mysqli_fetch_array 如何一次获取一行？

Question

My php code is as follows: This is part of a quiz where I am displaying one question and 4 multiple choices in html page via ajax jQuery. 我知道如何运行 while 循环并一个接一个地显示所有数据，但我如何一次只显示一个问题？

因此，在回答了一个问题后，我想查看下一个问题。 是否可以运行一个计数器并一次提取一个结果和下一个结果等等..？

<?php 
header("Access-Control-Allow-Origin: *");
require 'db.php';
// making empty variable
$createTable = "";

        $test_id=$_POST["test_id"];
        $sql=mysqli_query($con,"select * from mst_question where test_id='$test_id' ");
    $counter = 0;

while($row=mysqli_fetch_array($sql))
        {   
    $counter++;
        $createTable .= '<div class="text-subhead-2 text-center" style="background-color:#42A5F5">Question ';
        $createTable .= $counter;
        $createTable .= ' of 25</div>';
        $createTable .= '<div class="panel panel-default paper-shadow" data-z="0.5">';

        $createTable .= '<div class="panel-body">';
        $createTable .= '<p class="text-body-2">';
        $createTable .= $row['que_desc'];
        $createTable .= '</p>';
       $createTable .= '</div>';
        $createTable .= '</div>';

        $createTable .= '<div class="text-subhead-2 text-light">Your Answer</div>';
        $createTable .= '<div class="panel panel-default paper-shadow" data-z="0.5">';
        $createTable .= '<div class="panel-body">';
        $createTable .= '<div class="radio radio-success">';
        $createTable .= '<input type="radio" name="radio';
        $createTable .= $counter;
        $createTable .= '" id="radio1';
        $createTable .= $counter;
        $createTable .= '" value="';
        $createTable .= $row['ans1'];
        $createTable .= '" >';
        $createTable .= '<label for="radio1';
        $createTable .= $counter;
        $createTable .= '">';
        $createTable .= $row['ans1'];
        $createTable .= '</label>';
        $createTable .= '</div>';
        $createTable .= '<div class="radio radio-success">';
        $createTable .= '<input type="radio" name="radio';
        $createTable .= $counter;
        $createTable .= '" id="radio2';
        $createTable .= $counter;
        $createTable .= '" value="';
        $createTable .= $row['ans2'];
        $createTable .= '" >';
        $createTable .= '<label for="radio2';
        $createTable .= $counter;
        $createTable .= '">';
        $createTable .= $row['ans2'];
        $createTable .= '</label>';
        $createTable .= '</div>';
        $createTable .= '<div class="radio radio-success">';
        $createTable .= '<input type="radio" name="radio';
        $createTable .= $counter;
        $createTable .= '" id="radio3';
        $createTable .= $counter;
        $createTable .= '" value="';
        $createTable .= $row['ans3'];
        $createTable .= '" >';
        $createTable .= '<label for="radio3';
        $createTable .= $counter;
        $createTable .= '">';
        $createTable .= $row['ans3'];
        $createTable .= '</label>';
        $createTable .= '</div>';
        $createTable .= '<div class="radio radio-success">';
        $createTable .= '<input type="radio" name="radio';
        $createTable .= $counter;
        $createTable .= '" id="radio4';
        $createTable .= $counter;
        $createTable .= '" value="';
        $createTable .= $row['ans4'];
        $createTable .= '" >';
        $createTable .= '<label for="radio4';
        $createTable .= $counter;
        $createTable .= '">';
        $createTable .= $row['ans4'];
        $createTable .= '</label>';
        $createTable .= '</div>';
        $createTable .= '</div>';
        $createTable .= '</div>';

                        }

    echo $createTable;
    mysqli_close($con);
    ?>

Answer 1

首先，您的代码很危险，因为可以通过 sql 注入攻击。 您始终应该使用参数绑定。

最简单的方法是传递存储在 mst_question 中的问题的 id 并通过 WHERE 子句（如 test_id）选择一个。

//...
$test_id=$_POST["test_id"];
$questionId = filter_var($_POST['question_id'],FILTER_VALIDATE_INT);
if (!$questionId){
   die('done');
}

$stmt= mysqli_prepare($con,"select * from mst_question where test_id='$test_id' AND id=?");
mysqli_stmt_bind_param(**$stmt**, 'd',$questionId);
mysqli_stmt_execute(**$stmt**);
// work with $stmt. 
// f.e. your loop but now there will be only one execution
mysqli_stmt_close($stmt);
//...
$createTable .= '<input type="hidden" name="nextQuestionId" value="'.$nextQuestionId.'"/>';
//...

使用输入字段，您将返回下一个问题的 id，该问题可以在 javascript 代码中的 url 参数中传递。

如果您担心测验作弊者，您可以通过散列 nextQuestionId 来提高安全性。

//...
$stmt = mysqli_prepare($con,"select * from mst_question where test_id='$test_id' AND sha1(CONCAT('slat_',id))=?");
//...
$createTable .= '<input type="hidden" name="nextQuestionId" value="'.sha1('salt_'.$nextQuestionId).'"/>';
//...

这不是最好的解决方案，但需要对代码进行最少的更改。

我想建议切换到 PDO - 与数据库交互的非常友好和强大的方式。 看一个例子。