Question

从SQL数据库中动态获取变量并将其显示在网页上的表单，用户从该表单中选择任何选项后，它应存储在一些变量中。

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

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

$query = mysql_query("SELECT * FROM `microsoftq`  ORDER BY RAND() LIMIT 1 ");

    $rows1 = mysql_fetch_array($query);
    $q1 = $rows1['QNo'];
    $qus1 = $rows1['Question'];
    $a1 = $rows1['Opt1'];
    $b1 = $rows1['Opt2'];
    $c1 = $rows1['Opt3'];
    $d1 = $rows1['Opt4'];
    $ans1 = $rows1['Ans'];
    echo " <b>Question:-<br></b>$qus1 <br><br>";
    echo " <input type=radio name = 'answer$q1' ></input>$a1 &nbsp &nbsp<br>"; 
    echo " <input type=radio name = 'answer$q1' ></input>$b1 &nbsp &nbsp<br>"; 
    echo " <input type=radio name = 'answer$q1' ></input>$c1 &nbsp &nbsp <br>"; 
    echo " <input type=radio name = 'answer$q1' ></input>$d1 <br><br> "; 
<input type="submit" id="submit_id" name="SUBMIT" value="SUBMIT">
</form>

我正在尝试将变量存储在这种类型中，工作但是从数据库中随机选择值，其中一些也缺失了，...

<?php
if (isset($_POST['SUBMIT'])) 
{

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

    $opt1=$_POST["answer$q1"]; // problem is here 

    $id1=array("$q1","$opt1");
    if($ans1==$opt1)
    {
    $val1="ct";
    }
    else
    {
    $val1="wg";
    }

mysql_query("insert into stu1 values('$id1[0]','$id1[1]','$val1')")
or die(mysql_error());
?>

所以有人会帮助我解决这个问题，...

将动态值从FORM存储到PHP变量

0 个答案: