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Editorial Team
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Asked: 2026-05-29T16:13:45+00:00

I have 2 tables relating to a survey. When user answers each set of

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I have 2 tables relating to a survey. When user answers each set of questions and then click the submit button, it will loop each answer according to the form submitted in order to check within the database first, if the CustomerID and QuestionID have been found, then do the Update. If not found, do the Insert instead.

QUESTIONS table

  • QuestionID (PK)
  • QuestionText

ANSWERS table

  • AnswerID (PK)
  • CustomerID (FK)
  • QuestionID (FK)

  • AnswerText

    <html>
....
<form action="/db.php" method="POST">
 <?php echo $questiontext[1]; ?><input type="text" name="answer1" id="answer1">
 <?php echo $questiontext[2]; ?><input type="text" name="answer2" id="answer2">
 <?php echo $questiontext[3]; ?><input type="text" name="answer3" id="answer3">
 <?php echo $questiontext[4]; ?><input type="text" name="answer4" id="answer4">
 <?php echo $questiontext[5]; ?><input type="text" name="answer5" id="answer5">
 <?php echo $questiontext[6]; ?><input type="text" name="answer6" id="answer6">

 <input type="submit" name="submit" id="submit" value="Submit">
</form>
...
</html>

db.php

    <?php
    if(isset($_POST['submit'])) {
      $cusid = intval($_POST['CustomerID']);
      $answer1 = $db->real_escape_string($_POST['answer1']);
      $answer2 = $db->real_escape_string($_POST['answer2']);
      $answer3 = $db->real_escape_string($_POST['answer3']);
      $answer4 = $db->real_escape_string($_POST['answer4']);
      $answer5 = $db->real_escape_string($_POST['answer5']);
      $answer6 = $db->real_escape_string($_POST['answer6']);

      $sql = "INSERT INTO Answers (CustomerID, QuestionID, AnswerText) VALUES     
      ('".$cusid."','".$quesid."','".$answer."')";
      $res = $db->query($sql) or die ('Error: ' . mysqli_error($db));
    }
    ?>

My questions are:

  1. How to update each answer(1-6) one by one and then insert into the database if CustomerID and QuestionID have not been found by using array and SQL query, if found, then just update?
  2. How could I reference the QuestionID in order to related with AnswerText in HTML and PHP?
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1 Answer

  1. Editorial Team

    This is just an idea for you. Hope you can understand it. Make sure you replace database driven functions with your $db object.

    if(isset($_POST['submit'])) {
    $sql = "SELECT QuestionID FROM Questions ORDER BY QuestionID ";
    $res = $db->query($sql);
    $qus = 1;
    while ($row = mysqli_fetch_array($res , MYSQLI_ASSOC)) {
    {
        $questionID = $row['QuestionID'] ;
        $answer = $db->real_escape_string($_POST['answer' . $qus ]);


        $sql = "SELECT AnswerID  FROM Answers WHERE CustomerID='$cusid' AND QuestionID='$questionID' ";
        $resAns = $db->query($sql);
        $num_rows = $resAns->num_rows; // This should be replace with your $db object record count obtaining method
        if($num_rows == 1)
        {
            $sql = "UPDATE Answers SET AnswerText = '$answer' WHERE CustomerID='$cusid' AND QuestionID='$questionID'  ";
            // Execute your update query
        }
        else
        {
            $sql = "INSERT INTO Answers (CustomerID, QuestionID, AnswerText) VALUES     
  ('".$cusid."','".$questionID."','".$answer."')";
            // Execute your insert statement
        }
        $qus ++;
    }
}
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