Lua

Unzip Files Matching a Pattern (such as *.xml)

See more Zip Examples

Example code to unzip Files Matching a Pattern (such as *.xml) Lua

local chilkat = require("chilkat")

-- This example works with the .zip located at http://www.example-code.com/data/test.zip

-- This example assumes the Chilkat API to have been previously unlocked.
-- See Global Unlock Sample for sample code.

-- First, let's download the .zip
local http = chilkat.Http.new()
local success = http:download("http://www.example-code.com/data/test.zip", "qa_test/test.zip")
if not success then
    print(http.last_error_text)
    return
end

local zip = chilkat.Zip.new()

success = zip:open_zip("qa_test/test.zip")
if not success then
    print(zip.last_error_text)
    return
end

-- Show what's in the .zip by getting the directory as XML.
local zipContentsXml = zip:get_directory_as_xml()
if not zip.last_method_success then
    print(zip.last_error_text)
    return
end

print(zipContentsXml)

-- In this case, the contents of our .zip look like this:
-- <?xml version="1.0" encoding="utf-8" ?>
-- <zip_contents>
--     <dir name="test">
--         <dir name="abc">
--             <dir name="123">
--                 <file>ecc_public.pem</file>
--                 <file>starfish.jpg</file>
--             </dir>
--             <file>comodo-certificates.pem</file>
--             <file>pigs.xml</file>
--         </dir>
--         <file>COMODORSACertificationAuthority.crt</file>
--         <file>hamlet.xml</file>
--     </dir>
-- </zip_contents>

-- ------------------------------------------------------------------------
-- If we wish to unzip only .xml files, use "*.xml" for the pattern\
-- Unzip to the "qa_output" directory relative to our current working directory.
-- (You may unzip to an absolute path by passing an absolute path, such as "/someDir" or
-- "C:/someDir", etc.)
-- This will unzip only the .xml files.  The relative directory structure within
-- the .zip will be created in the local filesystem under the unzip directory
-- path passed in the first argument, which in this case is "qa_output".
local count = zip:unzip_matching("qa_output", "*.xml", true)
if count < 0 then
    print(zip.last_error_text)
    return
end

print("unzipped " .. count .. " matching file(s).")

-- ------------------------------------------------------------------------
-- If we wish to unzip only files under the "123" directory, we can do this:
count = zip:unzip_matching("qa_output2", "*/123/*", true)
if count < 0 then
    print(zip.last_error_text)
    return
end

print("unzipped " .. count .. " matching file(s).")

-- ------------------------------------------------------------------------
-- If no asterisk character is used, then the match must be exact.
-- For example, to unzip "test/hamlet.xml" we must pass the exact path
-- as it is in the .zip.  We cannot just pass "hamlet.xml".
count = zip:unzip_matching("qa_output3", "test/hamlet.xml", true)
if count < 0 then
    print(zip.last_error_text)
    return
end

print("unzipped " .. count .. " matching file(s).")

-- ------------------------------------------------------------------------
-- If we want to unzip any file named "hamlet.xml" found in any sub-directory...
count = zip:unzip_matching("qa_output4", "*/hamlet.xml", true)
if count < 0 then
    print(zip.last_error_text)
    return
end

print("unzipped " .. count .. " matching file(s).")