Question

尝试编写一个显示SQL数据库中所有公共内容的id和filepath的函数。这是我到目前为止所做的，但它未能运行。

<?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "pricosha";

    // Create connection
    $connection = mysql_connect($servername, $username, $password, $dbname);

    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 

    //Added two extra variables, username and content_name
    $query = "SELECT id, username, file_path, content_name FROM Content";
    $result = mysql_query($query);

    //Loop through the results of the query
    while($row = mysql_fetch_array($result)){ 
        echo "ID: " . $row["id"]. " Username: " . $row["username"]. " File Path: " . $row["file_path"]. " Content Name: " . $row["content_name"]. "<br>";
    }

    $conn->close();
    ?>

Answer 1

试试这个

$conn = mysql_connect($servername, $username, $password, $dbname);

Answer 2

用于创建与mysql的连接的变量是 $ connection ，而为此使用 $ conn ：

$conn->connect_error

和此：

$conn->close();

现在尝试将 $ conn 更改为 $ connection ，或者只是更改此内容：

$connection = mysql_connect($servername, $username, $password, $dbname);

到此：

$conn = mysql_connect($servername, $username, $password, $dbname);

到目前为止，函数mysql_connect，mysql_query和其他函数已被弃用。您需要将其更改为mysqli。试试这个。

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "pricosha";

$connection = new mysqli($servername, $username, $password, $dbname) or die(mysqli_errno());

$query = "SELECT id, username, file_path, content_name FROM Content";
$result = mysqli_query($connection, $query);

while($row = mysqli_fetch_assoc($result)){
    echo "ID: " . $row["id"]. " Username: " . $row["username"]. " File Path: " . $row["file_path"]. " Content Name: " . $row["content_name"]. "<br>";
}
?>

尝试编写一个在SQL数据库中显示数据的函数

2 个答案: