php - 尝试在一个查询中将数据插入 MySQL 表，但更改一个特定列

Question

我正在尝试 INSERT使用 SQL 语句将一些数据放入我的表中。一个用户完成了一个有 10 个问题的调查，然后我想 INSERT使用一条 SQL 语句将这 10 个答案存入数据库。所以基本上唯一会改变的 2 个字段是 questionID和 answer柱子？请注意，我刚刚在 <form> 中给出了 2 个问题作为示例信息标签。

问卷:

<!-- Attempt 2 of questionnaire -->
<form action="scripts/submit-survey.php" method="post">
    <?php
        $questionnaire = "Welcome Questionnaire";
        $questionID = 1;
        $stmt = $conn->prepare ("SELECT `questionnaire`.`questionnaireName`, `questionnaireQuestions`.`questionID`,`question`.`question` FROM `questionnaire` INNER JOIN `questionnaireQuestions` ON `questionnaire`.`questionnaireID` = `questionnaireQuestions`.`questionnaireID` INNER JOIN `question` ON `questionnaireQuestions`.`questionID` = `question`.`questionID` WHERE `questionnaire`.`questionnaireName` = ? AND `question`.`questionID` = ?");
        $stmt->bind_param("si", $questionnaire, $questionID);
        $stmt->execute();
        $result = $stmt->get_result();
        while($row = $result -> fetch_assoc())  
            {
    ?>
    <p><?php echo $row['question']; ?></p>
    <?php
            }
    ?>
    <label><input type="radio" name="q1" value="1"> 1</label>
    <br>
    <label><input type="radio" name="q1" value="2"> 2</label>
    <br>
    <label><input type="radio" name="q1" value="3"> 3</label>
    <br>

    <?php
        $questionnaire = "Welcome Questionnaire";
        $questionID = 2;
        $stmt = $conn->prepare ("SELECT `questionnaire`.`questionnaireName`, `questionnaireQuestions`.`questionID`,`question`.`question` FROM `questionnaire` INNER JOIN `questionnaireQuestions` ON `questionnaire`.`questionnaireID` = `questionnaireQuestions`.`questionnaireID` INNER JOIN `question` ON `questionnaireQuestions`.`questionID` = `question`.`questionID` WHERE `questionnaire`.`questionnaireName` = ? AND `question`.`questionID` = ?");
        $stmt->bind_param("si", $questionnaire, $questionID);
        $stmt->execute();
        $result = $stmt->get_result();
        while($row = $result -> fetch_assoc())  
            {
    ?>
    <p><?php echo $row['question']; ?></p>
    <?php
            }
    ?>
    <label><input type="radio" name="q2" value="1"> 1</label><br>
    <label><input type="radio" name="q2" value="2"> 2</label><br>
    <label><input type="radio" name="q2" value="3"> 3</label><br>
    <input type="submit" value="Submit">
</form>

submit-survey.php

<?php
  require 'db.php';
  session_start();
  $seshID = $_SESSION['studentID'];

    $a1 = $_POST['q1'];
    $a2 = $_POST['q2'];
    $a3 = $_POST['q3'];
    $a4 = $_POST['q4'];
    $a5 = $_POST['q5'];
    $a6 = $_POST['q6'];
    $a7 = $_POST['q7'];
    $a8 = $_POST['q8'];
    $a9 = $_POST['q9'];
    $a10 = $_POST['q10'];
    $answer_bank = array(
      $a1, $a2, $a3, $a4, $a5, $a6, $a7, $a8, $a9, $10
    );
    $stmt = $conn->prepare ("INSERT INTO `studentAnswer` (`studentAnswerID`, `studentID`, `questionID`, `answer`) VALUES (NULL, ? , ? ,?)");
          //echo $i;
          //echo ${"a". $i};
            $stmt->bind_param("iii", $seshID , $questionID , $a);
            foreach ($answer_bank as $a_b) {
              list($a) = $a_b;
              $stmt->execute();
          }
            $result = $stmt->get_result();

?>

数据库表:

Answer 1

$questionID 缺少 1 个值。

尝试

$questionID = 1;
foreach ($answer_bank as $a) {
     $stmt->execute();
     $questionID++;
}

使用别名改进您的查询

$stmt = $conn->prepare ("SELECT 
          `qn`.`questionnaireName`,
          `qQ`.`questionID`,
          `q`.`question`
      FROM `questionnaire` `qn`
      INNER JOIN `questionnaireQuestions` `qQ` ON `qn`.`questionnaireID` = `qQ`.`questionnaireID`
      INNER JOIN `question` `q` ON `qQ`.`questionID` = `q`.`questionID` 
      WHERE `qn`.`questionnaireName` = ? AND `q`.`questionID` = ?");