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php里面 button 怎么传值?

WBOY
WBOYOriginal
2016-06-23 14:17:332845browse

<input type="button" name="button" value="审核" style="width:80px; height:25px; cursor:pointer;" onMouseMove= "this.style.background= '#e1d4c7'" onMouseOut= "this.style.background= ''" onclick="window.location.href='gzd_manage_select_list.php?Gid=<?=$Gid;?>'" />

这个怎么传值onclick="window.location.href='gzd_manage_select_list.php?Gid==$Gid;?>'",onclick里面怎么把Gid==$Gid;?>的只传到gzd_manage_select_list.php页面去啊,谢谢!


回复讨论(解决方案)

这不是已经传了吗?url带参数

echo $_GET['Gid'];

这不是已经传了吗?url带参数
没有啊,在gzd_manage_select_list.php页面没有得到值?

echo $_GET['Gid'];
这个是在gzd_manage_select_list.php页面写就会有值啦嘛,可是我这边却没有值啊?

当前页面 echo $Gid; 有值吗

<input type="button" name="button" value="审核" style="width:80px; height:25px; cursor:pointer;" onMouseMove= "this.style.background= '#e1d4c7'" onMouseOut= "this.style.background= ''" onclick="window.location.href='gzd_manage_select_list.php?Gid=<?echo $Gid;?>'" />



就是把'='改成'echo'

还是需要标签,用get或者post就可以了

gzd_manage_select_list.php 页面里写

if (!empty($_GET['Gid'])){     //如果 Gid 不为空  $Gid = $_GET['Gid'];         //设置变量$Gid 的值为传过来的Gid值  echo $Gid;                   //如果赋值成功,会有值输出}

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