Question

我需要测试我的查询是返回1还是0（所以如果有一个匹配输入的密钥）

这是我的代码：

 $key=$_POST['key'];
 $queryKey = mysql_query("SELECT COUNT(*) FROM `smf_invites` WHERE `key` = '$key'");
 $query = mysql_num_rows($queryKey);

 if( !empty ($key))
 {
    echo 'You have entered a key';
    if (!empty ($query))
    {
        echo 'A key is corresponding';
    }

 }

编辑： $ connect = new mysqli（“XXX”，“XXX”，“XXX”，“smf”）;

if (mysqli_connect_errno()) 
{
    printf("Connection failed : %s\n", mysqli_connect_error());
    exit();
}
else
{
    echo 'Connected to database';
}
$key=$_POST['key'];
if( !empty ($key))
{
 echo 'You have entered a key';
 $key = mysqli_real_escape_string($_POST['key']);
 $queryKey = mysqli_query("SELECT 1 FROM `smf_invites` WHERE `key` =   '$key'");
if (mysqli_num_rows($queryKey))
{
    echo 'A key is corresponding';
}
}
else
{
    echo 'No keys entered';
}

现在我的代码给了我像

这样的错误

Warning: mysqli_real_escape_string() expects exactly 2 parameters, 1 given in D:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\registration.php on line 32

Warning: mysqli_query() expects at least 2 parameters, 1 given in D:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\registration.php on line 33

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, null given in D:\Program Files (x86)\EasyPHP-DevServer-14.1VC11\data\localweb\registration.php on line 34

Answer 1

此代码存在多个问题

如果未设置$_POST['key']，您将收到PHP通知
$queryKey = mysql_query("SELECT COUNT(*) FROM `smf_invites` WHERE `key` = '$key'");将始终返回一行。因此，检查返回的行数将无效。

以下是可以解决您的问题的更新代码。

 if( isset($_POST['key']))
 {
    echo 'You have entered a key';
    $key = mysql_real_escape_string($_POST['key']);
    $queryKey = mysql_query("SELECT 1 FROM `smf_invites` WHERE `key` = '$key'");
    if (mysql_num_rows($queryKey))
    {
        echo 'A key is corresponding';
    }
 }

Please, don't use mysql_* functions in new code。它们不再被维护and are officially deprecated。请参阅red box？转而了解prepared statements，并使用PDO或MySQLi - this article将帮助您确定哪个。如果您选择PDO here is a good tutorial。

您的代码也对SQL injections

Answer 2

与John的回答相结合

根据您新编辑的代码：

mysqli_需要为函数传递DB连接。

变化：

$key = mysqli_real_escape_string($_POST['key']);

为：

$key = mysqli_real_escape_string($connect, $_POST['key']);

或（在那里使用变量$key代替。）

然后是：

$queryKey = mysqli_query("SELECT 1 FROM `smf_invites` WHERE `key` =   '$key'");

为：

$queryKey = mysqli_query($connect, 
"SELECT 1 FROM `smf_invites` WHERE `key` =   '$key'") 
or die(mysqli_error($connect));

故障排除/调试

将error reporting添加到文件的顶部，这有助于查找错误。

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

另外，or die(mysqli_error($connect))

旁注：错误报告应仅在暂存时完成，而不是生产。

Answer 3

我为你创建了一个类，并举例说明了这一点。

<?php
class MyDb {
    protected $link;
    public function __construct() {
        if (!$this->link = mysql_connect('localhost', 'user', 'password')) {
            echo 'Could not connect to mysql';
            exit;
        }

        if (!mysql_select_db('table_name', $this->link)) {
            echo 'Could not select database';
            exit;
        }
    }

    public function exist($id) {
        if (is_int($id)) {
            $sql = "SELECT COUNT(*) FROM smf_invites WHERE `key` = '$id'";
            $result = mysql_query($sql, $this->link);
            return  mysql_num_rows($result);
        } else {
            return false;
        }
    }
}


$x = $_POST['key'] = 1;

if(!empty($_POST['key'])) {
    $db = new MyDb();
    if($db->exist($x)) {
        echo "exist";
    } else {
        echo "no exist";
    }
}

如何返回计数功能？

3 个答案: