Question

我正在学习PHP，我正在尝试使用MySQLi准备好的语句。我知道我的SQL正好返回我想要的东西，它在PHPmyAdmin中返回就好了，只是不在PHP中。

    $stmt=mysqli_stmt_init($mysql);
$query = "SELECT name,version,category FROM `software` WHERE id =?";

$stmt = mysqli_stmt_prepare($mysql, $query);

mysqli_stmt_bind_param($stmt, 'i', $sid);

mysqli_stmt_execute($stmt);

mysqli_stmt_bind_result($stmt, $name, $version, $category);

mysqli_stmt_fetch($stmt);

if (empty($name)){die("No results found.");};

echo "<center><h1><b>" . $name . "</b></h1><br />";

我知道$ sid = 1，因为我可以回应它并设置，我也可以删除绑定参数并设置？ 1和相同的结果。

非常感谢任何帮助，谢谢！

Answer 1

$stmt=mysqli_stmt_init($mysql);
$query = "SELECT name,version,category FROM `software` WHERE id =?";

$stmt = mysqli_stmt_prepare($mysql, $query);

mysqli_stmt_bind_param($stmt, 'i', $sid);

mysqli_stmt_execute($stmt);

mysqli_stmt_bind_result($stmt, $name, $version, $category);

///try the while
while (mysqli_stmt_fetch($stmt)) {
    printf ("%s (%s)\n", $name, $version, $category);
}

MySQLi准备好的声明没有返回

1 个答案: