无法将mysqli_result类的对象转换为int，并且所有条目均返回true [英] Object of class mysqli_result could not be converted to int and all entries return true

问题描述

我正在进行一个简单的登录.一种逻辑是，如果用户名不存在，则他们会收到一条错误消息，并将其重定向到注册页面. 但是，当我运行代码时，仍然收到上述错误，无论我以用户名输入什么，它仍然会返回true.

I am working on a simple login. One piece of logic is that if the username does not exist then they get an error message and they get redirected to a registration page. However when I run my code I keep getting the error described above and no matter what I enter as a username it still comes back as true.

   $res = $mysqli->query("SELECT Count(`userID`) FROM `users` WHERE `UserName` = '$username'");
    if($res ==0){
        return false;
    }
    else
        return true;
}

我在开发数据库时控制数据库，因此知道其中包含什么用户名.我的印象是，如果我输入数据库中的用户名，则计数将为1，因此将返回true，同样，如果我输入的名称不在数据库中，则计数将为0，因此为$ res会是假的吗?

I control the database as I'm developing it and thus know what usernames are in it. I was under the impression that if I put in a username that was in the database the count would be 1 and thus would return true, likewise if I entered a name that is not in the database then the count would be 0 and thus $res would be false?

推荐答案

您仍然要数数！ mysqli :: query的返回值不是整数，它是结果对象.

You still have to count! The return value of mysqli::query is NOT an integer, it is a result object.

if ($res->num_rows == 0) {
    return false;
} else {
    return true;
}

冗长地创建返回值被认为不是最佳的编码风格.

It is considered not the best coding style to verbosely create the return value.

return ($res->num_rows > 0); // will be true if result is bigger than 0

这会导致一个问题:如果返回两行或更多行该怎么办?

And this leads to the question: What if there are TWO OR MORE rows returned?

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