MySQLi準備聲明不返回

Question

我正在學習PHP，我正在嘗試使用MySQLi準備的語句。我知道我的SQL返回正是我想要的，它返回PHPmyAdmin就好，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，因爲我可以重複這一點，它設置，我還可以刪除綁定PARAMS和剛剛成立的？到1和相同的結果。

任何幫助真的不勝感激，謝謝！

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); 
}