Question

JACKPOT页面：

Username: <input type="text"     name="user_name">

<select name="number1">
<option value="0">0</option>
<option value="100">1</option>
<option value="200">2</option>
<option value="300">3</option>
<option value="400">4</option>
<option value="500">5</option>
<option value="600">6</option>
<option value="700">7</option>
<option value="800">8</option>
<option value="900">9</option>
</select>

<select name="number2">
<option value="00">0</option>
<option value="10">1</option>
<option value="20">2</option>
<option value="30">3</option>
<option value="40">4</option>
<option value="50">5</option>
<option value="60">6</option>
<option value="70">7</option>
<option value="80">8</option>
<option value="90">9</option>
</select>

<select name="number3">
<option value="0">0</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
<option value="6">6</option>
<option value="7">7</option>
<option value="8">8</option>
<option value="9">9</option>
</select>



<input type="submit" value="Submit">

</form>

结果页面：

<?php

$date = new DateTime();

define('DB_NAME', 'test');
define('DB_USER', 'root');
define('DB_PASSWORD', '');
define('DB_HOST', 'localhost');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if (!$link) 
{
    die('Could not connect: '. mysql_error());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) 
{
    die('Can\'t use '.DB_NAME.': '.mysql_error());
}

$data = mysql_query("SELECT * FROM config") or die(mysql_error());  
$number1 = mysql_real_escape_string($_POST['number1']);
$number2 = mysql_real_escape_string($_POST['number2']);
$number3 = mysql_real_escape_string($_POST['number3']);
$username = mysql_real_escape_string($_POST['user_name']);

while($info = mysql_fetch_array( $data ))
{
    if($info['jackpot_code'] == ($_POST['number1'] + $_POST['number2'] + $_POST['number3']))
    {
        echo 'CORRECT!';
    }   
    else if(($info['jackpot_code'] - ($_POST['number1'] + $_POST['number2'] + $_POST['number3'])) <= 30 && ($info['jackpot_code'] - ($_POST['number1'] + $_POST['number2'] + $_POST['number3'])) >= 0 or ($info['jackpot_code'] - ($_POST['number1'] + $_POST['number2'] + $_POST['number3'])) >= -30 && ($info['jackpot_code'] - ($_POST['number1'] + $_POST['number2'] + $_POST['number3'])) <= 0){
        $code = $_POST['number1'] + $_POST['number2'] + $_POST['number3'];
        echo "Guessed code: $code<br>";
        die('The code u guessed is at the range of +/- 30 to jackpot.');
    } else
    {
        die('CODE FALSE!');
    }
}

$number = $_POST['number1'] + $_POST['number2'] + $_POST['number3'];

echo "<br>Guessed code: $number";

$sql = "INSERT INTO jackpot_log (username, code, time) VALUES ('".$username."','".$number."','".$date->getTimestamp()."')";

if (!mysql_query($sql)) 
{
    die('Error: '. mysqli_error($con));
}
?>

这是我的代码，它完全有效。问题是：我想从我的数据库的“配置”表中更改我的“jackpot_code”。有没有办法随机化这个表，并在每次成功尝试后生成一个新的3位数代码。

这是我的数据库：我需要做的就是在每次正确尝试后随机化“jackpot_code”。 CLICK TO OPEN THE PHOTO OF DATABASE!

Answer 1

试试这个......

 $random =rand(100,999);

Answer 2

如果您的数据库已满，并且您希望获得随机数据，请使用

SELECT * FROM config ORDER BY rand() LIMIT 1

或者如果您想要生成新的，只需在回显“正确”之前添加它

INSERT INTO config (jackpot_code) VALUES (rand())

并更改选择

SELECT * FROM config ORDER BY id ASC LIMIT 1

Answer 3

成功插入命令后使用以下代码

如果jackpot_code为int数据类型

 <?php 


   mysql_query("UPDATE `config` SET `jackpot_code`= ".rand(100,990)." WHERE `id`=1");

?>

如果jackpot_code为varchar数据类型

 <?php 


   mysql_query("UPDATE `config` SET `jackpot_code`= '".rand(100,990)."' WHERE `id`=1");

?>

数据库中的随机数据

3 个答案: