Php 从数据库中获取答案

我正在制作一个在线测试脚本，您可以在其中的输入元素中输入答案。提交测试后，我希望将数据库的答案与输入的答案进行比较，以说明是否错误，但我使用的脚本不起作用！：

问题就在这里！ 在数据库中，我有50个现成答案中的4个（还不是全部），当我回答4个正确或错误时，返回的是不正确的答案。它列出了所有答案，无论它们在页面中是正确的还是不正确的，但它工作不正常，无论我做什么输入，它都会显示所有答案，直到49个都说不正确，然后出于某种原因，50个说正确

这是我的剧本：

 <?php
$con=mysqli_connect("localhost","dstsbsse","pass","user");
if (mysqli_connect_errno($con))
  {
  echo "ERROR - Failed to connect to MySQL Server. Please contact an Administrator at    English In York: " . mysqli_connect_error();
  }

//Set variables to hold output data and total score.

$output="";
$score=0;

//for-next loop.  This means "Set n to value one.  Every time through the loop (between {}) increase n by one.  Do this while n is less than or equal to 50"

for($n=1;$n<=50;$n++)
    {
    $sql="SELECT a$n FROM answer WHERE 1";
    // $sql="SELECT * FROM answer WHERE name='a$n'";  //sql is specific to your table of course - you will need to change this.
    $result = $con->query($sql); // perform the query
    $row = $result->fetch_assoc();  //load the result into the array $row
    $key="a".$n;                     //concatenate to generate the $_POST keys
    if($row['answer']==$_POST[$key]) //compare the data from the table with the answer
        {
        //answer is correct
        $score++;
        $output.="Answer $n is correct</BR>"; //add responses to the output string
        }
        else
        {
        $output.="Answer $n is incorrect</BR>";
        }
    }
$output.="Total score: $score/50";  //add the score
echo $output;  //echo to screen.

获取查询，如：

SELECT a1 FROM answer

将返回
$row['a1']
，而不是
$row['answer']

因此，您应该使用列名，而不是表名，它仍在返回，好像它是错误的。可能是数据库吗？
SELECT a1 FROM answer