当我尝试从MySQL检索信息时，我的页面崩溃 - My page crashes when I try to retrieve information from MySQL

My page crashes when I try to retrieve information from MySQL

本文关键字：我的崩溃信息检索 MySQL | 更新日期: 2024-03-31

我的PHP脚本有一些问题。

我有一个PHP函数，它存储在一个PHP文件中，我正试图从另一个脚本运行该PHP函数。

所以用代码来解释我自己：

类似.inc.php：

function post_exists($id) {
    $host = "example";
    $username = "example";
    $password = "example";
    $database = "example";
    $connection = new mysqli($host, $username, $password, $database);
    $id = $connection->real_escape_string($id);
    $query = $connection->query("SELECT COUNT(`id`) AS `count` FROM `posts` WHERE `id` = '$id'");
    while ( $row = $objQuery->fetch_object() ) {
            if ( $row->count == 1 ) return true;
}
}

profile.php：

include ( 'like.inc.php' );
    if (post_exists(70) === true) {
        echo 'Exists!';
    }

ID为70的帖子已经存在，所以它应该回显exists！但它只是崩溃了我的一半页面。所以也许它没有正确加载？

如有任何帮助，我们将不胜感激！

有$objQuery的地方必须有$query。尝试使用此代码：

 function post_exists($id) {
        $host = "example";
        $username = "example";
        $password = "example";
        $database = "example";
        $connection = new mysqli($host, $username, $password, $database);
        $id = $connection->real_escape_string($id);
        $query = $connection->query("SELECT COUNT(`id`) AS `count` FROM `posts` WHERE `id` = '$id'");
        while ( $row = $query->fetch_object() ) {
                if ( $row->count == 1 ) return true;
    }
    }

您可以用不同的方式编写函数，如下所示：

function post_exists($id) {
    $host = "example";
    $username = "example";
    $password = "example";
    $database = "example";
    $connection = new mysqli($host, $username, $password, $database);
    $id = $connection->real_escape_string($id);
    // Instead of getting count, just get the row and 
    // do the count with PHP
    $query = $connection->query("SELECT * FROM `posts` WHERE `id` = '$id' LIMIT 1");
    if($query){
        return $query->num_rows > 0;
    }
    return false;
}