mysqli为什么会发生这种情况? [英] mysqli why does this happens?
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问题描述
我有两个后续的mysqli语句,第二个返回:
I have two subsequent mysqli statements, and the second returns:
致命错误:在...中的非对象上调用成员函数bind_param()
Fatal error: Call to a member function bind_param() on a non-object in ...
为什么会这样?这是否意味着我需要打开两个不同的连接? 有什么方法可以避免这种情况(我喜欢将SQL连接详细信息保存在一个文件中)?
Why this happens? Does this means that I need to open two different connection? Is there any way to avoid this (I love keeping the SQL connection details in one file)?
代码在这里:
$db = new mysqli("localhost", "root", "", "database");
$stmt = $db->prepare("UPDATE posts SET vote_".$_POST['vote']." = vote_".$_POST['vote']." + 1 WHERE id=?");
$stmt->bind_param('s', $_POST['id_post']);
$stmt->execute();
$stmt->close();
$stmt = $db->prepare("INSERT INTO votes (kind, users_id, posts_id) VALUES (?, ?, ?)");
$stmt->bind_param('sss',$_POST['vote'],$_POST['id_user'],$_POST['id_post']);
$stmt->execute();
$stmt->close();
推荐答案
检查 mysqli :: prepare .如果为假,则应该使用 mysqli :: error .
Check the return value of mysqli::prepare. If it is FALSE, you should get the details for the occured error with mysqli::error.
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