我有一个表单,当我点击提交时我不希望页面刷新,这就是为什么我添加AJAX来实现这一点,你可以看到。问题是它不起作用。
<form id="formFooter" action="" method="post">
<h3>Select your trademark</h3>
<select class="form-control" name="trademark">
<option></option>
<option>©</option>
<option>™</option>
<option>®</option>
</select>
<h3>Your company name</h3>
<input class="form-control" type="text" name="companyName" placeholder="Your company name" />
<h3>Background Color</h3>
<input class="form-control" placeholder="(e.g. 00ff00)" type="text" name="backgroundColor">
<h3>Font Color</h3>
<input class="form-control" placeholder="(e.g. 00ff00)" type="text" name="fontColor">
<h3>Opacity</h3>
<input class="form-control" placeholder="(Pick a value between 0 and 1 e.g. 0.3)" type="text" name="opacity">
<br/>
<br/>
<button class="form-control" id="run" type="submit" name="submit">Generate footer</button>
</form>
<div id="showData"> </div>
<script type="text/javascript">
$('#run').on("click", function (e) {
var formData = new FormData($('#myForm')[0]);
$.ajax({
url: "script.php",
type: 'POST',
data: formData,
success: function (data) {
$('#showData').html(data);
},
cache: false,
contentType: false,
processData: false
});
return false;
});
</script>
这是script.php:
<?php
function footerPreview ()
{
echo "<h3>Preview:</h3>";
date_default_timezone_set('UTC');
$trademark = $_POST["trademark"];
$company = $_POST["companyName"];
$date = date("Y");
//style
$backgroundColor = $_POST['backgroundColor'];
$fontColor = $_POST['fontColor'];
$opacity = $_POST['opacity'];
echo "<div id='generated_footer_date' style='background-color:$backgroundColor; color:$fontColor; opacity: $opacity; ' >$trademark $date $company </div>";
}
// generate result for the head
function rawHead()
{
$head = htmlspecialchars('<head>
<meta charset="utf-8">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<link href="https://fonts.googleapis.com/css?family=Raleway:200" rel="stylesheet">
</head>',ENT_QUOTES);
echo "<pre><h4>Put this code inside your head tags</h4>$head</pre>";
}
// generate result for the body
function rawBody ()
{
$body1of5 = htmlspecialchars('<div id="footer_date">',ENT_QUOTES);
$body2of5 = $_POST["trademark"];
$body3of5 = date("Y");
$body4of5 = $_POST["companyName"];
$body5of5 = htmlspecialchars('</div>',ENT_QUOTES);
echo "<pre><h4>Put this code inside your body tags</h4>$body1of5 $body2of5 $body3of5 $body4of5 $body5of5 </pre>";
}
// generate result for the CSS
function rawCSS ()
{
$opacity = $_POST['opacity'];
$backgroundColor = $_POST['backgroundColor'];
$fontColor = $_POST['fontColor'];
echo
"<pre>
<h4>Put this code in your websites stylesheet</h4>
color:$fontColor;
background-color:$backgroundColor;
opacity:$opacity;
width:100%;
text-align:center;
padding-top:15px;
height:50px;
font-family: 'Raleway', sans-serif;
right: 0;
bottom: 0;
left: 0;
position:fixed;
</pre>";
}
// Generate eveything by one click
if(isset($_POST['submit']))
{
footerPreview();
rawHead();
rawBody();
rawCSS();
}
?>
当我点击提交时没有任何反应。我希望在不刷新的情况下在同一页面上生成script.php。
答案 0 :(得分:4)
您可以将Ajax请求简单化为:
首先,不需要在此处使用FormDate,因为您的file
中没有任何<form>
输入,因此您可以在请求中使用serialize()
数据:< / p>
var formData = $("#myForm").serialize();
其次,您只是在PHP中打印HTML,这意味着您只需要打印html,因此您可以在此处使用dataType=HTML
:
dataType: "html",
第三,还有一件事可以帮助您进行调试,在print_r($_POST)
文件中添加script.php
并检查控制台。
修改后的请求:
$(document).ready(function(){
$("#run").click(function(){
var formData = $("#myForm").serialize();
$.ajax({
type: "POST",
url: "script.php",
data: formData,
dataType: "html",
success: function(response)
{
$('#showData').html(response);
},
beforeSend: function()
{
//any loader
}
});
return false;
});
});
<强>更新强>
从你的评论:是的,它提交后显示。它显示了这个:数组 ([trademark] =&gt; [companyName] =&gt; [backgroundColor] =&gt; [fontColor] =&gt; [不透明度] =&gt; ) - Kevin Aartsen 6分钟前
查看此数组,submit
的结果中没有$_POST
,因此您有两个选项可以更改此内容:
1)您可以使用count()
功能检查if(count($_POST) > 0)
。
2)或者您可以使用<input type='submit' name='submit'>
代替<button type='submit' name='submit'>
答案 1 :(得分:1)
$(document).ready(function() {
$('#run').on("click", function (e) {
e.preventDefault();
alert('inside ajax call');
var formData = new FormData($('#myForm')[0]);
$.ajax({
url: "script.php",
type: 'POST',
data: formData,
success: function (data) {
$('#showData').html(data);
alert('ajax call success');
},
cache: false,
contentType: false,
processData: false
});
return false;
});
});
&#13;
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.0/jquery.min.js"></script>
<form id="formFooter" action="" method="post">
<h3>Select your trademark</h3>
<select class="form-control" name="trademark">
<option></option>
<option>©</option>
<option>™</option>
<option>®</option>
</select>
<h3>Your company name</h3>
<input class="form-control" type="text" name="companyName" placeholder="Your company name" />
<h3>Background Color</h3>
<input class="form-control" placeholder="(e.g. 00ff00)" type="text" name="backgroundColor">
<h3>Font Color</h3>
<input class="form-control" placeholder="(e.g. 00ff00)" type="text" name="fontColor">
<h3>Opacity</h3>
<input class="form-control" placeholder="(Pick a value between 0 and 1 e.g. 0.3)" type="text" name="opacity">
<br/>
<br/>
<button class="form-control" id="run" type="submit" name="submit">Generate footer</button>
</form>
<div id="showData"> </div>
&#13;
尝试上面的代码并在它适用时删除警告:)