Question

这是我第一次尝试使用mysqli prepared statement保护我的代码免受sql注入。所以请保持温和，用简单的语言解释一下，这样我才能理解。

现在我正在使用以下代码，我认为我是对的，但它会抛出这些错误，我根本不理解。

这是错误：

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

这是代码：

$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);
}

我将非常感谢你的帮助。

Answer 1

似乎你有一个缺少的参数，你应该有13个参数和13 ?检查密码后的两个参数。（我拿出signupdate）尝试以下内容：

$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));

获取有关上一个错误的更多详细信息

Answer 2

password是MySQL中的函数名。函数名称（如保留字）必须用反引号括起来用作字段名称。

就个人而言，我会说在所有数据库，表格和列名称周围添加反引号。

在MySQL中使用“裸”名称类似于在PHP中使用裸字符串。当然，$foo = bar;将工作，但它依赖于bar不是常数。好吧，在MySQL中，你依赖的是你的列名而不是保留字。一样。使用反引号！

警告：mysqli_stmt_bind_param（）期望参数1是mysqli_stmt，给定布尔值？

2 个答案: