I`m trying to cut a string (Name of a file) where I have to get a variable in the name.
But the problem is, I have to put it in a shell variable, until now it is ok.
Here is the example of what i have to do.
NAME_OF_THE_FILE_VARIABLEiWANTtoGET_DATE
NAMEfile_VARIABLEiWANT_DATE
NAME_FILE_VARIABLEiWANT_DATE
The position of the variable I want always can change, but it will be always 1 before last. The delimiter is the "_".
Is there a way to count the size of the array to get size-1 or something like that?
OBS: when i cut strings I always use things like that:
VARIABLEiWANT=`echo "$FILENAME" | cut 1 -d "_"`
awk -F'_' '{print $(NF-1)}' file
or you have a string
awk -F'_' '{print $(NF-1)}' <<< "$FILENAME"
save the output of above oneliner into your variable.
IFS=_ read -a array <<< "$FILENAME"
variable_i_want=${array[${#array[#]}-2]}
It's a bit of a mess visually, but it's more efficient than starting a new process. ${#array[#]} is the number of elements read from FILENAME, so the indices for the array range from 0 to ${#array[#]}-1.
As of bash 4.3, though, you can use a negative index instead of computing it.
variable_i_want=${array[-2]}
If you need POSIX compatibility (no arrays), then
tmp=${FILENAME%_${FILENAME##*_}} # FILENAME with last field removed
variable_i_want=${tmp##*_} # last field of tmp
Just got it... I find someone using a cat function... I got to use it with the echo... and rev. didn't understand the rev thing, but I think it revert the order of the delimiter.
CODIGO=`echo "$ARQ_NAME" | rev | cut -d "_" -f 2 | rev `
Related
I have a variable called $folder_name which contains the string
Release_2019_Config_V6_Standalone_PJ6678_Test
which is the name of a folder.
I'm trying to extract PJ6678 from the folder name.
I know the folder name will put the user id (the text I need) between the 5th and 6th underscore, I don't know what text/symbols will be present after the 6th underscore.
I'm using Bash script, i'd really appreciate the help if someone could help with this functionality as i'm completely lost trying to use sed (after reading for hours i'm assuming this is the correct tool for the job?
Here is a Bash only solution:
#!/bin/bash
INPUT="Release_2019_Config_V6_Standalone_PJ6678_Test"
IFS='_' read -ra IN <<< "$INPUT"
echo ${IN[5]}
Or use cut:
cut -d '_' -f 6 <<< "Release_2019_Config_V6_Standalone_PJ6678_Test"
Or use awk:
awk -F "_" '{ print $6 }' <<< "Release_2019_Config_V6_Standalone_PJ6678_Test"
If you want pure-bash solution, you can use tokenize the file name, and pick up the 5th element
IFS=_ read -a token <<< "$folder_name"
id=${token[5]}
Eliminating dependency and performance hit from launching additional programs per folder name.
Try this command:
echo $a | awk -F'_' '{print $6}'
Here, _ is the delimiter and $a is a variable that holds the value.
For completeness, here's a pure-shell solution that doesn't rely on bash extensions like arrays.
$ folder_name=Release_2019_Config_V6_Standalone_PJ6678_Test
$ tmp=${folder_name#*_*_*_*_*_} # Because we know how many _ to strip
$ echo ${tmp%_*}
PJ6678
Because the # operator strips the shortest prefix, this won't allow * to match any _ itself; if it did, we could shorten the prefix by making the underscore match one of the literal _ in the pattern instead.
With sed or awk, how could I match and print only organization and repo from
git#github.org.com/organization/repo.git ?
My intention is to assign both values to variables to using in something that I'm building to create local folders for organizations and repositories.
The desired result would be organization/repo only.
Currently in order to proceed i'm using cut -f2 -d':' | cut -f1 -d' but I'm ashamed of push this .
Since you need to create 2 shell variables it may be better to prefer a BASH solution.
You can use read with IFS='./'::
IFS='./:' read -ra arr <<< 'git#github.org.com/organization/repo.git'
echo "${arr[3]}/${arr[4]}"
organization/repo
IFS='./:' splits input string on DOT and /
and :colon
read -a populates an array from input text
I have some data that looks like this
"thumbnailUrl": "http://placehold.it/150/adf4e1"
I want to know how I can get the trailing part of the URL, I want the output to be
adf4e1
I was trying to grep when starting with / and ending with " but I'm only a beginner in shell scripting and need some help.
