隨機挑選變量並將其存儲在SQL數據庫中

Question

我正在爲我的公司進行在線aptitude測試，它將從數據庫中挑選20個隨機問題並將其顯示在網頁上以供回答。

問題是，它無法正常在數據庫中存儲的值（雖然存儲的問題和答案到數據庫的越來越混亂了），請任何一個能幫助我解決這個問題，

下面的代碼正從候選答案（簡單的演示一樣拿起只有3個隨機出題），..

<form id="form1" name="quest" method="POST" action="" style="margin-left:60px;"> 


<?php 

    $connect = mysql_connect("localhost","root","") 
    or die(mysql_error()); 
    $sel=mysql_select_db("demo"); 

$query = mysql_query("SELECT * FROM `questions` ORDER BY RAND() LIMIT 3 "); 


    $rows = mysql_fetch_array($query); 
    $q1 = $rows['QNo']; 
    $qus1 = $rows['Question']; 
    $a = $rows['Opt1']; 
    $b = $rows['Opt2']; 
    $c = $rows['Opt3']; 
    $d = $rows['Opt4']; 
    $ans = $rows['Ans']; 


    echo " <b>Question:-<br></b>$qus1 <br>"; 
    echo " <input type=radio name = 'answer$q1' value = '$a'></input>$a &nbsp &nbsp"; 
    echo " <input type=radio name = 'answer$q1' value = '$b'></input>$b &nbsp &nbsp"; 
    echo " <input type=radio name = 'answer$q1' value = '$c'></input>$c &nbsp &nbsp "; 
    echo " <input type=radio name = 'answer$q1' value = '$d'></input>$d <br><br> "; 



    $rows = mysql_fetch_array($query); 
    $q2 = $rows['QNo']; 
    $qus2 = $rows['Question']; 
    $a = $rows['Opt1']; 
    $b = $rows['Opt2']; 
    $c = $rows['Opt3']; 
    $d = $rows['Opt4']; 
    $ans = $rows['Ans']; 


    echo " <b>Question:-<br></b>$qus2 <br>"; 
    echo " <input type=radio name = 'answer$q2' value = '$a'></input>$a &nbsp &nbsp"; 
    echo " <input type=radio name = 'answer$q2' value = '$b'></input>$b &nbsp &nbsp"; 
    echo " <input type=radio name = 'answer$q2' value = '$c'></input>$c &nbsp &nbsp "; 
    echo " <input type=radio name = 'answer$q2' value = '$d'></input>$d <br><br> "; 




    $rows = mysql_fetch_array($query); 
    $q3 = $rows['QNo']; 
    $qus3 = $rows['Question']; 
    $a = $rows['Opt1']; 
    $b = $rows['Opt2']; 
    $c = $rows['Opt3']; 
    $d = $rows['Opt4']; 
    $ans = $rows['Ans']; 


    echo " <b>Question:-<br></b>$qus3 <br>"; 
    echo " <input type=radio name = 'answer$q3' value = '$a'></input>$a &nbsp &nbsp"; 
    echo " <input type=radio name = 'answer$q3' value = '$b'></input>$b &nbsp &nbsp"; 
    echo " <input type=radio name = 'answer$q3' value = '$c'></input>$c &nbsp &nbsp "; 
    echo " <input type=radio name = 'answer$q3' value = '$d'></input>$d <br><br> "; 

?> 


<input type="submit" id="submit_id" name="SUBMIT" value="SUBMIT"> 
</form> 

下一頁一部分存入數據庫，..

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

if (isset($_POST['SUBMIT'])) 
{ 


$opt1=$_POST["answer1"]; 
$opt2=$_POST["answer2"]; 
$opt3=$_POST["answer3"]; 

$username=$_GET['username']; // getting this value from last webpage pls dont worry about this 


    $connect = mysql_connect("localhost","root","") 
    or die(mysql_error()); 
    $sel=mysql_select_db("demo"); 


mysql_query("insert into $username values('$qus1','$opt1')") 
or die(mysql_error()); 
mysql_query("insert into $username values('$qus2','$opt2')") 
or die(mysql_error()); 
mysql_query("insert into $username values('$qus3','$opt3')") 
or die(mysql_error()); 


print "<script>window.close('techtest.php'); window.location = \"final.html\";</script>"; 

} 

?> 

Answer 1

有什麼問題號碼？這就是問題

請儘量不要重複代碼段。因此，它重寫這樣的事情：

<form id="form1" name="quest" method="POST" action="" style="margin-left:60px;"> 
<?php 

function input_option($QNo,$Opt) 
{ 
    echo "<input type=radio name='answer$QNo' value='$Opt'>$Opt</input>&nbsp&nbsp"; 
} 

$count = 3; 
$connect = mysql_connect("localhost","root","") or die(mysql_error()); 
$sel  = mysql_select_db("demo"); 
$query = mysql_query("SELECT * FROM `questions` ORDER BY RAND() LIMIT $count"); 

while ($row = mysql_fetch_assoc($query)) 
{ 
    extract($row); 
    echo "<b>Question:-<br></b>$Question <br>". 
     input_option($QNo,$Opt1). 
     input_option($QNo,$Opt2). 
     input_option($QNo,$Opt3). 
     input_option($QNo,$Opt4); 
} 

?> 
    <input type="submit" id="submit_id" name="SUBMIT" value="SUBMIT"> 
</form> 

現在我沒有看到任何東西錯頁，讓我們看看其他的代碼。我不認爲我們知道問題的數量，請記住數據庫中存在多個問題，並隨機選擇三個。他們可以有任何號碼。因此，下面的代碼段必須處理：

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

if (isset($_POST['SUBMIT'])) 
{ 
    $connect = mysql_connect("localhost","root","") or die(mysql_error()); 
    $sel  = mysql_select_db("demo"); 
    $username = $_GET['username']; 
    foreach ($_POST as $post) 
    if (substr($post,0,6) == 'answer') 
    { 
    $question = substr($post,6); 
    $option = ${$post}; 
    $sql  = "insert into $username values('$question','$option')"; 
    mysql_query($sql) or die(mysql_error()); 
    } 
} 

?> 
<script> 
    window.close('techtest.php'); 
    window.location = \"final.html\"; 
</script> 

正如你所看到的，我真的使用問題編號。請注意，編寫此示例代碼時，安全性未被考慮。黑客有足夠的空間。