javascript - 使用 ajax 从 mysql 添加和检索记录

Question

如图所示，我的 mysql 中有两个表，我希望系统在不刷新页面的情况下添加和检索评论。

我在这个函数中涉及三个 php 页面，它们是“DB.php”、“comment.php”和“action.php”代码如图所示:

数据库.php

    <?php
     $conn = mysql_connect('localhost','Practical4','1234') or die (mysql_error);
     $db=mysql_select_db('Practical4', $conn) or die (mysql_error);
     ?>

评论.php

           <----------------ajax script-------------------->

<script type="text/javascript" src="jquery.min.js"></script>
<script type="text/javascript">
$(function() {
    $(".submit_button").click(function() {
        var textcontent = $("#content").val();
        var dataString = 'content='+ textcontent;
        if(textcontent=='')
        {
            alert("Enter some text..");
            $("#content").focus();
        }
        else
        {
            $("#flash").show();
            $("#flash").fadeIn(400).html('<span class="load">Loading..</span>');
            $.ajax({
                type: "POST",
                url: "action.php",
                data: dataString,
                cache: true,
                success: function(html){
                    $("#show").after(html);
                    document.getElementById('content').value='';
                    $("#flash").hide();
                    $("#content").focus();
                }  
            });
        }
        return false;
    });
});
</script>
<div>

         <-----retrieve hotel id from hotel table-------->
<?php
$conn=mysqli_connect('localhost','Practical4','1234') or die('Not connected');

$database=mysqli_select_db($conn,'Practical4') or die('Database Not connected');

$id=$_GET['id'];

$query = "select * from hotel where name='$id'";
$data=mysqli_query($conn,$query);
while($rows=mysqli_fetch_array($data)){

    $name=$rows['name'];
    $price=$rows['price'];
    $duetime=$rows['dueTime'];
    $address=$rows['location'];
}
?>

  <---------------post form------------------->
<form method="post" name="form" action="">
<h3>Add Comment for <?php echo $name;?><h3>
<input type="text" name="name" id="name" value="<?php echo $name;?>" hidden > <br>
<textarea cols="30" rows="2" name="content" id="content" maxlength="145" >
</textarea><br />
<input type="submit" value="Post" name="submit" class="submit_button"/>

</form>
</div>
<div class="space"></div>
<div id="flash"></div>
<div id="show"></div>

Action .php

    <?php
include('DB.php');
$check = mysql_query("SELECT * FROM comment order by commentID desc");
if(isset($_POST['content']))
{
$content=mysql_real_escape_string(trim($_POST['content']));
$name=mysql_real_escape_string(trim($_POST['name']));

mysql_query("insert into comment(content,name) values ('$content','$name')");
$fetch= mysql_query("SELECT content FROM comment order by commentID desc where name = '$name'");
$row=mysql_fetch_array($fetch);
}
?>

<div class="showbox"> <?php echo $row['content']; ?> </div>

当我运行这个程序时，当我插入评论时页面没有任何显示，谁能帮我解决这个问题？非常感谢!!

Answer 1

部分改动如下:

评论.php

          <!-- ajax script -->

<script type="text/javascript" src="jquery.min.js"></script>
<script type="text/javascript">
$(function() {
    $(".submit_button").click(function() {
        var textcontent = $("#content").val();
        var name = $("#name").val();
        var dataString = 'content='+ textcontent + '&name='+name;
        if(textcontent=='')
        {
            alert("Enter some text..");
            $("#content").focus();
        }
        else
        {
            $("#flash").show();
            $("#flash").fadeIn(400).html('<span class="load">Loading..</span>');
            $.ajax({
                type: "POST",
                url: "action.php",
                data: dataString,
                cache: true,
                success: function(html){
                    $("#show").after(html);
                    document.getElementById('content').value='';
                    $("#flash").hide();
                    $("#content").focus();
                }  
            });
        }
        return false;
    });
});
</script>
<div>

         <!-- retrieve hotel id from hotel table -->
<?php
include('DB.php');

$id=$_GET['id'];

$query = mysql_query("select * from hotel where name='$id'");
while($rows=mysql_fetch_array($query)){

    $name=$rows['name'];
    $price=$rows['price'];
    $duetime=$rows['dueTime'];
    $address=$rows['location'];
}
?>

  <!-- post form -->
<form method="post" name="form" action="">
<h3>Add Comment for <?php echo $name;?><h3>
<input type="text" name="name" id="name" value="<?php echo $name;?>" hidden > <br>
<textarea cols="30" rows="2" name="content" id="content" maxlength="145" >
</textarea><br />
<input type="submit" value="Post" name="submit" class="submit_button"/>

</form>
</div>
<div class="space"></div>
<div id="flash"></div>
<div id="show"></div>

Action .php

<?php
include('DB.php');
$check = mysql_query("SELECT * FROM comment order by commentID desc");
if(isset($_POST['content']))
{
$content=$_POST['content'];
$name=$_POST['name'];

mysql_query("insert into comment (content,name) values ('$content','$name')");
echo '<div class="showbox">'.$content.'</div>';
}
?>

您的代码失败的原因:

• name 未添加到 dataString 中导致 name 未在 post 中发送
• 一些 mysql 错误