PHP bind_param() 问题；

Question

你好，我对 PHP 不是很熟悉，当我尝试执行查询时出现此错误。

警告：

Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement in /Users/site/userpanel.php on line 10

代码：

$query="SELECT SUM(gebruiker_id) AS totalitems FROM inventory WHERE gebruiker_id = 1";
$stmt = $db->prepare($query);
$stmt->bind_param("s", $items);
$stmt->execute();
$result = $stmt->bind_result($col1);

提前致谢！

Answer 1

你必须使用这样的占位符：

$query="SELECT SUM(gebruiker_id) AS totalitems FROM inventory WHERE gebruiker_id = ?";
$stmt = $db->prepare($query);
$stmt->bind_param("s", $items);
$stmt->execute();
$result = $stmt->bind_result($col1);

Answer 2

有时不需要绑定：

$query="SELECT SUM(gebruiker_id) AS totalitems FROM inventory WHERE gebruiker_id = 1";
$stmt = $db->prepare($query);
//$stmt->bind_param("s", $items);
$stmt->execute();
$result = $stmt->bind_result($col1);

或

$query="SELECT SUM(gebruiker_id) AS totalitems FROM inventory WHERE gebruiker_id = 1";
$stmt = $db->query($query);
$result = $stmt->bind_result($col1);

但是，如果您尝试将 $items 绑定到 gebruiker_id，请按照 @rizier123 的回答进行操作。

PHP bind_param() 问题；

PHP issue with bind_param();

php

mysqli