unzip multiple zip files

How to merge multiple log files?The following uses the standard CLF format log (APACHE) as an example:The log format of Apche is as follows:% H % L % u % t/"% R/" %> S % BExample:111.222.111.222--[03/APR/2002: 10: 30: 17 + 0800] "Get/index.html HTTP/1.1" 200 419 The simplest idea is to read the logs one by one and sort them by the time field in the log.Cat log1 log2 log3 | sort-K 4-T ""Note:-T "": the log f

Recently, I had to download multiple files from the server and compress them at the same time. I checked the Internet for a long time without a line. Finally, I came up with the following method for reference only. /*** Compressed download folder * @ param namelist download file List * @ param path download path * @ param zipname compressed file name */public void zipDownloadFile (HttpServletResponse resp

Some projects in order to better user experience, will download the file into a compressed file, direct download, so as to avoid the click to download files. There are many ways to compress files on the Internet, there are third-party loading DLL files, this paper mainly introduces DotNetZip compression method.DotNetZip's dll:http://download.csdn.net/detail/lilin

) { if (fileupload1.hasfile) { try { Fileupload1.saveas (Server.MapPath ("upload") + "\" + fileupload1.filename); Label1.Text = "Client path:" + FileUpload1.PostedFile.FileName + 1, asp.net fileupload multiple file Upload example With FileUpload for multiple file uploads, you can handle each file separately as a flyer file, and you can use the Httpfilecollection class to capture all the

ASP. NET MVC4 uses uploadify. js to upload multiple files, mvc4uploadify. js Page code: 1. Introduce js and css files 2. Code in the body Background code: Public ActionResult loadFileInfo () {StringBuilder sb = new StringBuilder (); DirectoryInfo theFolder = new DirectoryInfo (Server. MapPath ("~ /UploadFile/"); DirectoryInfo [] dirInfo = theFolder. getDirector

//file nameString[] names={"one.jpg", "two.jpg", "three.jpg", "four.jpg"}; //Four file streamsFileInputStream INPUT1 =NewFileInputStream (NewFile ("FileName path")); FileInputStream Input2=NewFileInputStream (NewFile ("FileName path")); FileInputStream INPUT3=NewFileInputStream (NewFile ("FileName path")); FileInputStream Input4=NewFileInputStream (NewFile ("FileName path")); Fileinputstream[] Inputs={INPUT1,INPUT2,INPUT3,INPUT4}; //Zip packaging Pict

, because of changes in requirements, it is also based on memory but requires a single compression and decompression of multiple files. The above method is not applicable. In the process of my online search to zipforg can solve this problem. So I downloaded the zipforge function Compressitmultifilebyzipforge (str_needtocompress_filenamelist:tstringlist; str_zip_file_name:string): Boolean; var My_ms, My_des