这是我和你们玩耍的代码.... CodePen.io
一旦游戏使用#play-again
,我似乎无法显示$('#play-again').show()
按钮。
对此有任何想法???
更新添加了原始HTML链接
答案 0 :(得分:0)
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="container">
<div class="banner">
<p>tic-tac-toe</p>
<button id="play-again"> <i class="fa fa-repeat"></i></button>
</div>
<div class="board">
<div class="row" id="row-1">
<div class="col" id="col-1-1">
<div class="content"></div>
</div>
<div class="col" id="col-1-2">
<div class="content"></div>
</div>
<div class="col" id="col-1-3">
<div class="content"></div>
</div>
</div>
<div class="row" id="row-2">
<div class="col" id="col-2-1">
<div class="content"></div>
</div>
<div class="col" id="col-2-2">
<div class="content"></div>
</div>
<div class="col" id="col-2-3">
<div class="content"></div>
</div>
</div>
<div class="row" id="row-3">
<div class="col" id="col-3-1">
<div class="content"></div>
</div>
<div class="col" id="col-3-2">
<div class="content"></div>
</div>
<div class="col" id="col-3-3">
<div class="content"></div>
</div>
</div>
</div>
</div>
问题是,当我更改时,您必须将button#play-again
放在p
之外,这对我有用