PHP的响应可以正常工作,但在div的开头有“undefined”文本

gwo2fgha  于 2024-01-05  发布在  PHP
关注(0)|答案(1)|浏览(131)

所以,我想用id“listberita”填充div,并使用来自aScore response的HTML代码。
下面是PHP代码,其中包含ID为“listberita”的div

  1. <!DOCTYPE html>
  2. <html lang="en">
  3. <head>
  4. <meta charset="UTF-8" />
  5. <meta name="viewport" content="width=device-width, initial-scale=1.0" />
  6. <title>Jabary - Website Budaya Jabar</title>
  7. <link href="https://cdn.jsdelivr.net/npm/[email protected]/dist/css/bootstrap.min.css" rel="stylesheet" integrity="sha384-EVSTQN3/azprG1Anm3QDgpJLIm9Nao0Yz1ztcQTwFspd3yD65VohhpuuCOmLASjC" crossorigin="anonymous">
  8. <script src="https://kit.fontawesome.com/f8535c9b97.js" crossorigin="anonymous"></script>
  9. <link rel="stylesheet" href="style.css">
  10. <link rel="stylesheet" href="berita.css">
  11. </head>
  12. <body>
  13. <!-- Navbar -->
  14. <?php include 'php/navbar.php'; ?>
  15. <div class="container my-5">
  16. <div class="kategoricaption mb-5">
  17. <div class="row">
  18. <div class="col">
  19. <h1 class="text-center fw-bold judulkategori">Berita</h1>
  20. <hr class="mx-auto" style="width:10%; background-color: #f49f16;">
  21. </div>
  22. </div>
  23. </div>
  24. <div class="card-group vgr-cards" id="listberita">
  25. </div>
  26. </div>
  27. <!-- footer -->
  28. <?php include 'php/footer.php'; ?>
  29. <script src="https://code.jquery.com/jquery-3.6.0.js" integrity="sha256-H+K7U5CnXl1h5ywQfKtSj8PCmoN9aaq30gDh27Xc0jk=" crossorigin="anonymous"></script>
  30. <script src="js/bootstrap.bundle.min.js "></script>
  31. <script src="berita.js"></script>
  32. </body>
  33. </html>

字符串
下面是我的aqjs代码

  1. $(this).ready(function() {
  2. getNews()
  3. function getNews() {
  4. $.ajax({
  5. type: "GET",
  6. url: "_php/getBerita.php",
  7. dataType: "JSON",
  8. success: function(response) {
  9. var kode;
  10. $.each(response, function(i, obj) {
  11. console.log(kode)
  12. kode += '<div class="card kartu pb-3"><div class="card-img-body"><img class="card-img" src="img/Berita/' + obj.gambar_berita + '" alt="Card imagecap"></div><div class="card-body"><h4 class="card-title">' + obj.nama_berita + '</h4><p class="card-text">' + obj.keterangan_berita.substring(0, 250) + '....</p><a href="isiBerita.php?id=' + obj.id + '#disqus_thread" class="btn btn-success mt-5">Read More</a></div></div>'
  13. $('#listberita').html(kode);
  14. })
  15. }
  16. });
  17. }
  18. })


这里是我的php代码,其中一个请求(php_/getBerita.php)

  1. <?php
  2. include '../koneksi.php';
  3. $result = $conn->query("SELECT * from tbl_berita");
  4. while($row=$result->fetch_assoc()){
  5. $data[]=$row;
  6. }
  7. echo json_encode($data);
  8. ?>


上面的代码是工作,它的返回数据我想要的.但有一个问题. Here is the problem
如何去掉div开头的“undefined”?

gudnpqoy

gudnpqoy1#

var kode;更改为var kode = "";
当你声明变量而没有初始化它时,它将是undefined。然后你的循环将文本附加到undefined变量上。这可能就是原因。

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