使用mysqli的排名系统

Question

我想检查用户的等级是多少，它以整数 0、1、2 或 3 的形式存储在我的数据库中。

我希望它像 echo $my_rank 那样被回显； #与其说 0、1、2 或 3，不如说 User 代表 0，Moderator 代表 1，Admin 代表 2，Owner 代表 3。

这是我的代码：

$my_rank = $_SESSION['rank'] = array("owner","administrator","moderator","user");

$rank = array (
    '0' => 'user',
    '1' => 'moderator',
    '2' => 'administrator',
    '3' => 'owner'
);

config.php

<?php

# Error Reporting
error_reporting(E_ERROR | E_WARNING | E_PARSE | E_NOTICE);

# No Error Reporting
//error_reporting(0);

# Message Responses
$success = array();
$danger = array();
$warning = array();
$info = array();

# Define the Database
define('DB_SERVER','localhost');
define('DB_USERNAME','blah');
define('DB_PASSWORD','blah');
define('DB_DATABASE','blah');

# Connect to the database
$con = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_DATABASE);
if (!$con) {
    $danger[] = mysqli_connect_error();
} else {
    //echo "It worked";
}
?>

Answer 1

可以使用 sql 语句查询数据库。您正在寻找的是用户表上的 select：

// array which holds the info
$rank = array ('user', 'moderator', 'administrator', 'owner');
// set up db connection
$con = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_DATABASE);
if (!$con) {
    $danger[] = mysqli_connect_error();
} else { 
    //echo "It worked";

    // Baklap4's Addition:

    // SQL statement to retrieve the row where username is piet
    $sql = "SELECT * FROM Users WHERE username = `piet`";

    // Execute the query and check if there is a result.
    if ($result = $con->query($sql)) { 
        // if there are results loop through them.
        while ($row = mysqli_fetch_assoc($result)) {
            // print for each rank a different rank name.
            echo $rank[$row['rank']];
        } 
        // free resultset
        $result->close();
    } 
    // close connection
    $con->close();
    //clean up the temp variables 
    unset($obj); 
    unset($sql); 
    unset($result);
}

这将从您的数据库中检索用户 Piet 并打印排名（在数据库中列出）。然后它将取消设置在 while 循环中创建的所有临时值，因此您不能再次使用它们。

您应该做的是将硬编码值 piet 更改为变量。人们可以去 http://localhost/scriptname.php?username=piet 检索相同的结果。

要使其成为变量，请按如下方式更改 sql 行：

$sql = "SELECT * FROM Users WHERE username = $_GET['username']";

Answer 2

嗯...根据你给我的信息，似乎只需要一个查询，从数据库中获取数据。然后我们就可以回显了。

在您的数据库连接的 else 块中，我们可以执行以下操作：

# Connect to the database
$con = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_DATABASE);
if (!$con) {
    $danger[] = mysqli_connect_error();
} else {
    //echo "It worked";

    // HopefulLlama code below:

    // Construct our SQL query, which will retrieve data from the database.
    $sql="SELECT * FROM Users";
    $result=mysqli_query($con,$sql);

    // Loop through all results return by our query
    while ($row = mysqli_fetch_assoc($result)) {
        // $row['rank'] will contain our Integer of 0, 1, 2, or 3
        echo $row['rank']

        // This will access your $rank array, and use the index retrieved from the database as an accessor.
        echo $rank[$row['rank']]
    }
}

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