問題描述

對于實際存在的電子郵件，我有一個簡單的準備聲明:

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';
}

結果:在應該回顯 hello

我在控制臺中運行了相同的查詢，并使用與上述相同的電子郵件得到了結果.

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

我也使用簡單的 mysqli 查詢進行了測試:

I tested using a simple mysqli query as well:

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

}

結果:我所期望的hello

還有 $result 的 num_rows 是 1.

Also $result's num_rows is 1.

為什么準備好的語句的num_row不大于0?

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

推薦答案

當您通過 mysqli 執行語句時，結果實際上并不在 PHP 中，直到您獲取它們——結果由數據庫引擎保存.所以mysqli_stmt對象無法知道執行后立即有多少結果.

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.

像這樣修改你的代碼:

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

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

查看手冊

這不適用于您的特定示例，但如果您的結果集很大，$stmt->store_result() 將消耗大量內存.在這種情況下，如果您只關心是否至少返回了一個結果，請不要存儲結果；相反，只需檢查結果元數據是否不為空:

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;

查看手冊

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