小编Pio*_*ros的帖子

我有一个代理权表,与地区,国家,地区,城市,行业,专业,行业,专业和次专业有着多对多的关系.

办公室和联系人之间也存在一对多的关系.

基本上我想要做的是创建一个显示这些表中特定记录的结果表.

我已经在表中的每个外键上创建了索引.这是我的查询:

SELECT agencies.agency, 
       agencies.website_url, 
       agencies.status, 
       agencies.SIZE, 
       agencies.id, 
       officedata.id, 
       contactdata.name, 
       contactdata.surname, 
       contactdata.job_title, 
       contactdata.email, 
       contactdata.mobile 
FROM   agencies 
       LEFT JOIN (SELECT agencies_industries.agency_id, 
                         agencies_industries.industry_id 
                  FROM   agencies_industries) AS industrydata 
         ON agencies.id = industrydata.agency_id 
       LEFT JOIN (SELECT agencies_professions.agency_id, 
                         agencies_professions.profession_id 
                  FROM   agencies_professions) AS professiondata 
         ON agencies.id = professiondata.agency_id 
       LEFT JOIN (SELECT agencies_sectors.agency_id, 
                         agencies_sectors.sector_id 
                  FROM   agencies_sectors) AS sectordata 
         ON agencies.id = sectordata.agency_id 
       LEFT JOIN (SELECT agencies_seniorities.agency_id, 
                         agencies_seniorities.seniority_id 
                  FROM   agencies_seniorities) AS senioritydata 
         ON agencies.id = senioritydata.agency_id 
       LEFT JOIN (SELECT agencies_zones.agency_id, 
                         agencies_zones.zone_id 
                  FROM   agencies_zones) AS zonesdata 
         ON …

每当我尝试以这种方式定义2个脚本时:

echo $this->Html->script(array('jquery', 'prototype'));

只是prototype工作.

当我这样做时:

echo $this->Html->script(array('prototype', 'jquery'));

只是jquery工作.

我该如何修复它以便两者都有效？