Question

我尝试获取ID = gmail = $ gmail并将其用于UPDATE表

但是我无法使用id和showin错误未定义的属性：mysqli_result :: $ fetch_assoc

函数addverificationcode（$ gmail，$ random） {

    $connection = mysqli_connect(DataBaseManager::HOST, DataBaseManager::USER, DataBaseManager::PASSWORD, DataBaseManager::DATABASENAME);
    mysqli_set_charset($connection, "utf8");
    $sqlQuery = "SELECT id FROM users WHERE gmail='$gmail'";
    $result = mysqli_query($connection, $sqlQuery);
    if ($result->num_rows > 0) {
            $sqlCommand = "UPDATE users
                      SET verificationcode  = '$random' 
                        WHERE id = $result->fetch_assoc()";
    }

    if(mysqli_query($connection, $sqlCommand)){
        return true;
    }else{
        echo("Error description: " . mysqli_error($connection));
        return false;
    }


    }

Answer 1

在对象中调用方法时必须使用curly braces：

        $sqlCommand = "UPDATE users
                  SET verificationcode  = '$random' 
                    WHERE id = {$result->fetch_assoc()}";

一种替代方法是使用串联：

        $sqlCommand = "UPDATE users
                  SET verificationcode  = '$random' 
                    WHERE id = " . $result->fetch_assoc();

请注意，在这种情况下，您可以将两个SQL语句合并为一个，例如update .. where id = (select id from ...)。

还请注意，您发布的代码容易受到SQL注入攻击的攻击。参见How can I prevent SQL injection in PHP?

Answer 2

fetch_assoc为您提供记录的数组:) 试试这个代码

 if ($result->num_rows > 0) {
   while($res = mysqli_fetch_array($result)){
            $id = $res['id'];     // you are breaking the array here and storing the array index in a new variable
            $sqlCommand = "UPDATE users
                      SET verificationcode  = '$random' 
                        WHERE id = $id";
       }
    }

现在它将可以使用。祝你好运：）

无法显示SELECT结果sql

2 个答案: