PHP get_result（）返回一個空的結果集

Question

所以，我有一份聲明，我已經成功地製備，約束，並執行：

SELECT * FROM users WHERE email = ? AND pass = SHA1(?) 

看起來返回一行，符合市場預期。但是，當我致電@$stmt->get_result()時，爲什麼$ result變量爲空？提前致謝。

$num_rows = mysqli_num_rows($stmt->get_result()); 

if ($num_rows == 1) { 

    // Fetch the result set 
    $result = $stmt->get_result(); 

    //if result empty echo false 
    if(empty($result)) { 
     echo "result is empty"; 
    } 
} 

Answer 1

只是把兩點意見一起，制定一個豆蔻....

<?php 
$result = $stmt->get_result(); 
$num_rows = mysqli_num_rows($result); 

if ($num_rows == 1) { 
    // already fetched the mysqli_result 
    // now lets fetch the one record from that result set 
    $row = $result->fetch_assoc(); 
    // ....do something with row 
} 
else { 
    // either 0 ...or more than 1 row 
    foo(); 
} 

但是，你甚至可以擺脫調用到mysqli_num_rows（）（所以它也可以在的情況下， unbuffered queries）

$result = $stmt->get_result(); 
$row = $result->fetch_assoc(); 
if (!$row) { 
    // no such record 
    foo(); 
} 
else { 
    // ....do something with row 

    // might want to check whether there are more matching records 
    // given the context there shouldn't, but ... 
}