好吧,我已经做了两天了——我的代码有点草率&杂乱无章,但我浏览了数百个问题、网站等,寻找答案或简单的解释;不幸的是,我的尝试仍然没有成功。
我正在用PHP/HTML构建一个"测验"游戏——网站引用了一个数据库,特别是一个标有"答案"的表格,其中包含以下信息:
- ID: Auto-Increment
- Question: Varchar
- Answer: Varchar
- Comment: Varchar
现在,网站上的一些信息——一旦用户登录,他/她就可以"玩"游戏;游戏只是一个HTML表单,上面显示一个随机的"答案表"问题。该表单有4个用户输入,但只需要两个。让我进入代码详细信息,然后我会问我的问题:
我的index.php页面(包含游戏表单)目前是:
<?php # index.php
session_start();
//check session first
if (!isset($_SESSION['email'])){
include ('../includes/header.php');
}else
{
session_start();
include ('../includes/header.php');
require_once ('../../mysql_connect.php');
$query = "SELECT * FROM answers ORDER BY RAND() LIMIT 1";
$result = @mysql_query ($query);
$num = mysql_num_rows($result);
if ($num > 0) { // If it ran OK, display all the records.
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
?>
<div class="newGame">
<h2>Are you a Question Master?<hr /></h2>
<h3 style="color:#000">Find Out Now!</h3>
</div>
<br />
<div class="newGameContain">
<form action="gameSubmit.php" method="post" autocomplete="off">
<h2><? echo $row["Question"]."<hr />"; ?></h2>
<h3>Enter Player Answers</h3>
<p><input type="text" placeholder="Player 1" name="player1" value="<? echo $_POST['player1']; ?>" /> <input type="text" placeholder="Player 2" name="player2" value="<? echo $_POST['player2']; ?>" /></p>
<p><input type="text" placeholder="Player 3" name="player3" value="<? echo $_POST['player3']; ?>" /> <input type="text" placeholder="Player 4" name="player4" value="<? echo $_POST['player4']; ?>" /></p>
<p><input type="submit" class="submitButton" /> <input type="reset" class="resetButton" value="Reset" /> </p>
<input type="hidden" name="ID" value="<?php echo $row["ID"]; ?>" />
<input type="hidden" name"Answer" value="<?php echo $row['Answer']; ?>" />
<input type="hidden" name="submitted" value="TRUE" />
</form>
<p></p>
</div>
<br />
<?php
} //end while statement
} //end if statement
mysql_close();
//include the footer
include ("../includes/footer.php");
}
?>
然后我的gameSubmit.php页面(表单操作)看起来是这样的——我只会给出一个快照,而不是整个内容:
<?php # index.php
session_start();
//check session first
if (!isset($_SESSION['email'])){
include ('../includes/header.php');
}else
{
session_start();
include ('../includes/header.php');
require_once ('../../mysql_connect.php');
$query = "SELECT * FROM answers ORDER BY RAND() LIMIT 1";
$result = @mysql_query ($query);
$num = mysql_num_rows($result);
if ($num > 0) { // If it ran OK, display all the records.
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
?>
<? if (isset($_POST['submitted'])){
$correct1Msg = "<div class='correct1Msg'><p style='color:#000;font-family:Arial, Helvetica, sans-serif;'>Player 1 entered the <span id='answerUnder'>correct answer</span>.</p></div><p></p>";
$correct2Msg = "<div class='correct2Msg'><p style='color:#000;font-family:Arial, Helvetica, sans-serif;'>Player 2 entered the <span id='answerUnder'>correct answer</span>.</p></div><p></p>";
$incorrect1Msg = "<div class='incorrect1Msg'><p style='color:#F00;font-family:Arial, Helvetica, sans-serif;'>Player 1 entered the <span id='answerUnder'>incorrect answer</span>.</p></div><p></p>";
$incorrect2Msg = "<div class='incorrect2Msg'><p style='color:#F00;font-family:Arial, Helvetica, sans-serif;'>Player 2 entered the <span id='answerUnder'>incorrect answer</span>.</p></div><p></p>";
$player1Answer = $_POST['player1'];
$player2Answer = $_POST['player2'];
$player3Answer = $_POST['player3'];
$player4Answer = $_POST['player4'];
$questionID = $row['ID'];
if ($questionID == "1" && $player1Answer != "Red"){
echo $incorrect1Msg;
}elseif ($questionID == "2" && $player1Answer != "4"){
echo $incorrect1Msg;
}else {
echo $correct1Msg;
}
if ($questionID == "1" && $player2Answer == "Red"){
echo $correct2Msg;
}elseif ($questionID == "2" && $player2Answer == "4"){
echo $correct2Msg;
}else{
echo $incorrect2Msg;
}
}
?>
<?php
} //end while statement
} //end if statement
mysql_close();
//include the footer
include ("../includes/footer.php");
}
?>
值得注意的是,gameSubmit.php页面也有相同的消息,以及player3Answer&播放器4电源。
所以我的问题是…
如果用户登录并打开index.php页面,系统会提示他/她"echo$row["Question"]"(这是一个使用$query="SELECT*from answers ORDER BY RAND()LIMIT 1"从MySQL数据库中提取的问题);-然后,用户继续在每个玩家各自的文本输入中输入答案。一旦用户单击提交按钮,表单将重定向到gameSubmitt.php-一旦加载,if(isset($_POST['submitted']){启动并交叉检查每个用户的回答并显示相应的消息。
目前,我的表单重定向到gameSubmit.php,然而,它没有引用上一个问题的正确答案——因此,幸运的是,在对答案进行"评分"时,会出现相同的答案。
为了在表单操作页面上实现输入验证,我需要做什么/需要更正什么?
再一次,我只是想随机检索一个问题,并在提交时用正确的答案检查输入的答案——我还希望我的代码能够检索正确的答案,而不是我必须键入每个答案,这样,如果添加了记录,我就不必更新代码。
感谢您的时间和帮助,非常感谢!(现在是期末考试周,我的压力再大不过了)
- 罗克曼德
只需将POST元素从索引页传递给gameSubmit.php,并带有问题id。
在索引页中添加隐藏元素,如。。
<input type="hidden" name="questionId" value="<?php echo $row['id']; ?>">
因此,您可以使用$_POST['questionId']
在pageSubmit.php中获取问题id