I came up with a quick and dirty solution, using grep (with perl regex) and cut:
$ cat file
"thumbnailUrl": "http://placehold.it/150/adf4e1"
"anotherUrl": "http://stackoverflow.com/questions/3979680"
"thumbnailUrl": "http://facebook.com/12f"
"randortag": "http://google.com/this/is/how/we/roll/3fk19as1"
$ cat file | grep -o '/\w*"$' | cut -d'/' -f2- | cut -d'"' -f1
adf4e1
3979680
12f
3fk19as1
We could kill this with a thousand little cuts, or just one blow from Awk:
awk -F'[/"]' '{ print $(NF-1); }'
Test:
$ echo '"thumbnailUrl": "http://placehold.it/150/adf4e1"' \
| awk -F'[/"]' '{ print $(NF-1); }'
adf4e1
Filter thorugh Awk using double quotes and slashes as field separators. This means that the trailing part ../adf4e1" is separated as {..}</>{adf4e1}<">{} where curly braces denote fields and angle brackets separators. The Awk variable NF gives the 1-based number of fields and so $NF is the last field. That's not the one we want, because it is blank; we want $(NF-1): the second last field.
"Golfed" version:
awk -F[/\"] '$0=$(NF-1)'
If the original string is coming from a larger JSON object, use something like jq to extract the value you want.
For example:
$ jq -n '{thumbnail: "http://placehold.it/150/adf4e1"}' |
> jq -r '.thumbnail|split("/")[-1]'
adf4e1
(The first command just generates a valid JSON object representing the original source of your data; the second command parses it and extracts the desired value. The split function splits the URL into an array, from which you only care about the last element.)
You can also do this purely in bash using string replacement and substring removal if you wrap your string in single quotes and assign it to a variable.
#!/bin/bash
string='"thumbnailUrl": "http://placehold.it/150/adf4e1"'
string="${string//\"}"
echo "${string##*/}"
adf4e1 #output
You can do that using 'cut' command in linux. Cut it using '/' and keep the last cut. Try it, its fun!
Refer http://www.thegeekstuff.com/2013/06/cut-command-examples
I am trying to extract "first 12 of last 24 character" from a line, i.e.,
for a line:
species,subl,cmp= 1 4 1 s1,torque= 0.41207E-09-0.45586E-13
I need to extract "0.41207E-0".
(I have not written the code, so don't curse me for its formatting. )
I have managed to do this via:
var_s=`grep "species,subl,cmp= $3 $4 $5" $tfile |sed -n '$s/.*\(........................\)$/\1/p'|sed -n '$s/\(............\).*$/\1/p'`
but, is there any more readable way of doing this, rather then counting dots?
EDIT
Thanks to both of you;
so, I have sed,awk grep and bash.
I will run that in loop, for 100's of file.
so, can you also suggest me which one is most efficient, wrt time?
One way with GNU sed (without counting dots):
$ sed -r 's/.*(.{11}).{12}/\1/' file
0.41207E-09
Similarly with GNU grep:
$ grep -Po '.{11}(?=.{12}$)' file
0.41207E-09
Perhaps a python solution may also be helpful:
python -c 'import sys;print "\n".join([a[-24:-13] for a in sys.stdin])' < file
0.41207E-09
I'm not sure your example data and question match up so just change the values in the {n} quantifier accordingly.
Simplest is using pure bash:
echo "${str:(-24):12}"
OR awk can also do that:
awk '{print substr($0, length($0)-23, 12)}' <<< $str
OUTPUT:
0.41207E-09
EDIT: For using bash solution on a file:
while read l; do echo "${l:(-24):12}"; done < file
Another one, less efficient but has the advantage of making you discover new tools
`echo "$str" | rev | cut -b 1-24 | rev | cut -b 1-12
You can use awk to get first 12 characters of last 24 characters from a line:
awk '{substr($0,(length($0)-23))};{print substr($0,(length($0)-10))}' myfile.txt
I am trying to make a very simple shell script that frees up some memory space. For this, I am using command
$ ps xopmem o%c
to output a list of running processes with its percentual memory usage. Now, I want to filter out only those lines whose value (in the first field) is larger than, say 5.0. How can I do that? I preferably use SED.
P.S.: I Googled around this topic on the Internet but found nothing. Any help will be appreciated.
sed cannot do numerical comparisons, you need to move one tool up, i.e. awk:
ps xopmem o%c | awk '$1 > 5.0'
That is, when the first field is greater than 5, execute the default block: { print $0 }.
One way. -n swith avoids printing automatically and the regexp matches any spaces at the beginning and any number bigger than 5.0. The p instruction prints lines that matches the previous regexp.
ps xopmem o%c | sed -ne '/^ *[5-9]\.[0-9]/ p'
ps xopmem o%c | sort -nr | \
while read line
do v=$(echo $line | cut -d' ' -f1);
[[ $v < 5 ]] && break;
echo $line;
done