MySQLi＆amp; PHP - 我看不到的明显的东西

Question

有人可以帮我这个吗？这让我很生气。我看不出有什么不对......

$stmt_query = "SELECT aboutme_profile FROM ProfilesAboutMe WHERE aboutme_profile = ? LIMIT 1";
$stmt = mysqli_prepare($sql_connect, $stmt_query);
mysqli_stmt_bind_param($stmt, "i", $return_profile_id);
mysqli_stmt_execute($stmt);
echo mysqli_stmt_num_rows($stmt); // Always shows 0
echo $return_profile_id; // Shows correct value

if (mysqli_stmt_num_rows($stmt) == 0)
{
    $stmt_query = "INSERT INTO ProfilesAboutMe (aboutme_profile, aboutme_text) VALUES (?, ?)";
    $stmt = mysqli_prepare($sql_connect, $stmt_query);
    mysqli_stmt_bind_param($stmt, "is", $return_profile_id, $aboutme);
    mysqli_stmt_execute($stmt);
}

两个查询都没问题，因为我已将它们复制并粘贴到phpMyAdmin中，并且在手动绑定params时它们运行正常。

  

警告：mysqli_stmt_bind_param（）期望参数1为mysqli_stmt，布尔值在

中给出

SELECT aboutme_profile FROM ProfilesAboutMe WHERE aboutme_profile = 123456789 LIMIT 1

当我在PHP中编辑查询以硬编码aboutme_profile值时，我得到0结果。我在MySQL中运行相同的查询并得到1个结果。