[英]MySQLi prepared statement not returning
I am learning PHP and I am trying to use prepared statements with MySQLi.我正在学习 PHP,我正在尝试在 MySQLi 中使用准备好的语句。 I know my SQL returns exactly what I want, it returns in PHPmyAdmin just fine, just not in PHP.
我知道我的 SQL 返回的正是我想要的,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 />";
I know $sid = 1, because I can echo that and it is set, I can also remove the bind params and just set the ?我知道 $sid = 1,因为我可以回应它并设置它,我也可以删除绑定参数并只设置 ? to 1 and same result.
到 1 和相同的结果。
Any help would really be appreciated, thank you!任何帮助将不胜感激,谢谢!
$stmt=mysqli_stmt_init($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);
}
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