MySQL COUNT 在 PHP 中不返回 False 或 True [英] MySQL COUNT not returning False or True in PHP

问题描述

所以我正在尝试检查用户是否已经喜欢了某个帖子.

So I'm trying to check if a user already liked a post.

代码:

function previously_liked($id) {
    if (isset($_SESSION['user_login'])) {
        $user = $_SESSION["user_login"];
    }
    else {
        $user = "";
    }

    $connection = new mysqli($host, $username, $password, $database);

    $id = $connection->real_escape_string($id);

    $query = $connection->query("SELECT COUNT(`id`) AS `count` FROM `likes` WHERE `id` = '$id' AND `user_id` = '".$connection->escape_string($user)."'");

    while ( $row = $query->fetch_object()->count ) {
            if ( $row->count == 0 ) return false;
            else return true;
}
}

问题是，它不会返回 false 或 true.在我的另一个 PHP 页面上，我试图像这样运行这个函数:

The thing is, it's not returning false or true. On my other PHP page I'm trying to run this function like this:

if (previously_liked(70) === true) {
    $aliked = "You've already liked this!";
}

echo $aliked;

我看到的唯一错误是 Notice: Undefined variable: aliked in profile.php on line 391

欢迎任何帮助！谢谢！

推荐答案

你可以这样做来发现代码中的错误:

You can do this things to finds error in your code:

echo $user; //before MySQL query

在替换.$connection->escape_string($user)"后，在 MySQL 控制台或 PHPmyAdmin 等应用程序中查询您的 MySQL 语句.到您的 user_id.如果它获取的行比您的 MySQL 语句正确，并且 $_SESSION["user_login"] 值有问题.

Query your MySQL statement in MySQL console or app like PHPmyAdmin after replacing ".$connection->escape_string($user)." to your user_id. If it fetch row than your MySQL statement is correct and there is something wrong with $_SESSION["user_login"] value.

可选地，您还可以回显返回的行以检查输出数据.

Optionally, you can also echo the returning rows to check the output data.

否则检查您的查询语句，如列名、表名、数据库等

Otherwise check your query statement as column names, table name, database etc.

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