Question

我的PHP脚本存在一些问题。

我有一个存储在PHP文件中的PHP函数，我正在尝试从另一个脚本运行该PHP函数。

所以用代码解释自己：

like.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!';
    }

存在ID为70的帖子，因此它应该回显Exists！但它只是崩溃了我的一半页面。也许它没有正确加载？

任何帮助都将受到高度赞赏！

Answer 1

$objQuery必须$query的位置。试试这段代码：

 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;
    }
    }

Answer 2

你可以用不同的方式编写函数：

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;
}

当我尝试从MySQL检索信息时，我的页面崩溃了

2 个答案: