小编Lab*_*aTT的帖子

当我点击提交时,我希望将所有表单数据发布到process.php.然后在process.php上我想回显出POST数据,最后将结果div中的所有内容替换为process.php中的内容.

<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 …