Hi all!
Here is my problem : I have a string like the following :
20030613170404;BAN_CAV ; starting script Loader.sh on ; 13/06/2003 at ; 17;04;03
I want to eraze all characters located after "Loader.sh", because there are unuseful.
I tried to use sed...but it didnt work....i guess i... (1 Reply)
Hopefully someone can help out here. This is probably fairly basic, but I've searched and tried several variations of the solutions presented in these forums, so I'll go ahead and ask.
How can I locate a string in a file, delete the characters after the string and then replace the string with a... (2 Replies)
Hi There!
I have the following string
which i need to convert to
i.e. between each occurence of the delimiter ('|' in this case), i need to delete all characters from the '|' to the ':' so that |10,9:12/xxx| becomes |12/xxx|
How can i do this using sed?
Thanks in advance! (13 Replies)
Hi all,
I have the following lines
<b>A gtwrhwrthwr text hghthwrhtwrtw </b><font color='#06C'>; text text (text)
<b>B gtwrhwrthwr text hghthwrhtwrtw </b><font color='#06C'>; text text (text)
<b>J gtwrhwrthwr text hghthwrhtwrtw </b><font color='#06C'>; text text (text)
and I would like to... (5 Replies)
Hi
I would like to batch delete the "note" entry from bib files. The start would be defined by "note ={" and the end by "}." (see example bib entry below).
I tried the following command which does not have any effect:
cat input.bib| sed -e 's/note = {.*}.//' > output.bib
Any help would... (2 Replies)
helloo
I wonder if there's a way to cut characters out of a string and keep only
the last 2 by using sed.
For example if there's the todays' date:
2012-05-06
and we only want to keep the last 2 characters which are the day.
Is there a quick way to do it with sed? (2 Replies)
Hi,
I have a xml file (Config.xml)
<Header name="" TDate="" PDate="">
<Config>
{"config" { "Nation" "Pri:|Sec:"}}
</Config>
</Header>
Now I wanted to printed all the strings between "". I tried the following
cat Config.xml | sed -n 's/.*\.*//p'
... (8 Replies)
Hi,
I have a text file with some lines like this:
/MEDIA/DISK1/23568742.MOV
/MEDIA/DISK1/87456321.AVI
/MEDIA/DISK2/PART1/45753131.AVI
/IMPORT/44452.WAV
...
I want to remove the last 12 characters in each line that it ends "AVI". Should look like this:
/MEDIA/DISK1/23568742.MOV... (12 Replies)
Hi,
I hope you can help me out please?
I need to replace from character 8-16 with AAAAAAAA and the rest should stay the same after character 16
gtwrhtrd11111111rjytwyejtyjejetjyetgeaEHT
wrehrhw22222222hytekutkyukrylryilruilrGEQTH
hrwjyety33333333gtrhwrjrgkreglqeriugn;RUGNEURGU
... (4 Replies)
I have this fastq file:
@M04961:22:000000000-B5VGJ:1:1101:9280:7106 1:N:0:86
GGGGGGGGGGGGCATGAAAACATACAAACCGTCTTTCCAGAAATTGTTCCAAGTATCGGCAACAGCTTTATCAATACCATGAAAAATATCAACCACACCA
+test-1
GGGGGGGGGGGGGGGGGCCGGGGGFF,EDFFGEDFG,@DGGCGGEGGG7DCGGGF68CGFFFGGGG@CGDGFFDFEFEFF:30CGAFFDFEFF8CAF;;8... (10 Replies)
Discussion started by: Xterra
10 Replies
LEARN ABOUT LINUX
zegrep
ZGREP(1) General Commands Manual ZGREP(1)NAME
zgrep - search possibly compressed files for a regular expression
SYNOPSIS
zgrep [ grep_options ] [ -e ] pattern filename...
DESCRIPTION
Zgrep invokes grep on compressed or gzipped files. These grep options will cause zgrep to terminate with an error code:
(-[drRzZ]|--di*|--exc*|--inc*|--rec*|--nu*). All other options specified are passed directly to grep. If no file is specified, then the
standard input is decompressed if necessary and fed to grep. Otherwise the given files are uncompressed if necessary and fed to grep.
If the GREP environment variable is set, zgrep uses it as the grep program to be invoked.
EXIT CODE
2 - An option that is not supported was specified.
AUTHOR
Charles Levert (charles@comm.polymtl.ca)
SEE ALSO grep(1), gzexe(1), gzip(1), zdiff(1), zforce(1), zmore(1), znew(1)ZGREP(1)