问题描述

对于实际存在的电子邮件，我有一个简单的准备声明:

I have a simple prepared statement for an email that actually exists:

$mysqli = new mysqli("localhost", "root", "", "test");
if (mysqli_connect_errno()) {
    printf("Connect failed: %s
", mysqli_connect_error());
    exit();
}

$sql = 'SELECT `email` FROM `users` WHERE `email` = ?';
$email = 'example@hotmail.com';

if ($stmt = $mysqli->prepare($sql)) {
    $stmt->bind_param('s', $email);
    $stmt->execute();

    if ($stmt->num_rows) {
        echo 'hello';
    }

    echo 'No user';
}

结果:在应该回显 hello

我在控制台中运行了相同的查询，并使用与上述相同的电子邮件得到了结果.

I ran the same query in the console and got a result using same email as above.

我也使用简单的 mysqli 查询进行了测试:

I tested using a simple mysqli query as well:

if ($result = $mysqli->query("SELECT email FROM users WHERE email = 'example@hotmail.com'")) {
    echo 'hello';

}

结果:我所期望的hello

还有 $result 的 num_rows 是 1.

Also $result's num_rows is 1.

为什么准备好的语句的num_row不大于0?

Why is the prepared statment's num_row not greater than 0?

推荐答案

当您通过 mysqli 执行语句时，结果实际上并不在 PHP 中，直到您获取它们——结果由数据库引擎保存.所以mysqli_stmt对象无法知道执行后立即有多少结果.

When you execute a statement through mysqli, the results are not actually in PHP until you fetch them -- the results are held by the DB engine. So the mysqli_stmt object has no way to know how many results there are immediately after execution.

像这样修改你的代码:

$stmt->execute();
$stmt->store_result(); // pull results into PHP memory

// now you can check $stmt->num_rows;

查看手册

这不适用于您的特定示例，但如果您的结果集很大，$stmt->store_result() 将消耗大量内存.在这种情况下，如果您只关心是否至少返回了一个结果，请不要存储结果；相反，只需检查结果元数据是否不为空:

This doesn't apply to your particular example, but if your result set is large, $stmt->store_result() will consume a lot of memory. In this case, if all you care about is figuring out whether at least one result was returned, don't store results; instead, just check whether the result metadata is not null:

$stmt->execute();
$hasResult = $stmt->result_metadata ? true : false;

查看手册

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