小编mck*_*mck的帖子

使用 7.4.1，当我加载带有 videojs 的页面时，Chrome devtools 向我显示此错误：

Refused to create a worker from      
'blob:https://dev.culturediscovery.com/51e9879d-fa81-4044-9117-        
7328c0df4dd6' because it violates the following Content Security Policy directive: "default-src * data: 'unsafe-eval' 'unsafe-inline'". Note that 'worker-src' was not explicitly set, so 'default-src' is used as a fallback.
(anonymous) @   video.min.js:1830
(anonymous) @   video.min.js:2
(anonymous) @   video.min.js:2

谁能帮我弄清楚如何处理这个问题？

我试图绕过处理连接,mysql和coldfusion.以下查询在没有最后条件的情况下工作.

        <cfquery name="GetWeekends">
        SELECT w.id, w.weekend_type, w.community_id, w.start_date, w.end_date,
          w.language,
          c.community_id, c.location, c.language, c.state, c.country
        FROM _weekends w
        INNER JOIN _communities c
        ON w.community_id=c.community_id
        WHERE w.weekend_type = 1 AND w.start_date > Now() AND           
        #DateFormat(w.start_date, "m")# = '#form.home_by_month#'
        ORDER BY w.start_date ASC
        </cfquery>

它正在奄奄一息

  #DateFormat(w.start_date, "m")#

告诉我变量[W]不存在.对不起,我正在学习,因为我去了这里......

我正在执行两个查询，以根据同一数据库中两个表的数据构建一个表。我现在的代码在下面，但是我知道我正在以这种方式创建不必要的负载。我一直在尝试加入表格以得到相同的结果，但是没有运气。有输入吗？

<cfquery name="GetWeekends">
   SELECT id, weekend_type, community_id, start_date, end_date, language
   FROM _weekends
   WHERE weekend_type = 1 and start_date > Now()
   ORDER BY start_date ASC
</cfquery>  

<cfloop query="GetWeekends">                
    <cfquery name="GetCommunity">
        SELECT community_id, location, language, state, country
        FROM _communities
        WHERE community_id = #getweekends.community_id#
    </cfquery>                  
    <tr>
        <td>#DateFormat(start_date, "mm/dd/yyyy")#</td>
        <td>#GetComm.location#</td>
        <td>#GetComm.state#</td>
        <td>#GetComm.country#</td>
        <td>#GetComm.language#</td>
   </tr>
</cfloop>