怎么输出跳转不对的呢

Question

$gofrom=$_SERVER['HTTP_REFERER'];
//echo $gofrom;
//exit;
if($jx_yzm_from_post!= $jx_yzm_from_cookie){
    echo '<script language="JavaScript">alert("验证码错误!");window.location.href='"+$gofrom+"';</script>';
        exit;
}gofrom 这样跳写错哪里了呢

Answer 1

echo '<script language="JavaScript">alert("验证码错误!");window.location.href='"+$gofrom+"';</script>';
改成：
echo '<script language="JavaScript">alert("验证码错误!");window.location.href="'.$gofrom.'";</script>';

Answer 2

代码修改如下即可：
$gofrom=$_SERVER['HTTP_REFERER'];
//echo $gofrom;
//exit;
if($jx_yzm_from_post!= $jx_yzm_from_cookie){
echo '<script language="JavaScript">alert("验证码错误!");window.location.href="'.$gofrom.'.";</script>';
exit;
}

Answer 3

href='"+$gofrom+"'
改成href='".$gofrom."'

Answer 4

$gofrom=$_SERVER['HTTP_REFERER'];if($jx_yzm_from_post!= $jx_yzm_from_cookie)
{echo '<script>alert("验证码错误!");location.href='".$gofrom."';</script>';
exit;}