使用全局变量来动态更改URL [英] Use Global Variables to dynamically change URL
问题描述
所以这将很难解释，
我有一个A文件夹，并且有一个B文件夹。我正在为他们使用一个共同的数据库。
这两个文件夹都有以下文件，其中包含以下文件=> config.php，index.php，upload.php，control_admin.php，admin.php和view.php 。有一些基本的数据库，CSS和JS文件，其中B与A共享（没有任何问题）。
现在A是项目A，B是项目B.
对于每个项目，我都有URL => a.example.com& b.example.com
在a.example.com/index.php上传一个像project_name = Apple Version = 1这样的图像文件。
在b.example.com/index.php上传一个像project_name = Banana Version = 1这样的特征的图像文件。
这些图像分别上传到A和B中的图像文件夹。
现在查看这些图片，我可以到
a.example.com \view。 php？$ version = 1，可以看到图片。
同样适用于b.example.com
我在.htaccess中修改了这个内容，以创建一个友好的URL，它将我带到
$ b $ .example.com \1，并根据需要显示图像。
现在出现这个问题。每个view.php文件都有以下图片链接：
< img src =<？php echohttp： //\".$GLOBALS['url'].\"\".$GLOBALS['img_folder'].\"/\".\"new？> style ='margin：0px auto;显示：块;最大宽度：100％;高度：自动;'>
我从每个文件的config.php取这些全局变量。
对于A：
<？php $ GLOBALS ['project_name'] ='Apple'; ？>
<？php $ GLOBALS ['url'] ='a.example.com/'？>
<？php $ GLOBALS [folder] ='FolderA'; ？>
<？php $ GLOBALS ['img_folder'] ='AppleImages'; ？>
对于B：
<？php $ GLOBALS ['project_name'] ='香蕉';？>
<？php $ GLOBALS ['url'] ='b.example.com/'?>
<？php $ GLOBALS [folder] ='FolderB'; ？>
<？php $ GLOBALS ['img_folder'] ='BananaImages'; ？>
现在的问题是，如果我上传A的版本1，然后上传B的版本1，请转到a.example.com/1或b.example.com/1的友好网址链接...我在B上传文件后无法看到A处的网址，因为该网址很难指向B因为这是最近上传到同一个数据库。
如果我上传到a.example.com，然后上传到b .example.com，具有相同版本的文件...我可以在各自的URL上看到这两个图像
a.example.com/2和b.example.com/ 2
我希望我有道理。提前致谢。
https://drive.google.com/drive/folders/0B5nDbj-RhZJMa3drNFlfRzVaLXc?usp=sharing
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view.php：line 64
$ project_name_download_form = $ _GET ['project_name'];
除非您使用像这样的网址访问view.php：
a.example.com \view.php？version = 1& project_name = apple
您的 $ project_name_download_form 将有一个空字符串值，并且您的SQL查询变为：
SELECT demo_id，demo_name，demo_version，demo_details，file
FROM demo
WHERE demo_name =''
AND demo_version ='1'
所以2行动，将第64行改为：
$ project_name_download_form = $ GLOBALS ['project_name'];
然后直接检查数据库以确保demo_name列的值已被设置只是空白）。
So this is going to be really hard to explain,
I have an A folder and I have a B folder. I am using a common database for both of them.
Both folder have following files, inside each of them => config.php, index.php, upload.php, control_admin.php, admin.php and view.php. There are some basic database, CSS and JS files which B shares with A(not of any problem).
Now A is for project A and B is for project B.
for each of the projects, I have URLs => a.example.com & b.example.com
At a.example.com/index.php I upload an image file which characteristics like project_name = Apple Version = 1.
At b.example.com/index.php I upload an image file which characteristics like project_name = Banana Version = 1.
These images are uploaded to image folders in A and B respectively.
Now to view these images, I could have gone to
a.example.com\view.php?$version=1 and could have seen the pic.
similarly for b.example.com
I have modified this in .htaccess to make a friendly URL which takes me to
a.example.com\1 and displays the image as required.
Now comes the problem. Each of the view.php file have the following image link
<img src = "<?php echo "http://".$GLOBALS['url']."".$GLOBALS['img_folder']."/".$new ?>" style = 'margin: 0px auto; display:block; max-width: 100%;height: auto;'>
I take these global variables from config.php of each file.
For A:
<?php $GLOBALS['project_name'] = 'Apple';?>
<?php $GLOBALS['url'] = 'a.example.com/'?>
<?php $GLOBALS["folder"] = 'FolderA'; ?>
<?php $GLOBALS['img_folder'] = 'AppleImages'; ?>
For B:
<?php $GLOBALS['project_name'] = 'Banana';?>
<?php $GLOBALS['url'] = 'b.example.com/'?>
<?php $GLOBALS["folder"] = 'FolderB'; ?>
<?php $GLOBALS['img_folder'] = 'BananaImages'; ?>
Now the issue is, if I upload version 1 of A and then version 1 of B and then I go to the link which is friendly URL from a.example.com/1 or b.example.com/1... I cannot see the one at A after I upload a file at B. Because the URL is hard pointing at B as it is the recent upload in the same database.
What can I do to the URL to make sure, if I upload to a.example.com and then to b.example.com, files with same version... I can see both images at their respective URLs
a.example.com/2 and b.example.com/2
I hope I made sense. Thanks in advance.
https://drive.google.com/drive/folders/0B5nDbj-RhZJMa3drNFlfRzVaLXc?usp=sharing
I looked through the google drive link you provided.
view.php : line 64
$project_name_download_form = $_GET['project_name'];
Unless you are accessing view.php with a url like this:
a.example.com\view.php?version=1&project_name=apple
Your $project_name_download_form will have an empty string value and your SQL query becomes:
SELECT demo_id, demo_name, demo_version, demo_details, file
FROM demo
WHERE demo_name = ''
AND demo_version = '1'
So 2 actions, change line 64 to:
$project_name_download_form = $GLOBALS['project_name'];
And then check the database directly to make sure the demo_name column has values being set in it (and not just blank).
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