PHP - 的mysqli :: prepare返回false [英] PHP - mysqli::prepare returning false

问题描述

我有这个非常简单的PHP code：

I have this really simple PHP code:

$mysqli = new mysqli('localhost', 'xxx', 'xxxxx', 'xxx');

$query = "SELECT * FROM questions WHERE id = ?";
$stmt = $mysqli->prepare($query);
$stmt->bind_param('d', $_GET['qid']);
$stmt->execute();
$stmt->bind_result($id, $content, $correct_ans, $lol);
$stmt->fetch();

//do sth with the data

$query = "SELECT * FROM answers WHERE question_id = ?";
$stmt = $mysqli->prepare($query);
$stmt->bind_param('d', $_GET['qid']);
$stmt->execute();
$stmt->bind_result($id, $content, $lol);

while($stmt->fetch())
{
    //do sth
}

基本上，它似乎无论我做什么，第二个的mysqli :: prepare（）将始终返回false，但的mysqli ::误差为空，出于某种原因。任何人都可以在这里看到一个错误？

Basically, it seems that no matter what I do the second mysqli::prepare() will always return false, however mysqli::error is empty for some reason. Can anyone see a mistake here?

PS。这不是一个重复的问题;是的，它已经被问： MySQLi的prepared语句返回false 但作者并没有刻意去分享他与大家的解决方案。

PS. It's not a duplicate question; yes, it has been already asked: MySQLi prepared statement returning false but the author didn't bother to share his solution with everybody.

编辑：我想我应该解释：只有第二prepare返回false，首先是精美绝伦。更令人奇怪的是，如果我删除code的第一线（即第一个prepare），第二个将不工作的问题...

I thought I should explain this: ONLY the second prepare returns false, the first is absolutely fine. What is even more weird is that if I remove the first lines of code (ie the first prepare) the second will work without problems...

EDIT2：嗯，我能说的是现在跆拳道。事实证明，如果我删除的任务，（我不做 $语句= prepare（）...）只是调用函数， $ mysqli-＆GT;错误列表不为空 - 它说：命令不同步，你现在不能运行此命令。如果我做的任务，它是空的......

Well, all I can say now is WTF. It turns out that if I remove the assignment, (I don't do $stmt = prepare()...) but just call the function, $mysqli->error list is not empty - it says "Commands out of sync; you can't run this command now". If I do the assignment, it's empty...

EDIT3：我切换到PDO和它完美的作品。直到我可以证明，否则，我会认为是库MySQLi马车。

I switched to PDO and it works perfectly. Until I can prove otherwise, I will assume that MySQLi is buggy.

推荐答案

第二次使用前的mysqli :: prepare（）您必须的免费的mysqli导致或的关闭当前的mysqli声明：

Before second usage mysqli::prepare() you must either free mysqli result or close current mysqli statement:

...
//do sth with the data

$mysqli->free(); // or $stmt->close();

$query = "SELECT * FROM answers WHERE question_id = ?";
$stmt  = $mysqli->prepare($query);
...

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