Php 随机选取变量并将其存储在SQL数据库中

我正在为我的公司进行在线能力倾向测试，该测试将从数据库中随机抽取20个问题，并显示在网页上进行回答

问题是，在将问题和答案存储到数据库时，它没有正确地将值存储在数据库中，它变得混乱，请任何人都可以帮助我解决这个问题

下面的代码是从候选人的简单演示中得到的答案，比如只挑选3个随机问题

<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>

下一部分是存储到数据库中

<?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>";

}

?>

问题号是多少？这就是问题所在 请尽量不要重复代码。因此，将其改写为如下内容：

<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>

现在我看不出该页面有任何其他错误，所以让我们看看另一段代码。我认为我们不知道问题的数量，记住数据库中有多个问题，你随机选择三个。他们可以有任何号码。下面的代码必须处理这个问题：

<?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，我确实使用了问题编号。请注意，在编写此示例代码时，安全性并不是一个考虑因素。有足够的空间来破解这个问题。

你需要获取带答案的问题编号，同时在问题显示上发布参数你需要添加你的问题编号，在发回你的答案时，你需要使用数组发布带答案/选项的问题编号，然后你可以获取值并相应地插入你的问题。我认为有问题这里$opt1=$_POST[answer1]$opt2=$_POST[answer2]$opt3=$_POST[回答3]；我没有从表格中正确地选择值，…这就是我所说的，你只要得到所有答案1，2，3你也应该得到相关的Qno，那么我认为它是正确的。。检查wt ur从表单发送和wt ur接收..您好，为了优化代码，您的代码正在显示问题，但未将值存储在数据库中，。。