Question

所以我的js：

$('#introorderbtn').click(function () {
    var display = document.getElementById("display").value;
    var skype = document.getElementById("skype").value;
    var music = document.getElementById("music").value;
    var ideas = document.getElementById("ideas").value;

    $.post('php/intro.php',{id:ID,gname:gname,name:Name,email:email,display:display,skype:skype,music:music,ideas:ideas}, function(data) {

    });

    jQuery.ajax({
        url: 'php/file.php',
        type: 'post',
        data: {id:ID,gname:gname,name:Name,email:email,display:display,skype:skype,music:music,ideas:ideas},
        success: function(data){
            $('#results').html(data);
            $('#display').val = "";
            $('#skype').val = "";
            $('#music').val = "";
            $('#ideas').val = "";
        }
    });

});

我的HTML

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div style="display = none" id="gal">
        <h1>Order a free Introduction</h1>
        <p>This is a 2D Intro: about 7 sec. long and choisable with an alpha-transition (so called "Opener").<br>They are in 1920x1080px and 60fps of course!<br> Most clients picked this product, do it like them!<br> When you are not sure about my art, take a look into the Gallery.</p>
        <form>
            <h1>Gimme some Data:</h1>
            <input id="display" type="text" placeholder="Name Displayed in the Intro">
            <input id="skype" type="text" placeholder="Your Skype ID">
            <input id="music" type="url" placeholder="Music (No Dubstep Trap stuff pls)">
            <textarea id="ideas" placeholder="Your Ideas: "></textarea>
            <input type="submit" id="introorderbtn">
        </form>
        <div id="results"></div>

这是我的php

<?php
$id = $_POST['id'];
$gname = $_POST['gname'];
$name = $_POST['name'];
$email = $_POST['email'];
$display = $_POST['display'];
$skype = $_POST['skype'];
$music = $_POST['music'];
$ideas = $_POST['ideas'];

echo("Test");
---msql connection that works 100%---
?>

php单独使用没有问题，但是当我尝试发布表单中的数据以及其他脚本中的更多字符串时，它并没有。成功功能正在运行，但它没有显示结果。

Answer 1

我看到的第一个问题是

dataType: JSON,

应该是

dataType: 'JSON',

或者只是将其关闭，因为jQuery应该能够自动解决它。

另外，我认为您需要更改

$('#results').html(data);

为：

$('#results').html(JSON.stringify(data));

或至少不是对象的东西。

当然，假设data实际上以json的形式返回，正如您所指出的那样。

jQuery.ajax无法正常工作

1 个答案: