mysqli,这是我的代码,php只在本地主机上工作 我试图让它在我们的网站上为一个小组项目工作,我以前从未学过php,所以我对它相当陌生 <?php $servername= "sql5.freemysqlhosting.net:3306"; $username="sql5404117"; $password="******"; $dbname="sql5404117"; $mysqli = new mysqli($ser

这是我的代码,php只在本地主机上工作

我试图让它在我们的网站上为一个小组项目工作,我以前从未学过php,所以我对它相当陌生

<?php
$servername= "sql5.freemysqlhosting.net:3306";
$username="sql5404117";
$password="******";
$dbname="sql5404117";
$mysqli = new mysqli($servername,$username,$password,$dbname);
    if($mysqli === false){
        die("ERROR: Could not connect. " . mysqli_connect_error());
    }
$query = "SELECT * FROM David";
if($result = mysqli_query($mysqli, $query)){
    if(mysqli_num_rows($result) > 0){
        echo "<table>";
            echo "<tr>";
                echo "<th>Album Number</th>";
                echo "<th>Song</th>";
                echo "<th>Artist</th>";
                echo "<th>Album</th>";
            echo "</tr>";
        while($row = mysqli_fetch_array($result)){
            echo "<tr>";
                echo "<td>" . $row['V1'] . "</td>";
                echo "<td>" . $row['V2'] . "</td>";
                echo "<td>" . $row['V3'] . "</td>";
                echo "<td>" . $row['V4'] . "</td>";
            echo "</tr>";
        }
        echo "</table>";
        // Free result set
        mysqli_free_result($result);
    } else{
        echo "No records matching your query were found.";
    }
} else{
    echo "ERROR: Could not able to execute $query. " . mysqli_error($mysqli);
}
 
// Close connection
mysqli_close($mysqli);
试试看

而不是

$servername= "sql5.freemysqlhosting.net:3306";
除非您连接到外部服务器,在这种情况下,您会显式地将其命名为公共地址

此外,正如@user3783243在注释中提到的,port是中的第5个参数,因此,如果您需要指定它,您可以这样做

$mysqli = new mysqli($servername, $username, $password, $dbname, $port);

注意:3306通常是默认的MySQL端口,因此通常不必指定它。

端口名是第5个参数,而不是主机名的一部分<代码>新mysqli(“127.0.0.1”,“用户”,“密码”,“数据库”,3306)
检查您的php版本的live server,我认为它太旧或继承了,请选择一个合适的php版本。您能提供一条错误消息和/或告诉我们什么不起作用吗?
$mysqli = new mysqli($servername, $username, $password, $dbname, $port);