小编Aeh*_*mlo的帖子

我想在视图中添加多个具有不同不透明度的阴影.阴影的规格如下:

• 模糊半径为1的Y偏移为4
• 模糊半径为10的Y偏移量为10
• Y偏移为2,模糊半径为4
• 模糊半径为1,展开为1(无偏移,可能必须是4个不同的阴影)

我可以用CALayers 来完成所有这些工作.这是我为此工作的代码(请注意我还没有打扰设置shadowPath,直到我得到多个阴影的东西才会工作):

layer.cornerRadius = 4
layer.masksToBounds = false
layer.shouldRasterize = true
let layer2 = CALayer(layer: layer), layer3 = CALayer(layer: layer), layer4 = CALayer(layer: layer)
layer.shadowOffset = CGSizeMake(0, 4)
layer.shadowRadius = 1
layer2.shadowOffset = CGSizeMake(0, 10)
layer2.shadowRadius = 10
layer2.shadowColor = UIColor.blackColor().CGColor
layer2.shouldRasterize = true //Evidently not copied during initialization from self.layer
layer3.shadowOffset = CGSizeMake(0, 2)
layer3.shadowRadius = 4
layer3.shouldRasterize = true
layer4.shadowOffset = CGSizeMake(0, 1)
layer4.shadowRadius = 1
layer4.shadowOpacity = 0.1 …

我在电子邮件地址上使用mysqli_real_escape_string(),它返回一个空字符串.它使用任何电子邮件地址执行此操作

<?php
//from previous page - submitted by user.
$_POST['email']="aehmlo@aehmlo.com";
$_POST['password']='mypass1234';




//Link, I can verify it works.
$mysql_info=array(
     "url"=>"url",
     "username"=>"username",
     "password"=>"password",
     "database"=>"database"
);
$link=mysqli_connect($mysql_info['url'],$mysql_info['username'],$mysql_info['password'],$mysql_info['database']);


//Now I attempt to sanitize the user input.
$email=mysqli_real_escape_string($link,$_POST['email']);
$password=sha1(mysqli_real_escape_string($link,$_POST['password']));
var_dump($email);
var_dump($password);?>

我的表的排序是"latin1_swedish_ci".

我有一个PHP脚本,它可以在函数调用中向多个收件人发送电子卡(以逗号分隔的电子邮件地址和mail()每个电子邮件地址分别添加一个).但是,在查看收到的电子邮件时,每个客户端都可以看到电子邮件发送到的其他地址,这让我相信它们都是通过一封电子邮件发送的,尽管有单独的mail()电话.这是我目前的代码:

<?php
$headers  = "From: ".$_POST['email']."\r\n"; 
$headers .= "Content-type: text/html\r\n";
$array=explode(",", $_POST['sendto']);
for ($i = 0; $i < count($array); ++$i) {
    mail(trim($array[$i]), "Happy Holidays!", $body, $headers);
}
?>

如何解决此问题,以便收件人只能在"收件人"字段中看到他们的电子邮件地址？谢谢!