Domain names like com/.cn/.org are good to solve.Similar. com.cn/.org.cn similar domain name does not have a direct solution, I would like to collect all the top-level domain name generation list and then use regular match, but found this.PublicsuffixBasic usage in the intro
In the Windows programming environment, we have an API function that can be invoked gethostbyname, and we had to implement similar functions through components in the ASP, such as Aspdns and so on, Now MS has provided us with a new programming interface on ASP.net, so that we can easily implement such a similar function. The reason is that we have a DNS class in asp.net that we can easily implement the features we mentioned above in this class.We can see that the following code is a very simple
This article mainly introduces the PHP access to the domain name of Google included specific implementation, need friends can refer to theThe code is as follows: function Get_index ($domain) {$url = http://www.google.com/search?source=hpbiw=1440bih=762q= Site%3a$domainaq=faqi=g10aql=oq= "; $html =file_get_contents ($url); Preg_match ('//ui ', $html, $index); For
In the Windows programming environment, we have an API function that can be invoked gethostbyname, which we had to implement through the component in the ASP
Similar functions, such as Aspdns, and so on, now Ms on the ASP.net to provide us with a new programming interface, so that we can easily achieve such a similar
The reason is that in the ASP.net gives us a DNS class, we can through this Class easy to implement what we mentioned above
Function.
We can see that the following code is a very si
This article mainly introduces a simple way to obtain file extensions in PHP, examples of the use of PHP to obtain file name extension skills, with a certain reference value, the need for friends can refer to the
The example in this article describes a simple way to get a file na
Thinkphp3.2 {code ...} in this case, I get the field name into an array $ arrKey, but I think this method is stupid and there should be a direct way to get it. is there any ??? Thinkphp 3.2
$use=D('classone');$db=$use->select(5);$arr = $db[0] ? $db[0] : '';foreach($arr as $key => $value){ $arrKey[] = $key;}
In
The example in this article describes a simple way to get a file name extension from PHP. Share to everyone for your reference. The implementation methods are as follows:
I hope this article will help you with your PHP program design.
1. Under XCODE5, get the program name (app name) in the following way:NSString *proname = [[[NSBundle Mainbundle] infodictionary] Objectforkey:@ "cfbundledisplayname "];NSLog (@ "Dicname = = =%@", [[NSBundle Mainbundle] infodictionary]);After printing you can see:Cfbundledisplayname = Testofbutton;Cfbundlename = Testofbutton;2. Under Xcode6,
One way to get the process name of the current user process in the kernel driver
In the kernel driver, you can use the psgetcurrentprocess function to obtain the eprocess structure address of the currently called driver process. many articles say that the process name is stored at the 0x174 offset of the eprocess stru
Thinkphp 3.2
$use=D('classone');$db=$use->select(5);$arr = $db[0] ? $db[0] : '';foreach($arr as $key => $value){ $arrKey[] = $key;}
This, I took the name of the field to get into an array $arrKey, but I think this method is rather stupid, there should be a direct access to the method, I ask there is no???
Reply content:
Thinkphp 3.2
$use=D('classone');$db=$use->select(5);$arr = $db[0] ? $db[0] :
1. The picture is placed in the SDcard,
Copy Code code as follows:
Bitmap imagebitmap = bitmapfactory.decodefile (path is the path to the picture, with the directory is/sdcard)
2. The picture is under the project's Res folder
Copy Code code as follows:
Get Application Object
ApplicationInfo appInfo = Getapplicationinfo ();
To get the ID of the picture (
1. Place the image in sdcard,
CopyCodeThe Code is as follows: bitmap imagebitmap = bitmapfactory. decodefile (PATH) (path is the image path, and the directory is/sdcard)
2. The image is in the res folder of the project.Copy codeThe Code is as follows: // obtain the application objectApplicationinfo appinfo = getapplicationinfo ();// Obtain the image ID (name is the image name, "drawable" is the directory
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