如何从MySQL列中提取图像以在Web应用程序上显示

Question

Sorry if I am not too clear..

Okay so what I have is a web app that shows all the fleet vehicles for my company.I am pulling the information from my database and what I want is to insert and image for the "support van" so that when it is displayed on my app that the image is pulled from the database, I am not sure about Blob data or anything so a clear instruction would be really helpful

here is my master query-Not the "support vans" query ill post that under this one

Maaster-query

 SELECT
   vi.id as 'VehicleId', 
   vi.class_type as 'VehicleClass',
   vi.registration_number as 'VehicleRegistrationNumber',
   vr.role_name as 'VehicleRole',
   vm.name as 'VehicleMake',
   vmo.name as 'VehicleModel',
   ud.name as 'Depot location'

 FROM 
   unify_rebuild.vehicle_information as vi
 LEFT JOIN 
   unify_rebuild.vehicle_role as vr
  ON
   vi.unit_role = vr.role_id
 LEFT JOIN
   unify_rebuild.vehicle_manufacturer as vm
  ON
   vi.make = vm.id
 LEFT JOIN
   unify_rebuild.vehicle_model as vmo
  ON
   vi.model = vmo.id
 LEFT JOIN
   unify_rebuild.unify_depot as ud
  ON
   vi.depot_current_location = ud.id

and here is the support van query-

 SELECT
   vi.id as 'VehicleId', 
   vi.class_type as 'VehicleClass',
   vi.registration_number as 'VehicleRegistrationNumber',
   vr.role_name as 'VehicleRole',
   vm.name as 'VehicleMake',
   vmo.name as 'VehicleModel',
   ud.name as 'Depot location'

 FROM 
   unify_rebuild.vehicle_information as vi
 LEFT JOIN 
   unify_rebuild.vehicle_role as vr
 ON
   vi.unit_role = vr.role_id
 LEFT JOIN
   unify_rebuild.vehicle_manufacturer as vm
 ON
   vi.make = vm.id
 LEFT JOIN
   unify_rebuild.vehicle_model as vmo
 ON
   vi.model = vmo.id
 LEFT JOIN
   unify_rebuild.unify_depot as ud
 ON
   vi.depot_current_location = ud.id

 WHERE vr.role_name='Support Van';

Any questions please leave a comment and thanks in advance for any help

Calvin

Answer 1

通过您的CMS你上传图片到一个文件夹，并把他们的地址，并将其保存在数据库中，并显示出你把URL从数据库和图像来源，像给图像的URL，我认为这个链接会为你有所帮助如何将文件名存储在数据库中，以及使用PHP将映像上传到服务器时的其他信息？

Answer 2

Well you can simply record the url of the image into DB as a new row (for ex. vi.imageUrl=' http://url-of-image.png ').

Then you call that row of the DB and you save it into a variable (you know how to extract values from the DB) for ex: $url_of_img_extracted_from_db is the variable of the image url, then simply you can call it into html with <img src="$url_of_img_extracted_from_db">

I know the code I posted is wrong, it is just an example, you know how to do it ;)