Php 无法通过输入为图像的jquery提交post变量
我想写一个插件,它将通过点击输入图像或任何其他方法将存储为输入(图像)id的变量传递给表单 =========================================================================================Php 无法通过输入为图像的jquery提交post变量,php,html,jquery-plugins,Php,Html,Jquery Plugins,我想写一个插件,它将通过点击输入图像或任何其他方法将存储为输入(图像)id的变量传递给表单 ========================================================================================= <?php echo $_POST['id']; ?> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf
<?php
echo $_POST['id'];
?>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Kunfu Panda</title>
<link href="css/main.css" rel="stylesheet" type="text/css" media="projection,screen" />
</head>
<body>
<div class="wrapper" >
<form action="panda.php" method="post" class="openid_provider" />
<input type="image" src="images/google_button.png" id="google" />
</form>
</div>
<script type="text/javascript" src="js/test.js"></script>
<script type="text/javascript" src="js/jquery-ui-1.8.19.custom.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('form').openid();
})
</script>
</body>
</html>
现在,点击谷歌图片提交表单,这是一个id为“google”的输入,但同一页面的表单操作无法识别$u POST['provider']变量
请帮帮我。试试这个:
<?php
var_dump($_POST);
?>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Kunfu Panda</title>
<link href="css/main.css" rel="stylesheet" type="text/css" media="projection,screen" />
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.js"></script>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.18/jquery-ui.js"></script>
<script type="text/javascript" src="js/test.js"></script>
</head>
<body>
<div class="wrapper" >
<form action="panda.php" method="post" class="openid_provider" />
<input type="image" src="images/google_button.png" id="google"/>
</form>
</div>
</body>
<script type="text/javascript">
$(document).ready(function(){
$('form').openid();
})
</script>
功夫熊猫
$(文档).ready(函数(){
$('form').openid();
})
js:
(函数($){
$.fn.openid=函数(){
$(“输入”)。单击(函数(e){
var provider=$(this.attr('id');//获取id
var myinput=$('').val(提供者);
$('form').append(myinput);
$('form').submit();
})
}
})(jQuery);
html/js示例中有许多错误:1)您有JQueryUI,但没有JQuery;2)您的test.js必须在JQuery之后加载;3)在错误的元素上调用submit;4)无法通过“provider”提交(),5)图像输入没有“name”属性,6)发布图像输入只会在用户单击图像的位置提供xy坐标。
<?php
var_dump($_POST);
?>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Kunfu Panda</title>
<link href="css/main.css" rel="stylesheet" type="text/css" media="projection,screen" />
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.js"></script>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.18/jquery-ui.js"></script>
<script type="text/javascript" src="js/test.js"></script>
</head>
<body>
<div class="wrapper" >
<form action="panda.php" method="post" class="openid_provider" />
<input type="image" src="images/google_button.png" id="google"/>
</form>
</div>
</body>
<script type="text/javascript">
$(document).ready(function(){
$('form').openid();
})
</script>
(function($){
$.fn.openid = function() {
$('input').click(function(e){
var provider = $(this).attr('id'); //get id
var myinput = $('<input type="hidden" name="provider"/>').val(provider);
$('form').append(myinput);
$('form').submit();
})
}
})(jQuery);