Question

我正在尝试通过ajax将表单中的数据发送到'mess'数据库表中，但我的控制台上没有错误，数据没有插入表中。

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="utf-8">
    <meta name="viewport" content="width=device-width, initial-scale=1">
    <link rel="stylesheet" href="css/style.css" type="text/css" />
    <script src="jquery-3.1.1.js"></script>
    <script src="script.js"></script>

// this is the form
<form action="<?= $_SERVER['PHP_SELF']; ?>" method="POST">
        <fieldset style="width:40%; margin:0 auto;">
        <label for="message" id="message" >Message</label><textarea cols="60" rows="8" id="message" name="message"
        required></textarea>

       <input type="hidden" id="" name="id" value="<?= $id ?>" />
        <input type="hidden" id="topic_id" name="topic_id" value="<?= $topic_id ?>" />
        <input type="hidden" id="user_one" name="user_one" value="<?= $user_one ?>" />
        <input type="hidden" id="user_two" name="user_two" value="<?= $user_two ?>" />
        <input type="submit" name="submitted" id="send" value="Send" />
        <span id="error"></span>
        </fieldset>

    </form>

//这是我的script.js

 $(document).ready(function(){
    /*post message via ajax*/
    $("#send").on("click", function(){
        var message = $.trim($("#message").val()),
            topic_id = $.trim($("#topic_id").val()),
            user_one = $.trim($("#user_one").val()),
            user_two = $.trim($("#user_two").val()),
            error = $("#error");

        if((message != "") && (topic_id != "") && (user_one != "") && (user_two != "")){
            error.text("Sending...");
            $.post("handlemessageSend.php",{message:message,topic_id:topic_id,user_one:user_one,user_two:user_two}, function(data){
                error.text(data);
                //clear the message box
                $("#message").val("");
            });
        }
    });

//这是我的handlemessageSend.php

    require_once("templates/connect.inc.php");
    //post message
    if(isset($_POST['message'])){
        $message = mysqli_real_escape_string($con, $_POST['message']);
        $topic_id = mysqli_real_escape_string($con, $_POST['topic_id']);
        $user_one = mysqli_real_escape_string($con, $_POST['user_one']);
        $user_two = mysqli_real_escape_string($con, $_POST['user_two']);


        //insert into `mess table`
        $q = mysqli_query($conn, "INSERT INTO mess(sender, reciever, message, topic_id) VALUES ('$user_one','$user_two','$message', '$topic_id')");
        if($q){
            echo "Posted";
        }else{
            echo "Error";
        }
    }

使用ajax无法正常工作将数据发送到数据库

0 个答案: