I am trying to utilize a grep lookahead to get a value at the end of a line for a project I'm working on. The main issue I'm having is that I'm not sure how to use a shell variable in the grep lookahead syntax in cshell
Here's the gist of what I'm trying to do.
There will be a dogfile.txt with several lines listing the names of dogs in the format below
genericDog2033, pomeranian
genericDog2034, greatDane
genericDog2035, Doberman
I wanted a way of retrieving the breed of the dog after the comma on each line so I thought a grep lookahead might be a good way of doing it. The project I'm working on isn't so hard-coded however, so I have no way of knowing what genericDog number I am searching for. There will be a shell variable in a greater while loop which will have access to the dog name.
For example if I set the dogNumber variable to the first dog in the file like so:
set dogNumber = genericDog2033
I then try to access the value of dogNumber in the grep lookahead
set dogBreed = `cat File.txt | grep -oP '(?<=$dogNumber ,)[^ ]*'`
The problem with the line above is that I think grep is looking for the literal string "$dognumber ," in the file which obviously doesn't exist. Is there some sort of wrapper I can put around the shell variable so cshell knows that dogNumber is a variable? I'm also open to other methods of doing this. Any help would be appreciated, this is the literal last line of code I need to finish my project and I'm at my wits end.
Variable expansion only happens inside double quotes ("), and not single quotes ('):
% set var = 'hello'
% echo '$var'
$var
% echo "$var"
hello
Furthermore, you have an error in your regexp:
(?<=$dogNumber ,)[^ ]*
In your data, the space is after the comma, not before.
% set dogNumber = genericDog2033
% set dogBreed = `cat a | grep -oP "(?<=$dogNumber, )[^ ]*"`
% echo $dogBreed
pomeranian
The easiest way to debug this is to not use variables at all in the first place, and simply check if the grep works:
% grep -oP "(?<=genericDog2034 ,)[^ ].*" a
[no output]
Then first make the grep work with static data, add the variable to make that work, and then put it all together by assigning it to a variable.
Hi i have a few archive of FW log and occasionally im required to compare them with a series of IP addresses (thousand of them) to get the date and time if the ip addresses matches. my current script is as follow:
#input the list of ip into array
mapfile -t -O 1 var < ip.txt while true
do
#check array is not null
if [[-n "${var[i]}"]] then
zcat /.../abc.log.gz | grep "${var[i]}"
((i++))
It does work but its way too slow and i would think that grep-ping a line with multiple strings would be faster than zcat on every ip line. So my question is is there a way to generate a 'long grep search string' from the ip.txt? or is there a better way to do this
Sure. One thing is that using cat is usually slightly inefficient. I'd recommend using zgrep here instead. You could generate a regex as follows
IP=`paste -s -d ' ' ip.txt`
zgrep -E "(${IP// /|})" /.../abc.log.gz
The first line loads the IP addresses into IP as a single line. The second line builds up a regex that looks something like (127.0.0.1|8.8.8.8) by replacing spaces with |'s. It then uses zgrep to search through abc.log.gz once, with that -Extended regex.
However, I recommend that you do not do this. Firstly, you should escape strings put into a regex. Even if you know that ip.txt really contains IP addresses (e.g. not controlled by a malicious user), you should still escape the periods. But rather than building up a search string and then escape it, just use the -Fixed strings and -file features of grep. Then you get the simple and fast one-liner:
zgrep -F -f ip.txt /.../abc.log.gz
I'm trying to execute Sox from a Lua script.
It works fine when I pass literals as arguments.
os.execute('"C:\\Sox\\sox.exe" -S C:\\SoX\\test.wav -r 22050 C:\\Sox\\SoX_out.wav')
or
os.execute [["C:\\Sox\\sox.exe" -S C:\\SoX\\test.wav -r 22050 C:\\Sox\\SoX_out.wav]]
however, what I'd like to do (as example), when I try this:
filename = "C:\\SoX\\test.wav"
os.execute('"C:\\Sox\\sox.exe" -S filename -r 22050 C:\\Sox\\SoX_out.wav')
I get:
C:\Sox\sox.exe FAIL formats: can't open input file `filename': No such file or directory
So my question is how to pass a string properly as command argument ?
In Lua, a string literal is exactly and only that string. Strings don't know anything about variables, the global environment, local variables, etc. They're just strings. The string "filename" in Lua will always be a string of 8 characters. It will not go out and try to find a variable named filename and extract something from it.
What you want is to build a string, from multiple strings. Part of the string will come from literals, and part will come from a variable. Lua has several tools for that. The simplest is the .. concatenation operator:
[["C:\Sox\sox.exe" -S ]] .. filename .. [[ -r 22050 C:\Sox\SoX_out.wav]]
This builds a new string from a string literal, the contents of the filename variable, and another string literal. The spaces you see at the end of the first literal and the beginning of the second are necessary, since Lua will not insert spaces between the two concatenated pieces.
For more complex cases, it's useful to just build a table of parameters and use table.concat to build a string out of them:
local params =
{
[[C:\Sox\sox.exe]],
"-S",
filename,
"-r 22050",
[[C:\Sox\SoX_out.wav]]
}
os.execute(table.concat(params, " "))
Note the lack of spaces in the string literals. This is because table.concat's second parameter is a string to be inserted between the entries in the array. So between each array element will be a space; we don't need to manually add them.
I am writing a csh script that will extract a line from a file xyz.
the xyz file contains a no. of lines of code and the line in which I am interested appears after 2-3 lines of the file.
I tried the following code
set product1 = `grep -e '<product_version_info.*/>' xyz`
I want it to be in a way so that as the script find out that line it should save that line in some variable as a string & terminate reading the file immediately ie. it should not read furthermore aftr extracting the line.
Please help !!
grep has an -m or --max-count flag that tells it to stop after a specified number of matches. Hopefully your version of grep supports it.
set product1 = `grep -m 1 -e '<product_version_info.*/>' xyz`
From the man page linked above:
-m NUM, --max-count=NUM
Stop reading a file after NUM matching lines. If the input is
standard input from a regular file, and NUM matching lines are
output, grep ensures that the standard input is positioned to
just after the last matching line before exiting, regardless of
the presence of trailing context lines. This enables a calling
process to resume a search. When grep stops after NUM matching
lines, it outputs any trailing context lines. When the -c or
--count option is also used, grep does not output a count
greater than NUM. When the -v or --invert-match option is also
used, grep stops after outputting NUM non-matching lines.
As an alternative, you can always the command below to just check the first few lines (since it always occurs in the first 2-3 lines):
set product1 = `head -3 xyz | grep -e '<product_version_info.*/>'`
I think you're asking to return the first matching line in the file. If so, one solution is to pipe the grep result to head
set product1 = `grep -e '<product_version_info.*/>' xyz | head -1`
How do I remove or address a specific occurrence of a character in sed?
I'm editing a CSV file and I want to remove all text between the third and the fifth occurrence of the comma (that is, dropping fields four and five) . Is there any way to achieve this using sed?
E.g:
% cat myfile
one,two,three,dropthis,dropthat,six,...
% sed -i 's/someregex//' myfile
% cat myfile
one,two,three,,six,...
If it is okay to consider cut command then:
$ cut -d, -f1-3,6- file
awk or any other tools that are able to split strings on delimiters are better for the job than sed
$ cat file
1,2,3,4,5,6,7,8,9,10
Ruby(1.9+)
$ ruby -ne 's=$_.split(","); s[2,3]=nil ;puts s.compact.join(",") ' file
1,2,6,7,8,9,10
using awk
$ awk 'BEGIN{FS=OFS=","}{$3=$4=$5="";}{gsub(/,,*/,",")}1' file
1,2,6,7,8,9,10
A real parser in action
#!/usr/bin/python
import csv
import sys
cr = csv.reader(open('my-data.csv', 'rb'))
cw = csv.writer(open('stripped-data.csv', 'wb'))
for row in cr:
cw.writerow(row[0:3] + row[5:])
But do note the preface to the csv module:
The so-called CSV (Comma Separated
Values) format is the most common
import and export format for
spreadsheets and databases. There is
no “CSV standard”, so the format is
operationally defined by the many
applications which read and write it.
The lack of a standard means that
subtle differences often exist in the
data produced and consumed by
different applications. These
differences can make it annoying to
process CSV files from multiple
sources. Still, while the delimiters
and quoting characters vary, the
overall format is similar enough that
it is possible to write a single
module which can efficiently
manipulate such data, hiding the
details of reading and writing the
data from the programmer.
$ cat my-data.csv
1
1,2
1,2,3
1,2,3,4,
1,2,3,4,5
1,2,3,4,5,6
1,2,3,4,5,6,
1,2,,4,5,6
1,2,"3,3",4,5,6
1,"2,2",3,4,5,6
,,3,4,5
,,,4,5
,,,,5
$ python csvdrop.py
$ cat stripped-data.csv
1
1,2
1,2,3
1,2,3
1,2,3
1,2,3,6
1,2,3,6,
1,2,,6
1,2,"3,3",6
1,"2,2",3,6
,,3
,,
,,