Question

我创建了一个博客平台，我有一个ajax更新页面，我可以选择要显示的文章及其评论和添加评论。当我发表评论时，它会记录用户信息，已发表评论的文章，但不会将评论值存储在数据库中。这是代码：

<div align="center">
  <h3>Comentarii:</h3>
  <form method="POST">
    <textarea rows="4" cols="50" name="comentariu" placeholder="Comenteaza">
        </textarea><br>
    <input type="submit" name="submit"><br>
    <hr>
  </form>
</div>

<?php
  $comnou = $_POST['comentariu'];
  $comuser = $_SESSION['user'];
  $conadaugacom = mysqli_connect("localhost", "root", "", "blog");
  $sqladaugacom = "insert into comentarii (continut_comentarii, 
  user_comentarii, articol_comentarii) values ('$comnou', '$comuser', '$ta')";
  mysqli_query($conadaugacom, $sqladaugacom);
  mysqli_close($conadaugacom);
?>

AJAX代码 - ＆gt;

function showUser(str) {
  if (str == "") {
    document.getElementById("txtHint").innerHTML = "";
    return;
  } else {
    if (window.XMLHttpRequest) {
      xmlhttp = new XMLHttpRequest();
    } else {
      xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
    }
    xmlhttp.onreadystatechange = function() {
      if (this.readyState == 4 && this.status == 200) {
        document.getElementById("txtHint").innerHTML = this.responseText;
      }
    };
    xmlhttp.open("GET", "getuser.php?q=" + str, true);
    xmlhttp.send();
  }
}

Answer 1

确保在 POST 请求之前从表单序列化数据。

HTML

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<div align="center">
  <h3>Comentarii:</h3>
  <form method="POST">
    <textarea rows="4" cols="50" name="comentariu" placeholder="Comenteaza">
        </textarea><br>
    <input type="submit" name="submit"><br>
    <hr>
  </form>
</div>

Jquery的

<script>
     $('input').submit(function(e){
        e.preventDefault();

        var data = $('form').serialize();
        $.ajax({
           url: 'your_url_to_post.php',
           data: data,
           success: function(response){

           },
           type: 'POST'
        });
     });
</script>

php文件

添加此行以检查 comentariu 非空

if(isset($_POST['comentariu'])){
     $comentariu = $_POST['comentariu];
}

不能从html textarea获取值到php变量

1 个答案: