大家好我需要一些关于ajax加载方法的帮助.基本上我需要使用Ajax load()在用户检查单选按钮后在主页面中显示jsp.我附上了需要显示jsp的页面图像以及我的jsp代码..请帮忙!
<div id="verification">
<p align=center class="4f1_title" ><u>Verification Decision</u></p>
<table border="0" cellpadding="0" cellspacing="0" align=center
width="100%">
<tbody>
<tr>
<td width="10%"></td>
<td width="8%">Passed <input id="passed" type="radio"
value="P" onclick="()"></td>
<td colspan="2" width="8%">Pending <input id="pending"
type="radio" value="H" onclick="()"></td>
<td width="9%">True Hit <input id="failed" type="radio"
value="F" onclick="()" disabled></td>
<td width="13%">Parcel Returned <input id="returned"
type="radio" value="S" onclick="()"></td>
<td width="23%">Referred to Law Enforcement <input id="law"
type="radio" value="L" onclick="()"></td>
<td width="8%">Retired <input id="retired" type="radio"
value="R" onclick="()" disabled></td>
<td width="12%">Return Reason <input id="ac"
type="radio" value="C" onclick="()"></td>
<td width="10%"></td>
</tr>
</tbody>
</table>
</div>
<br>
<div align=center>
<a href="javascript:handleClick()"><u>
<div id='showhidecomment'>Show Comments</div></u></a>
<div id='chkcomments'>
</div>
</div>
<br>
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尝试
$(function() { // when DOM is ready
$("#showhidecomment").click(function(){ // when #showhidecomment is clicked
$("#chkcomments").load("sample.jsp"); // load the sample.jsp page in the #chkcomments element
});
});
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并将您的html(链接部分)更改为
<div>
<div id='showhidecomment'>Show Comments</div>
<div id='chkcomments'></div>
</div>
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由于div元素在a元素内部无效.
更新 评论
我会data-url为这些元素添加一个自定义属性(指定要加载的页面)
<input id="passed" type="radio" value="P" data-url="some-page.jsp" />
<input id="law" type="radio" value="L" data-url="some-other-page.jsp" />
<input id="ac" type="radio" value="C" data-url="some-third-page.jsp" />
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然后对它们应用单个处理程序
$('#verification').on('change','input[type="radio"][data-url]', function(){
if (this.checked){
var url = $(this).data('url');
$("#chkcomments").load( url );
}
});
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