使用JQuery Ajax提交表单几次，无需重新加载页面

Question

我有一个表单，可以使用Ajax提交一次。因为回应也是一种形式，它取代了旧的。但是，当我然后按提交按钮表单不提交与Ajax和正常的方法。为什么？ 这里我jQuery的代码：

<script> 
    $(document).ready(function() { 
     $('#ajax_form').submit(function(e) { 
      var form = $('#ajax_form'); 
      var data = form.serialize(); 
      $.post('game/write.php', data, function(response) { 
       console.log(response); 
       $('#power').replaceWith(response); 
      }); 
      return false; 
     }); 
    }); 
</script> 

形式：

<div id="power"> 
<div class="span4"> 
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="1" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form> 
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="4" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form> 
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="7" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form> 
</div> 
<div class="span4"> 
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="2" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form> 
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="5" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form> 
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="8" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form> 
</div> 
<div class="span4"> 
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="3" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form> 
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="6" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form> 
    <form action="game/write.php" id="ajax_form" method="post"><input type="hidden" value="9" name="button"><button type="submit" class="btn btn-block btn-success"><img src="images/null.png"></button></form> 
</div> 

这对于自带也回来了，只有Buttoncolors是不同的。

Answer 1

你需要事件代表团 -

$(document).on('submit','#ajax_form',function(e) { 

试试这个 -

$(document).on('submit','#ajax_form',function(e) { 
    var form = $('#ajax_form'); 
    var data = form.serialize(); 
    $.post('game/write.php', data, function(response) { 
     console.log(response); 
     $('#power').replaceWith(response); 
    }); 
    return false; 
}); 

Answer 2

event.preventDefault（）

文档：http://api.jquery.com/event.preventDefault/

现在更新你的代码像波纹管

<script> 
$(document).ready(function() { 


    $('#ajax_form').click(function(event) {//use click instead of submit 

    event.preventDefault() // then use even.preventDefault() 

    var form = $('#ajax_form'); 
    var data = form.serialize(); 
    $.post('game/write.php', data, function(response) { 
     console.log(response); 
     $('#power').replaceWith(response); 
    }); 
    return false; 
}); 

}); 
</script> 

add event.preventDefault（）[如果调用此方法，则不会触发该事件的默认操作。 ]