是否有一种方法允许多个跨域使用Access-Control-Allow-Origin头?

我知道*,但它太开放了。我只讲几个域。

举个例子,是这样的:

Access-Control-Allow-Origin: http://domain1.example, http://domain2.example

我已经尝试了上面的代码,但它似乎不能在Firefox中工作。

是否可以指定多个域,还是只能指定一个域?


当前回答

对于多个域,在你的。htaccess:

<IfModule mod_headers.c>
    SetEnvIf Origin "http(s)?://(www\.)?(domain1.example|domain2.example)$" AccessControlAllowOrigin=$0$1
    Header add Access-Control-Allow-Origin %{AccessControlAllowOrigin}e env=AccessControlAllowOrigin
    Header set Access-Control-Allow-Credentials true
</IfModule>

其他回答

下面是如何回显Origin头,如果它匹配你的域与Nginx,这是有用的,如果你想为一个字体多子域:

location /fonts {
    # this will echo back the origin header
    if ($http_origin ~ "example.org$") {
        add_header "Access-Control-Allow-Origin" $http_origin;
    }
}

下面是一个Java web应用的解决方案,基于yesthatguy的回答。

我使用的是Jersey REST 1.x

配置web.xml以识别Jersey REST和CORSResponseFilter

<!-- Jersey REST config -->
<servlet>
  <servlet-name>JAX-RS Servlet</servlet-name>
  <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
  <init-param>
    <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
    <param-value>true</param-value>
  </init-param>
  <init-param>
    <param-name>com.sun.jersey.spi.container.ContainerResponseFilters</param-name>
    <param-value>com.your.package.CORSResponseFilter</param-value>
  </init-param>
  <init-param>
    <param-name>com.sun.jersey.config.property.packages</param-name>
    <param-value>com.your.package</param-value>
  </init-param>
  <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
  <servlet-name>JAX-RS Servlet</servlet-name>
  <url-pattern>/ws/*</url-pattern>
</servlet-mapping>

下面是CORSResponseFilter的代码

import com.sun.jersey.spi.container.ContainerRequest;
import com.sun.jersey.spi.container.ContainerResponse;
import com.sun.jersey.spi.container.ContainerResponseFilter;


public class CORSResponseFilter implements ContainerResponseFilter{

    @Override
    public ContainerResponse filter(ContainerRequest request,
            ContainerResponse response) {
        
        String[] allowDomain = {"http://localhost:9000","https://my.domain.example"};
        Set<String> allowedOrigins = new HashSet<String>(Arrays.asList (allowDomain));                  
        
        String originHeader = request.getHeaderValue("Origin");
        
        if(allowedOrigins.contains(originHeader)) {
            response.getHttpHeaders().add("Access-Control-Allow-Origin", originHeader);
                        
            response.getHttpHeaders().add("Access-Control-Allow-Headers",
                    "origin, content-type, accept, authorization");
            response.getHttpHeaders().add("Access-Control-Allow-Credentials", "true");
            response.getHttpHeaders().add("Access-Control-Allow-Methods",
                    "GET, POST, PUT, DELETE, OPTIONS, HEAD");
        }
        
        return response;
    }
}

PHP代码:

$httpOrigin = isset($_SERVER['HTTP_ORIGIN']) ? $_SERVER['HTTP_ORIGIN'] : null;
if (in_array($httpOrigin, [
    'http://localhost:9000', // Co-worker dev-server
    'http://127.0.0.1:9001', // My dev-server
])) header("Access-Control-Allow-Origin: ${httpOrigin}");
header('Access-Control-Allow-Credentials: true');

只能为Access-Control-Allow-Origin标头指定一个原点。但是您可以根据请求在响应中设置原点。不要忘记设置Vary标头。在PHP中,我将做以下工作:

/**
 * Enable CORS for the passed origins.
 * Adds the Access-Control-Allow-Origin header to the response with the origin that matched the one in the request.
 * @param array $origins
 * @return string|null returns the matched origin or null
*/
function allowOrigins($origins)
{
    $val = $_SERVER['HTTP_ORIGIN'] ?? null;
    if (in_array($val, $origins, true)) {
        header('Access-Control-Allow-Origin: '.$val);
        header('Vary: Origin');
        
        return $val;
    }
        
    return null;
}
    
if (allowOrigins(['http://localhost', 'https://localhost'])) {
   echo your response here, e.g. token
}

并非所有浏览器都使用HTTP_ORIGIN。HTTP_ORIGIN有多安全?对我来说,它在FF中是空的。 我有网站,我允许访问我的网站发送一个网站ID,然后我检查我的数据库记录与ID,并获得SITE_URL列值(www.yoursite.com)。

header('Access-Control-Allow-Origin: http://'.$row['SITE_URL']);

即使发送了一个有效的站点ID,请求也需要来自我的数据库中列出的与该站点ID相关的域。