使用JSON打印sql表

Question

I want to print all sql table and show them in html table. I am new in php, JSON and AJAX. I send username successfully and get result in php. I think there is a problem in JSON part or AJAX. Can anyone help me?

index.php

<div class="col-lg-6">
  <p id="usr" style="color:gray; font-size: 48px;"></p>
  <script type="text/javascript">

      var usr = document.getElementById("dom-target");
      var username = usr.textContent;
      username = username.trim().replace(/ /g, '%20');
      document.getElementById("usr").innerHTML = username;
      var sendtophp = "username="+username;

      $.ajax({
        type: "POST",
        url: "getcoursetable.php",
        data: sendtophp,
        dataType:"json",
        success: function(response) {
          console.log(response);
          var trhtml ='';
          document.getElementById("demo").innerHTML = response;
          $.each(response, function (i, item) {
              trHTML += '<tr><td>' + item.cname + '</td><td>' + item.subject + '</td><td>' + item.course + '</td><td>'+ item.grade + '</td></tr>';
          });
          $('#results').append(trHTML);
        }
      });

  </script>
  <table id="results"></table>
</div>

getcoursetable.php

<?php
  include_once "connection.php";

  if(isset($_POST["username"])){
    $nick = $_POST["username"];

    $prep = "SELECT * FROM `enrolledtable` WHERE nickname='$nick'";

    $results = mysqli_query($con, $prep);

    $jsonData = array();
    while ($row = $results->fetch_row()) {
        $jsonData[] = $row;
    }
    echo json_encode($jsonData);
      }
 ?>

Now, I can print data but not like a table, like that

<p id="demo">denemee,CS,300,B,denemee,CS,301,B ,denemee,CS,305,B ,denemee,CS,307,B,denemee,CS,408,A-,denemee,IE,208,B ,denemee,MATH,306,B</p>

Answer 1

I solve my problem. This can take table in php and create html table.

index.php

<div class="col-lg-6">
  <p id="usr" style="color:gray; font-size: 48px;"></p>
  <script type="text/javascript">

      var usr = document.getElementById("dom-target");
      var username = usr.textContent;
      username = username.trim().replace(/ /g, '%20');
      document.getElementById("usr").innerHTML = username;
      var sendtophp = "username="+username;

      $.ajax({
        type: "POST",
        url: "getcoursetable.php",
        data: sendtophp,
        dataType:"json",
        success: function(response) {
          var trhtml ='';
          $.each(response, function (i, item) {
              trhtml += '<tr><td>' + item[0] + '</td><td>' + item[1] + '</td><td>'+ item[2] + '</td></tr>';
          });
          $('.append').append(trhtml);
        }
      });

  </script>
  <table id="results">

    <tr>
    <th>Name</th>
    <th>Subject</th>
    <th>Course</th>
    <th>Grade</th>
    </tr>
    <tbody class="append">

    </tbody>
    </table>
</div>

getcoursetable.php

<?php
  header('Content-Type: application/json');
      include_once "connection.php";

      if(isset($_POST["username"])){
        $nick = $_POST["username"];

        $prep = "SELECT subject,course,grade FROM `enrolledtable` WHERE nickname='$nick'";

        $results = mysqli_query($con, $prep);

        $jsonData = array();
        while ($row = $results->fetch_row()) {
            $jsonData[] = $row;
        }
        echo json_encode($jsonData);
      }
 ?>