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php - 虽然命令不断将 x 值增加 2 而不是 1

问题描述

所以我的目标是获取选中单选按钮的值并将其与我拥有的数据库答案进行比较。但是，当我执行 while 循环时，$x 在执行 $x++ 时从 0 变为 2；主页和结果页面发布在下面。在此先感谢您的帮助。

主文件

<?php
error_reporting(E_ALL ^ E_WARNING);
$con = mysqli_connect("localhost","root","","mcq");

//connection check
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  };
  //get questions
  $questions = mysqli_query($con, "SELECT `ID`, `Question`, `Answer` FROM `main`");
  $answers = mysqli_query($con, "SELECT `ans1`, `ans2`, `ans3`, `ans4` FROM `answerlist`");
  echo("<form action='result.php' method='post'>");
  if ($questions->num_rows > 0){
    // output data of each row
    while(($row = $questions->fetch_assoc()) && ($row1 = $answers->fetch_assoc())) {

        echo ($row[ID].". ".$row[Question]."<br/>"."<input type='radio' name='id".$row[ID]."' value='".$row1[ans1]."'/>".$row1[ans1]."<br/>"."<input type='radio' name='id".$row[ID]."' value='".$row1[ans2]."'/>".$row1[ans2]."<br/>"."<input type='radio' name='id".$row[ID]."' value='".$row1[ans3]."'/>".$row1[ans3]."<br/>"."<input type='radio' name='id".$row[ID]."' value='".$row1[ans4]."'/>".$row1[ans4]);

    }
};
echo("<br/><button type='submit'>SUBMIT</button>");
echo("</form>");



?>

结果.php

<?php
$count = 0;
$con = mysqli_connect("localhost","root","","mcq");

//connection check
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
};

$questions = mysqli_query($con, "SELECT `ID`,`Answer` FROM `main`");

$qcount = count(mysqli_fetch_assoc($questions));
$x = 1;
while($x < $qcount){

$id = "id".$x;
$answerval = $_POST["$id"];

while($row = $questions->fetch_assoc()){
    print_r($row);
    $ans = $row['Answer'];
if($answerval == $ans){
    $count= $count + 1;
    echo("TRUE");

};

 };
     };
    echo("<br/>".$count);


?>

标签： phpmysql

解决方案

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