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Home Backend Development PHP Tutorial 关于JQUERY传递参数的问题

关于JQUERY传递参数的问题

Jun 23, 2016 pm 02:26 PM

大家好,小弟我想问个问题
场景:一个页面上面有2个按钮, ,两个按钮名称分别是数据库里面的信息名称,我希望通过点击这两个按钮,可以获取对应的数据库信息在下方的div层中显示,目前已经可以将该按钮的值给到jquery中,操作如下:
对应的标签信息

       for(循环){     	echo "<a>";                	echo "<i class=\"dapei_partleft\"></i>";                    echo "<span class=\"dapei_partmiddle\" title='$type_category_name' name='$type_category_id' >$type_category_name</span>";                    echo "<i class=\"dapei_partright\"></i>";               echo " </a>";}
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红色标注的就是整个每个标签对应的值。
jquery代码:
		$(function(){    		$('.dapei_partmiddle').each(function(){    			$(this).click(function(){    				$(this).html($(this).attr('name'))        		})    	})
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现在我想知道的时候,我如何通过点击某个标签,就把某个标签中的name给调出来,并且放到数据库select语句中
   以上求解,,在线等。谢谢


回复讨论(解决方案)

求助呀。。。。。。

求解啊。有大神帮帮忙吗??

思路:
通过点击按钮触发js事件,获取按钮的value,再通过ajax传递到服务端,根据传递按钮的参数,返回对应的查询结果(封装成一个字符串,记得要将返回的字符排好版式)给前端。
切换一次按钮,就返回查询到的结果组成串返回前端即可,具体代码实现,要看你的功底了。

思路:
通过点击按钮触发js事件,获取按钮的value,再通过ajax传递到服务端,根据传递按钮的参数,返回对应的查询结果(封装成一个字符串,记得要将返回的字符排好版式)给前端。
切换一次按钮,就返回查询到的结果组成串返回前端即可,具体代码实现,要看你的功底了。 能有例子我参考下吗?

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