[问题] PHP的接收档里$_POST没有JQUERY的值

楼主: reeselee (强爷)   2013-12-14 14:36:55
烦请各位先进指教
我用了TWZIPCODE,可是在PHP接受档$_POST里却抓不到传过来的值
不知道哪里出错了,我有PIINT看过$_POST里面的内容了,就是没有TWZIPCODE的那些名称
PHP版本:PHP Version 5.2.13
<script type="text/javascript"
src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
<script type="text/javascript" src="js/memberAdd.js"></script>
<table width="80%" border="0" cellspacing="0" cellpadding="0" style="margin:0
auto;" class="itable">
<form name="addmember" action="member.php?action=insert&s=2" method="POST">
........略.........
<tr>
<td height="24" align="right" valign="middle" nowrap="nowrap"
style="width:130px;padding-right:10px;" bgcolor="e0e0e0">
地址
</td>
<td valign="middle" align="left"nowrap="nowrap" bgcolor="#f5f5f5"
style="padding-left:10px;">
<div id="addr_1">
</div>
<input name="address" type="text" class="t01" id="address" size="50" value="">
</td>
</tr>
<tr>
........略.........
</tr>
</form>
</table>
<script type="text/JavaScript">
$(function(){
$('#addr_1').twzipcode({
countyName: 'addr_county',
districtName: 'addr_area',
zipcodeName: 'addr_zip',
css: ['addr-county', 'addr-area', 'addr-zip']
});
});
</script>

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