Question

我创建了一个结合了sql，php和html的测验网站。问题是从数据库中提取的。我的* .js代码和* .php代码在

之下

	<script>
$(document).ready(function(){
        $('body').on('click','input[type="radio"]', function(){
            var curr_id = $('.question').data('nextQuestion');
            var answer1 = $('#radio1').is(':checked');
            var answer2 = $('#radio2').is(':checked');
            var answer3 = $('#radio3').is(':checked');
            var answer4 = $('#radio4').is(':checked');
            getCorrectAnswer(curr_id, answer1, answer2, answer3, answer4);
			setTimeout(getQuestion.bind(this,curr_id, answer1, answer2, answer3, answer4), 1000);

    });

    function getQuestion(curr_id, answer1=false, answer2=false, answer3=false, answer4=false){
		console.log(curr_id);
        $.post("ajax.php",
        {
            next_id: parseInt(curr_id)+1,
            answer1: answer1,
            answer2: answer2,
            answer3: answer3,
            answer4: answer4,
        },
        function(data, status){
            $('#container_for_questions').html(data);
        });
    }

    function getCorrectAnswer(curr_id, answer1=false, answer2=false, answer3=false, answer4=false){
        $.post("ajax_get_correct_answer.php",
        {
            next_id: parseInt(curr_id),
            answer1: answer1,
            answer2: answer2,
            answer3: answer3,
            answer4: answer4,
        },
        function(data, status){
            $('#container_for_questions').html(data);
        });
    }

    getQuestion(-1);
});

</script>

<?php
// Start the session
session_start();

$con=mysqli_connect("localhost","root","","quiz"); // change here to your data
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

// Check the number of all questions, if next_id is more than last question, back to first or whatever you want;
$response=mysqli_query($con,"select * from questions");
$number_of_all_questions = mysqli_num_rows($response);

if($_POST['next_id'] == 0){
	// reset to default
	$_SESSION["correct_score"] = 0;
	$_SESSION["not_correct_score"] = 0;
}


if($number_of_all_questions <= $_POST['next_id']){
	// Quiz finished, show results
    echo"<div>
	<h2>Results:</h2>
	<p>Correct answers: {$_SESSION['correct_score']}</p>
	<p>Wrong answers: {$_SESSION['not_correct_score']}</p>

	</div>";



}else{

	// query next question
	$response=mysqli_query($con,"select * from questions WHERE id =(select min(id) from questions where id > {$_POST['next_id']})");
	?>

	<?php while($result=mysqli_fetch_array($response,MYSQLI_ASSOC)){ ?>

		<div id="question_<?= $result['id'] ?>" class='question' data-next-question="<?= $_POST['next_id'] ?>"> <!--check the class for plurals if error occurs-->
			<h2><?= $result['id'].".".$result['question_name'] ?></h2>
			<div class='align'>
				<input type="radio" value="1" id='radio1' name='1'>
				<label id='ans1' for='radio1'><?= $result['answer1'] ?></label>
				<br/>
				<input type="radio" value="2" id='radio2' name='2'>
				<label id='ans2' for='radio2'><?= $result['answer2'] ?></label>
				<br/>
				<input type="radio" value="3" id='radio3' name='3'>
				<label id='ans3' for='radio3'><?= $result['answer3'] ?></label>
				<br/>
				<input type="radio" value="4" id='radio4' name='4'>
				<label id='ans4' for='radio4'><?= $result['answer4'] ?></label>
			</div>
			<br/>
			<?php /*<input type="button" data-next-question="<?= $_POST['next_id'] ?>" id='next' value='Next!' name='question' class='butt'/> */?>
		</div>
	<?php }?>
<?php }?>
<?php mysqli_close($con); ?>

但是当我运行这个时，这就是我得到的错误：

“ErrorException Undefined index：next_id”。

at HandleExceptions-＆gt; handleError（8，'Undefined index：next_id'，   'C：\ xampp \ htdocs \ np \ app \ ajax_get_correct_answer.php'，28，   数组（'con'=＆gt;对象（mysqli）））我不太关于PHP。任何人都可以帮忙

Answer 1

替换下面

if($_POST['next_id'] == 0){
    // reset to default
    $_SESSION["correct_score"] = 0;
    $_SESSION["not_correct_score"] = 0;
}

要 //检查是否在ajax的Post中设置了next_id。如果将reset设置为default（correct_score，not_correct_score）

if(isset($_POST['next_id']) && $_POST['next_id'] == 0){
    // reset to default
    $_SESSION["correct_score"] = 0;
    $_SESSION["not_correct_score"] = 0;
}

您可以使用以下建议。

//在顶部声明变量

 $next_id=0;
 if(isset($_POST['next_id']) && $_POST['next_id'] != 0){
    $next_id=$_POST['next_id'];
    }

稍后您会看到$_POST['next_id']替换为$next_id

ErrorException未定义索引：php中的next_id

1 个答案: