php - Sql 查询存储在变量中并使用该变量使用 PHP 添加内容

Question

如何将 SQL 查询存储到变量以将其当前值加 1。我已经有一个代码，但它不会将 1 添加到选择查询 ['eCorrect'] 中的当前值，而是在提交表单后结果将为 2。它显示此错误 mysqli_result 类的对象无法转换为 int。我已将“eCorrect”字段设置为数据类型 INT。我不知道我的错误在哪里，请帮忙提前谢谢。

<form method="post" action="elQ2.php">
        <p>Question: sample question sample question sample question sample question</p>
        <input type="radio" name="Ctwo" value="wrong">Choice A<br>
        <input type="radio" name="Ctwo" value="wrong">Choice B<br>
        <input type="radio" name="Ctwo" value="correct">Choice C<br>
        <input type="radio" name="Ctwo" value="wrong">Choice D<br>
        <input type="submit" name="submit" value="Submit" class="pull-right">
        </form>
    </div>
</div>

这是我的 php 代码

<?php
$correctResult = $db->query("SELECT eCorrect FROM elearn WHERE eID=1");
$wrongResult = $db->query("SELECT eWrong FROM elearn WHERE eID=1");

$addC = $correctResult;
$addW = $wrongResult;
    if(isset($_POST['submit'])) {

if(!empty($_POST['Ctwo'])){
    if($_POST['Ctwo']=="correct")   {

                $eAns = $_POST['Ctwo'];
                $eAns= $addC + 1 ;
                if($update = $db->query("   UPDATE elearn SET eCorrect = '".$eAns."' WHERE eID=1 "))    {
            }else{
                die($db->error);
            }
                echo "Correct Answer";

    }else{
                //Wrong answer
                $eAns = $_POST['Ctwo'];
                $eAns = $addW + 1 ;
                if($update = $db->query("   UPDATE elearn SET eWrong = '".$eAns."' WHERE eID=1 "))  {
            }else{
                die($db->error);
            }   
                echo "Wrong answer";
            }
        } else{
            echo "You must choose an answer";
        }
}
?>

Answer 1

出现此问题的原因是 $addC 和 $addW 变量包含结果集。所以实际上您正在向结果集添加 +1。

尝试这样做:

$cRow = $db->query("SELECT eCorrect FROM elearn WHERE eID=1");
$correctResult = $cRow[0];

$wRow = $db->query("SELECT eWrong FROM elearn WHERE eID=1");
$wrongResult = $cRow[0];