unzip

I have seen only one question on here but it does not answer my question. I am running a typical LAMP server that has the most up to date PHP 5 and MYSQL 5 with Redhat Linux. I need to find a PHP only solution because my host does not allow me to use shell. Here is my code that extracts ZIPs that are not passworded from vBulletin uploads to another directory: if ($_GET['add'] == TRUE){ $zip = new ZipArchive; $res = $zip->open($SOURCE FOLDER); if ($res === TRUE) { $zip->extractTo('$DESTINATION FOLDER/'); $zip->close(); echo 'File has been added to the library successfuly'; //Add a flag to that

I have number of Zip Files {'File1.zip', 'File2.zip', 'File3.zip',..., 'FileN.zip'} in which each zip file contains a Data.csv file. I want to read the data in 'Data.csv' of each Zip file without having to extract the Zip files' contents. Is this possible..? Certainly Winzip / 7zip / Winrar do not have COM interface component which can invoke directly unlike word, excel an other application. Hence @Java is appropriate Idea is don't extract files physically , however create absolute path of file such that windows consider as physical presence of File (similar to ~tmp file) here the code

A password cannot be specified in unzip ( utils ) function. The other function I am aware of, getZip ( Hmisc ), only works for zip files containing one compressed file. I would like to do something like this to unzip all the files in foo.zip in Windows 8: unzip("foo.zip", password = "mypass") I found this question very useful but saw that no formal answers were posted, so here goes: First I installed 7z. Then I added "C:\Program Files\7-Zip\" to my environment path. I tested that the 7z command was recognized from the command line. I opened R and typed in system("7z x secure.7z -pPASSWORD")

I have file with names like ex.zip . In this example, the Zip file contains only one file with the same name(ie. `ex.txt'), which is quite large. I don't want to extract the zip file every time.Hence I need to read the content of the file(ex.txt) without extracting the zip file. I tried some code like below But i can only read the name of the file in the variable. How do I read the content of the file and stores it in the variable? Thank you in Advance fis=new FileInputStream("C:/Documents and Settings/satheesh/Desktop/ex.zip"); ZipInputStream zis = new ZipInputStream(new BufferedInputStream

1、把/home目录下面的mydata目录压缩为mydata.zip 　　zip -r mydata.zip mydata #压缩mydata目录 2、把/home目录下面的mydata.zip解压到mydatabak目录里面 　　unzip mydata.zip -d mydatabak 3、把/home目录下面的abc文件夹和123.txt压缩成为abc123.zip 　　zip -r abc123.zip abc 123.txt 4、把/home目录下面的wwwroot.zip直接解压到/home目录里面 　　unzip wwwroot.zip 5、把/home目录下面的abc12.zip、abc23.zip、abc34.zip同时解压到/home目录里面 　　unzip abc\*.zip 6、查看把/home目录下面的wwwroot.zip里面的内容 　　unzip -v wwwroot.zip 7、验证/home目录下面的wwwroot.zip是否完整 　　unzip -t wwwroot.zip 8、把/home目录下面wwwroot.zip里面的所有文件解压到第一级目录 　　unzip -j wwwroot.zip 主要参数 -c：将解压缩的结果 -l：显示压缩文件内所包含的文件 -p：与-c参数类似，会将解压缩的结果显示到屏幕上，但不会执行任何的转换 -t

How to go about unzipping a file in swift? In Objective-C, I used SSZipArchive and I loved it. As seen in the code below. I suspect if I decide to keep SSZipArchive, I will have to bridge an Objective-C file to my Swift file. Is there any updated third-party, or better yet Apple-Documentation to unzip a file in Swift? NSString *zipPath = [self.globalFileStrucure stringByAppendingPathComponent:@"zipfile.zip"]; [data writeToFile:zipPath options:0 error:&error]; BOOL unZipped = 0; unZipped = [SSZipArchive unzipFileAtPath:zipPath toDestination:self.globalFileStrucure]; Swift 2 (Update): So it

I have a file that I want to read that is itself zipped within a zip archive. For example, parent.zip contains child.zip, which contains child.txt. I am having trouble reading child.zip. Can anyone correct my code? I assume that I need to create child.zip as a file-like object and then open it with a second instance of zipfile, but being new to python my zipfile.ZipFile(zfile.open(name)) is silly. It raises a zipfile.BadZipfile: "File is not a zip file" on (independently validated) child.zip import zipfile with zipfile.ZipFile("parent.zip", "r") as zfile: for name in zfile.namelist(): if re