请教我的代码中的问题 ajax+php(CodeIgniter)
我用ajax将页面的值传递到后台,但是却无法获取,请教下我这个该怎么写呢?麻烦大牛们了。
我的页面代码为:
<script> <br /> $(function() { <br /> $("form[name='estimate']").submit(function(e) { <br /> e.preventDefault(); <br /> var btn = $("form[name='estimate']").find("input[type='submit']"), <br /> price = $(this).find(".price"); <br /> btn.button('loading'); <br /> price.html(""); <br /> $.ajax({ <br /> type: "POST", <br /> url: '<?php echo base_url(); ?>index.php?module=testprice&view=getprice', <br /> data: $(this).serializeArray(), <br /> success: function(data) { <br /> price.html("<h3>$" + data + " RMB<h4>" + ""); <br /> btn.button('reset'); <br /> }, <br /> error: function() { <br /> alert("Service Unavailable"); <br /> } <br /> }); <br /> }); <br /> <br /> </script>
估算
我的后台代码如下:
function getUserPrice() {
$tozipcode_t = $_POST('postal_code');
}
我这里无论怎样都无法获取值,请大牛帮忙分析一下,多谢多谢了
回复讨论(解决方案)
$tozipcode_t = $_POST['postal_code'];
十分感谢。我真的很多不会,以后多和您请教了。

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