Question

如何将sql查询存储到变量以将1添加到其当前值。我已经有了一个代码但它没有在select查询['eCorrect']中添加1当前值，而是在提交表单后结果为2。并且它显示此错误类mysqli_result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提前帮助谢谢。

<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

出现问题是因为$ addC和$ addW变量包含结果集。实际上，您正在为结果集添加+1。

尝试改为：

$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];

Sql查询存储在一个变量中并使用该变量来使用PHP添加一些东西

1 个答案: