Question

我正在尝试创建我的第一个AJAX调用。我所要做的就是向我的数据库发送一条消息，其中包含user_id，消息和日期。

截至目前，当我点击提交按钮时，甚至都没有发生任何事情。为什么这不提交，我不确定我是否正确创建了ajax调用。

我做错了什么？

我的ajax电话

 $(document).ready(function(){ 
            $("#$(".contactButton")").on("change", function(){
                $user = this.value;
                $.ajax({ 
                    url: "insert_announcements.php", 
                    type: "POST",
                    data: "username="+$user, 
                    success: function(text){ 
                        if(text == "Error!"){ 
                            alert("Unable to get user info!"); 
                            $(".announcement_success").fadeIn();
                            $(".announcement_success").show();
                            $('.announcement_success').html('Payment Status Changed!');
                            $('.announcement_success').delay(5000).fadeOut(400);
                            alert(data);
                        } else { 
                            var txtArr = text.split('|');
                        } 
                    }, 
                    error: function(xhr, textStatus, errorThrown){ 
                        alert(textStatus + "|" + errorThrown); 
                    } 
                });
            });
        });

我的表格

<div class="announcement_success"></div>
    <p>Add New Announcement</p>
        <form action="" method="POST" id="insert_announcements" >
            <textarea rows="4" cols="50" id="announcement_message "name="message" class="inputbarmessage" placeholder="Message" required></textarea>
            <label for="contactButton">
                <input type="button" class="contactButton" value="Add Announcement" id="submit">
            </label>
        </form>

PHP文件insert_announcements

$con = mysqli_connect("localhost", "", "", "");
$stmt2 = $con->prepare("INSERT INTO announcements (user_id, message, date) VALUES (?, ?, NOW())");
    if ( !$stmt2 || $con->error ) {
        // Check Errors for prepare
         die('Announcement INSERT prepare() failed: ' . htmlspecialchars($con->error));
    }
    if(!$stmt2->bind_param('isi', $announcement_user_id, $announcement_message)) {
        // Check errors for binding parameters
        die('Announcement INSERT bind_param() failed: ' . htmlspecialchars($stmt2->error));
    }
    if(!$stmt2->execute()) {
        die('Announcement INSERT execute() failed: ' . htmlspecialchars($stmt2->error));
    }
        echo "Announcement was added successfully!";
    else
    {
         echo "Announcement Failed!";
    }

Answer 1

Jquery Code - 点击Add announcement按钮时，会调用libunistring进行ajax调用。

   <form action="" method="POST" id="insert_announcements" >
        <textarea rows="4" cols="50" id="announcement_message"name="message" class="inputbarmessage" placeholder="Message" required></textarea>
        <input type="text" id="userid" value="" placeholder="user id">
        <label for="contactButton">
            <input type="button" class="contactButton" value="Add Announcement" id="submit">
        </label>
    </form>

这应该是你的HTML -

  $stmt2 = $con->prepare("INSERT INTO announcements (user_id, message, date) VALUES (mysqli_escape_string($_POST['userid']), mysqli_escape_string($_POST['msg']), NOW())");

现在代码为php -

{{1}}

通过AJAX调用将消息发送到数据库的问题

1 个答案: