好的,所以我一直在這工作了兩天 - 我的代碼有點草率&混亂,但我已經超過了數百個問題,網站等。等尋找答案或只是我理解的解釋;不幸的是,我的努力仍然不成功。PHP/MySQL「測驗」遊戲 - 表單輸入值
我建立在PHP/HTML一個「測驗」遊戲 - 網頁引用了一個數據庫,具體地說,提交討論標記爲「答案」,它擁有以下信息:
- ID: Auto-Increment
- Question: Varchar
- Answer: Varchar
- Comment: Varchar
現在,對於一些信息在網站上 - 一旦用戶登錄,他/她可以「玩」遊戲;遊戲只是一個HTML表單,它在上面顯示了一個隨機的「答案表」問題。該表格有4個用戶輸入,但只需要兩個。讓我進入代碼的細節,然後我會問我的問題:
我的index.php頁面(其中包含遊戲形式)是目前:
<?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 $row['Answer']; ?>" />
<input type="hidden" name="submitted" value="TRUE" />
</form>
<p></p>
</div>
<br />
<?php
} //end while statement
} //end if statement
mysql_close();
//include the footer
include ("../includes/footer.php");
}
?>
然後我gameSubmit.php頁面(表單操作)看起來是這樣的 - 我只會給一個快照,而不是整個事情:
<?php # index.php
session_start();
//check session first
if (!isset($_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 $incorrect2Msg;
}
}
?>
<?php
} //end while statement
} //end if statement
mysql_close();
//include the footer
include ("../includes/footer.php");
}
?>
作爲一個說明,該gameSubmit.php頁面也有相同的消息,如果... ... ELSEIF爲player3Answer &聲明player4Answer。
所以我的問題是...
如果用戶登錄並打開index.php頁面,他/她會被提示「回聲$行[」問題「]」(這是一個使用$ query =「SELECT * FROM answers ORDER BY RAND()LIMIT 1」從MySQL數據庫中提取問題; - 然後用戶在每個玩家各自的文本輸入中輸入一個答案,一旦用戶點擊提交按鈕,表單重定向到gameSubmit.php - 一旦加載,如果(isset($ _ POST ['submitted'])){啓動並交叉檢查每個用戶的答案並顯示相應的消息
目前,我的表單重定向到gameSubmit.php,但是,它沒有引用前面的問題fo正確的答案 - 因此它的純粹運氣在「評分」答案時出現相同的答案。
爲了在表單操作頁面上實現輸入驗證,我需要做什麼/需要糾正哪些內容?
再一次,我只是想隨機檢索一個問題,並提交檢查正確答案的輸入答案 - 我還想我的代碼能夠檢索正確的答案,而不是我必須輸入每個回答,這樣,如果記錄被添加,我不必更新代碼。
感謝您的時間和幫助,非常感謝! (它的最後一週,我不能更強調)
- Rockmandew
此外,看到一個工作示例 - http://kethcart.uwmsois.com/qm/htdocs/Home/index.php - 隨時註冊一個帳戶,並登錄到登錄主頁。 – rockmandew