所有我想要做的是当我点击提交我想要所有表单数据被发送到process.php ...然后在process.php我想回显出POST数据...然后替换一切结果div来什么在process.php做...jQuery POST表单数据
<script type="text/javascript">
$(document).ready(function(){
$("#myform").submit(function() {
$.ajax({
type: "POST",
dataType: "html",
cache: false,
url: "process.php",
success: function(data){
$("#results").html(data);
}
});
return false;
});
//$("#myform").submit(function() {
//$('#results').html("yay");
//}
// });
//});
});
</script>
<form name="myform" id="myform" action="" method="POST">
<!-- The Name form field -->
<label for="name" id="name_label">zoom</label>
<input type="text" name="zoom" id="zoom" size="30" value=""/>
<br>
</select>
<!-- The Submit button -->
<input type="submit" name="submit" value="Submit">
</form>
<!-- FORM END ---------------------------------------- -->
<!-- RESULTS START ---------------------------------------- -->
<div id="results">nooooooo<?PHP $_SESSION[''] ?><div>
<!-- <input type="image" name="mapcoords" border="0" src="mapgen.php"> ---- -->
<!-- RESULTS END ---------------------------------------- -->
那么,有什么问题? – Nacho 2011-04-24 19:53:41
你的代码似乎这样做(冷检)。问题是什么? – Halcyon 2011-04-24 19:54:05
这是什么问题? – 2011-04-24 20:04:12