php - 警告 : mysqli_stmt_bind_param() expects parameter 1 to be mysqli_stmt, bool 值给出？

Question

这个问题已经有答案了:

What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such (1 个回答)

已关闭 5 年前。

这是我第一次尝试使用mysqli prepared statement来保护我的代码免受sql注入(inject)。 。所以请温柔一点，用简单的语言解释一下，以便我能理解。

现在我正在使用以下代码，我认为我是对的，但它引发了这些错误，我根本不明白。

错误如下:

Warning: mysqli_stmt_bind_param() expects parameter 1 to be mysqli_stmt, boolean given in on line 92

Warning: mysqli_stmt_execute() expects parameter 1 to be mysqli_stmt, boolean given in on line 93

Warning: mysqli_stmt_close() expects parameter 1 to be mysqli_stmt, boolean given in  on line 96

代码如下:

$stmt = mysqli_prepare(
    $db_conx,
    "INSERT $storenameTable (firstname, lastname, username, address_1, address_2, postcode,  country, county, city, email, password, storeShop, signupdate) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)"
);
//after validation, of course
mysqli_stmt_bind_param($stmt, "issi", $firstname, $lastname, $username, $address_1, $address_2, $postcode, $country, $county, $city, $email, $hashedPass, $storenameTable);
mysqli_stmt_execute($stmt);     <//<<<<<<<< line 92
if (mysqli_affected_rows($db_conx))     <//<<<<<<<< line 93
{
    mysqli_stmt_close($stmt);  <//<<<<<<<< line 96
    //update was successful
    $id = mysqli_insert_id($db_conx);
}

非常感谢您的帮助。

Answer 1

看来你缺少一个参数，你应该有13个参数和13个？检查密码后的两个参数。 (我拿出了 signupdate)尝试以下操作:

$stmt = mysqli_prepare(
    $db_conx,
    "INSERT INTO $storenameTable (firstname, lastname, username, address_1, address_2, postcode,  country, county, city, email, password, storeShop) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)"
);
//after validation, of course
mysqli_stmt_bind_param($stmt, "issi", $firstname, $lastname, $username, $address_1, $address_2, $postcode, $country, $county, $city, $email, $hashedPass, $storenameTable);
mysqli_stmt_execute($stmt);     <//<<<<<<<< line 92
if (mysqli_affected_rows($db_conx))     <//<<<<<<<< line 93
{
    mysqli_stmt_close($stmt);  <//<<<<<<<< line 96
    //update was successful
    $id = mysqli_insert_id($db_conx);
}

您还可以使用 var_dump(mysqli_error($db_conx)); 获取有关最后一个错误的更多详细信息