我正在嘗試創建一個網站，其中 php 與后端 sql 連接。有 1 行用戶名 = 'user1'。然而，這個簡單的 PHP 代碼來獲取“user1”的詳細信息 -<?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", $val); // standard integer representation?>上面的代碼返回這個輸出 -$val 的值為 0，顯然有 1 行用戶名 = 'user1'。請幫忙，我需要在我的 php 腳本中獲取“user1”的詳細信息。
2 回答
揚帆大魚
TA貢獻1799條經驗 獲得超9個贊
您需要在任何地方正確使用連接變量名稱。（在$conn您的情況下，它是，但您使用了$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);
?>
慕婉清6462132
TA貢獻1804條經驗 獲得超2個贊
你做錯了什么，請更新并嘗試
$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);
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