PHP SQL查询未向前端返回正确的行数

Question

我正在尝试创建一个网站，其中php与后端sql连接。有1行的用户名='user1'。

但是，这个简单的PHP代码可以获取“ user1”的详细信息-

<?php
    /* Database credentials. Assuming you are running MySQL
    server with default setting (user 'root' with password 'root') */
    define('DB_SERVER', 'localhost');
    define('DB_USERNAME', 'root');
    define('DB_PASSWORD', 'root');
    define('DB_NAME', 'startup');

    /* Attempt to connect to MySQL database */
    $conn = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_NAME);

    // Check connection
    if($conn === false){
        die("ERROR: Could not connect. " . mysqli_connect_error());
    }
    echo '<p>Connected successfully</p>';

    $sql = "SELECT id, username, password FROM users WHERE username = ?";
    $stmt = mysqli_prepare($link, $sql);
    $param_username = "user1";
    mysqli_stmt_bind_param($stmt, "s", $param_username);

    mysqli_stmt_execute($stmt);
    // mysqli_stmt_store_result($stmt);
    echo '<p>done</p>';
    $val = mysqli_stmt_num_rows($stmt);
    printf("%%d = '%d'\n", $val); // standard integer representation
?>

以上代码返回此输出-

当明显有1行的用户名='user1'时，$ val的值为0。请提供帮助，我需要在我的PHP脚本中获取“ user1”的详细信息。

Answer 1

您需要在任何地方正确使用连接变量名。（在这种情况下，它是$conn，但您使用的是$link）

<?php
    /* Database credentials. Assuming you are running MySQL
    server with default setting (user 'root' with password 'root') */
    define('DB_SERVER', 'localhost');
    define('DB_USERNAME', 'root');
    define('DB_PASSWORD', 'root');
    define('DB_NAME', 'startup');

    $conn = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_NAME);

    /* check connection */
    if (!$conn) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }
    $param_username = "user1";
    $stmt = mysqli_prepare($conn, "SELECT id, username, password FROM users WHERE `username` = ?");
    mysqli_stmt_bind_param($stmt, "s", $param_username);

    /* execute prepared statement */
    mysqli_stmt_execute($stmt);

    printf("%d Row inserted.\n", mysqli_num_rows($stmt));

    /* close statement and connection */
    mysqli_stmt_close($stmt);

    /* close connection */
    mysqli_close($conn);
?>

Answer 2

您做错了什么，请对此进行更新并尝试

$sql = "SELECT id, username, password FROM users WHERE username = ?";
$stmt = mysqli_prepare($link, $sql);
$param_username = "user1";
mysqli_stmt_bind_param($stmt, "s", $param_username);
mysqli_stmt_execute($stmt);