Question

我正在尝试将数据插入数据库，但我在bind_param（）中收到此错误：

Fatal error: Only variables can be passed by reference in ... on line 48

我一直在网上搜索，但我不太明白这个错误意味着什么，因为我已经完成了这个插入，这是我第一次收到这个错误。

导致此错误的问题是什么？

下面是插入代码：

 $answersql = "INSERT INTO Penalty_Marks (PenaltyAnswer, PenaltyMarks, QuestionId) 
    VALUES (?, ?, ?)";

if (!$insertanswer = $mysqli->prepare($answersql)) {
    // Handle errors with prepare operation here
}

   $c = count($_POST['incorrect']);

    for($i = 0;  $i < $c; $i++ )
    {


$insertanswer->bind_param('iii', $_POST['incorrect'][$i], $_POST['answerMarks'][$i], $_POST['numQuestion'][$i]);

下面是检索详细信息的表单（下面的表单已经删除，以便于查看）：

<form id="PenaltyMarks" action="insertpenaltymarks.php" method="post">

    <table id='penaltytbl'>
    <?php

    foreach($ques_ans as $questionId => $inc_ans)
    {
        $q_row_span = count($inc_ans);
        $row_count = 0;
        $inc_ans = array_values($inc_ans);

    ?>

    <tr class="questiontd">
    <td>
    <input type="hidden" name="numQuestion" value="<?php echo$questionId?>" />
    </td>

    <td>
    <input type="hidden" class="hiddenincorrect" name="incorrect[]" value="<?php echo$inc_ans[$row_count];?>">
    </td>

    <td>
    <input  name="answerMarks[]" type="text" data-type="qmark" value='0'  />
    </td>

    </tr>
        <?php
            //remaining incorrect answers in separate row (if any) follows here
        if($row_count < $q_row_span - 1) 
        {
            for($i=($row_count + 1); $i<$q_row_span; $i++) { ?>     
                <tr>
                <td>
                <input type="hidden" class="hiddenincorrect" name="incorrect[]" value="<?php echo$inc_ans[$i];?>">
                </td>

                <td class="answermarkstd">
                <input  name="answerMarks[]" type="text" data-type="qmark" value='0'  />            
                </td>
                </tr>
        <?php
            }
        }
    }

    ?>
    </table>

    </form>

Answer 1

问题是bindParam的参数实际上必须是变量，而不是字符串或数组。这段代码应该有效：

$incorrect = $_POST['incorrect'][$i];
$answerMarks = $_POST['answerMarks'][$i];
$numQuestion = $_POST['numQuestion'][$i];
$insertanswer->bind_param('iii', $incorrect, $answerMarks, $numQuestion);

Answer 2

Try this:
Declare variables x,y,z at appropriate locations.

$answersql = "INSERT INTO Penalty_Marks (PenaltyAnswer, PenaltyMarks, QuestionId) 
    VALUES (?, ?, ?)";

    $x = $_POST['incorrect'][$i];
    $y = $_POST['answerMarks'][$i];
    $z = $_POST['numQuestion'][$i];

if (!$insertanswer = $mysqli->prepare($answersql)) {
    // Handle errors with prepare operation here
}

   $c = count($_POST['incorrect']);

    for($i = 0;  $i < $c; $i++ )
    {


$insertanswer->bind_param('iii', x,y,z);


Hope that solves it.

只有变量可以通过引用错误传递 - 如何修复

2 个答案: