抱歉,如果我不太清楚..
好吧,我有一个网络应用程序,显示我公司的所有车队车辆。我正在从数据库中提取信息,我想要的是插入“支持货车”的图像,以便当它出现时在我的应用程序上显示图像是从数据库中提取的,我不确定 Blob 数据或任何内容,因此清晰的说明将非常有帮助
这是我的主查询 - 不是“支持货车”查询,我将在此查询下发布
主查询
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
这是支援车查询-
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';
如有任何问题,请发表评论,并提前感谢您的帮助
卡尔文
最佳答案
通过您的 cms,您将图像上传到文件夹并获取其地址并将其保存在数据库中 并显示该图像,您从数据库中获取 URL,并在图像源中给出该图像的 URL 我认为这个链接对你有帮助 How to store file name in database, with other info while uploading image to server using PHP?
关于php - 如何从 MySQL 列中提取图像以显示在 Web 应用程序上,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/25847302/