gzip | 易学教程

python extract keywords from multiple .gz files

阅读更多关于 python extract keywords from multiple .gz files

问题 Question: How to search keywords from multiple files in Python(including compressed gz file and uncompressed file) I have multiple archived logs in a folder, the latest file is "messages",and the older logs will auto-compressed as .gz file. -rw------- 1 root root 21262610 Nov 4 11:20 messages -rw------- 1 root root 3047453 Nov 2 15:49 messages-20191102-1572680982.gz -rw------- 1 root root 3018032 Nov 3 04:43 messages-20191103-1572727394.gz -rw------- 1 root root 3026617 Nov 3 17:32 messages

web性能优化之GZIP压缩

阅读更多关于 web性能优化之GZIP压缩

从服务端优化来说，通过对服务端做压缩配置可以大大减小文本文件的体积，从而使加载文本的速度成倍的加快。目前比较通用的压缩方法是启用gzip压缩。它会把浏览器请求的页面，以及页面中引用的静态资源以压缩包的形式发送到客户端,然后在客户端完成解压和拼装.下图是使用相关工具检测没有开启网页压缩的效果。检测地址： http://www.whatsmyip.org/http-compression-test/?url=d3d3LmZlbmd6aWdhbWUuY29t windows下apache开启GZIP 实现步骤： 1.找到并打开apache/conf目录中的httpd.conf文件 2.httpd.conf中打开deflate_Module和headers_Module模块，具体做法为将如下两句前面的#去掉：　　LoadModule deflate_module modules/mod_deflate.so 　　LoadModule headers_module modules/mod_headers.so 3.在httpd.conf文件底部加入如下代码配置需要压缩的文件： <IfModule mod_deflate.c> SetOutputFilter DEFLATE AddOutputFilterByType DEFLATE text/plain

bash uncompress gz and store in variable

阅读更多关于 bash uncompress gz and store in variable

问题 I need to uncompress a .gz file and store it in a variable, so I can use it later. So, the idea is that I generate *.fastq.gz files, and I need to uncompress them and keep just the *.fastq file. Then, I would like to store its name in a variable, so I can call the file for further processing. Here, there is the code I am executing: input: $file.fastq.gz Where $file is the name of the file (it changes, as this code is inside a loop) reads=$(gunzip $file.fastq) echo $reads Does anybody know

zlib inflate returning a buffer error

阅读更多关于 zlib inflate returning a buffer error

问题 I'm trying to decompress a zip file using zlib (without using any extension or 3rd party). Initially, the src_len is 48756255, and the dest_len is 49209890. The first pass in the while loop is fine: err is Z_OK and the second pass through starts. On the second pass, no matter what I do I get a Z_BUF_ERROR from inflate. stream.total_out at this point is 49034460, so there is a bit remaining but stream.avail_in on the second pass is 0. In any case, I would expect inflate to give me Z_STREAM_END

Problem in compressing xml to gzip without using “encoding/xml” package

阅读更多关于 Problem in compressing xml to gzip without using “encoding/xml” package

问题 I am working on a program that compresses xml files to gzip using Go. but the program failed to generate files, however it does generate the output when I try to convert .txt file to gzip and I don't have to use encoding/xml package. Here is my program:- package main import ( "bytes" "compress/gzip" "fmt" "io" "log" "os" ) type Notes struct { To string `xml:"to"` From string `xml:"from"` Heading string `xml:"heading"` Body string `xml:"body"` } func main() { var buf bytes.Buffer zw := gzip

Using TSQL to Unzip a value

阅读更多关于 Using TSQL to Unzip a value

问题 How can I unzip a varbinary(max) value in a stored procedure? I'd like to implement or invoke a gunzip algorithm within TSQL code, preferably without enabling the CLR. 回答1: Look into sp_OACreate http://msdn.microsoft.com/en-us/library/ms189763.aspx sp_oamethod http://msdn.microsoft.com/en-us/library/ms174984.aspx You can write the value as a file using file system object, then use a command line gunzip program on it and the read the file using FSO back. Yes this is not practical but I am not

Issue in gzip compression

阅读更多关于 Issue in gzip compression

问题 I am using gzip to compress the loading files. For that i have written the following code in server.xml in tomcat. <Connector compressableMimeType="text/html,text/xml,text/json,text/javascript,text/css,text/plain,application/json,application/javascript" compression="on" compressionMinSize="2048" connectionTimeout="200000" noCompressionUserAgents="gozilla, traviata" port="8080" protocol="HTTP/1.1" redirectPort="8443"/> What i get is "Content-Encoding : gzip". But for some files it is not get

How to gzip an NSURLConnection request?

阅读更多关于 How to gzip an NSURLConnection request?

问题 Apparently, NSURLConnection automatically decodes gzipped responses. But, can it gzip encode its requests? If so, how? Or, does it do this automatically too? 回答1: An HTTP request body cannot be gzipped at the protocol level, as there is no way for the client to know whether the server supports that or not; it works for the server response because the client indicates in the request whether a gzipped response body can be accepted. (It could be done in the future something like how Expect: 100

urlopen trouble while trying to download a gzip file

阅读更多关于 urlopen trouble while trying to download a gzip file

问题 I am going to use the wiktionary dump for the purpose of POS tagging. Somehow it gets stuck when downloading. Here is my code: import nltk from urllib import urlopen from collections import Counter import gzip url = 'http://dumps.wikimedia.org/enwiktionary/latest/enwiktionary-latest-all-titles-in-ns0.gz' fStream = gzip.open(urlopen(url).read(), 'rb') dictFile = fStream.read() fStream.close() text = nltk.Text(word.lower() for word in dictFile()) tokens = nltk.word_tokenize(text) Here is the

mapreduce in java - gzip input files

阅读更多关于 mapreduce in java - gzip input files

问题 I'm using java , and i'm trying to write a mapreduce that will recieve as an Input a folder containing multiple gz files. I've been looking all over but all the tutorials that i've found exmplain how to process a simple text File, but haven't found anything that solves my problem. I've asked around at my workplace, but only got references to scala, which i'm not familier with. Any help would be appreciated. 回答1: Hadoop checks the file extension to detect compressed files. The compression