echo "3f1c4e4b6b16bbbd69b2ee476dc4f83aalert('欢迎光临');location='pub_main.php';2cacc6d41bbb37262a98f745aa00fbf0";
我验证身份信息后应该是跳出一个“欢迎光临”提示窗口,然后跳转到pub_main.php页面啊,为什么我跳转不了呢
回复讨论(解决方案)
前面报错了吧 ?
加个exit看看
提示正常不能跳转页面么?
是否路径有问题呢?我试了下,语句没问题的。
这句话没问题啊
查看网页源代码
估计是没有执行到
echo "3f1c4e4b6b16bbbd69b2ee476dc4f83aalert('欢迎光临');window.location.href='pub_main.php';2cacc6d41bbb37262a98f745aa00fbf0";
这样可以跳转。
都是高手高手高高手。
我也会
echo "3f1c4e4b6b16bbbd69b2ee476dc4f83aalert('欢迎光临');window.location.href='pub_main.php';2cacc6d41bbb37262a98f745aa00fbf0";试试这样,估计可以
echo "<script>alert('欢迎光临');location.href='pub_main.php';</script>";
楼上的正解了...

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