This image is the /etc/passwd fileLinux: What single sed command can be used to output matches from /etc/passwd that have Smith or Jones in their description (5th field) to a file called smith_jones.txt?
I wouldn't use sed, but it looks like you're referencing a standard /etc/passwd file, so something that may do what you're looking for is this:
cat /etc/passwd | awk -F ":" '{if ($5 ~ /Smith/ || $5 ~ /Jones/) print}'
So awk '{print $5}' is commonly used to print the 5th column of something piped to it, in this case the /etc/passwd file. However, as it's not tabular data, I've supplied -F argument with the delimiter ":" as that's what splits our values.
It's then a fairly easy if statement essentially saying, if this string contains Smith OR Jones in it somewhere, print it.
Related
Lets say i have a file called test and in this file contains some data:
jon:TX:34:red,green,yellow,black,orange
I'm trying to make it so it will only print the 4th field up until the comma and nothing else. But I need to leave the current FS in place because the fields are separated by the ":". Hope this makes sense.
I have been running this command:
awk '{ FS=":"; print $4 }' /test
I want my output to look like this.
jon:TX:34:red
or if you could even just figure out how i could just print the 4th field would be a good help too
red
It's overkill for your needs but in general to print the yth ,-separated subfield of the xth :-separated field of any input would be:
$ awk -F':' -v s=',' -v x=4 -v y=1 '{split($x,a,s); print a[y]}' file
red
Or
awk -F '[:,]' '{print $4}' test
output
red
It sounds like you are trying to extract the first field of the fourth field. Top level fields are delimited by ":" and the nested field is delimited by ",".
Combining two cut processes achieves this easily:
<input.txt cut -d: -f4 | cut -d, -f1
If you want all fields until the first comma, extract the first comma-delimited field without first cutting on colon:
cut -d, -f1 input.txt
if you want a purely regex approach :
echo 'jon:TX:34:red,green,yellow,black,orange' |
mawk NF=NF FS='.+:|,.+' OFS=
red
if you only want "red" without the trailing newline ("\n"), use RS/ORS instead of FS/OFS — (the % is the command prompt, i.e. no trailing \n):
mawk2 8 RS='.+:|,.+' ORS=
red%
if u wanna hard-code in the $4 :
gawk '$_= $4' FS=,\|: # gawk or nawk
mawk '$!NF=$4' FS=,\|: # any mawk
red
and if you only want the non-numeric text :
nawk NF=NF FS='[!-<]+' OFS='\f\b'
jon
TX
red
green
yellow
black
orange
If you have
jon:TX:34:red,green,yellow,black,orange
and desired output is
jon:TX:34:red
then just treat input as comma-separated and get 1st field, which might be expressed in GNU AWK as
echo "jon:TX:34:red,green,yellow,black,orange" | awk 'BEGIN{FS=","}{print $1}'
gives output
jon:TX:34:red
Explanation: I inform GNU AWK that , character is field separator (FS), for each line I print 1st column ($1)
(tested in GNU Awk 5.0.1)
I need to grep some pattern and further i need to print some output within that. Currently I am using the below command which is working fine. But I like to eliminate using multiple pipe and want to use single awk command to achieve the same output. Is there a way to do it using awk?
root#Server1 # cat file
Jenny:Mon,Tue,Wed:Morning
David:Thu,Fri,Sat:Evening
root#Server1 # awk '/Jenny/ {print $0}' file | awk -F ":" '{ print $2 }' | awk -F "," '{ print $1 }'
Mon
I want to get this output using single awk command. Any help?
You can try something like:
awk -F: '/Jenny/ {split($2,a,","); print a[1]}' file
Try this
awk -F'[:,]+' '/Jenny/{print $2}' file.txt
It is using muliple -F value inside the [ ]
The + means one or more since it is treated as a regex.
For this particular job, I find grep to be slightly more robust.
Unless your company has a policy not to hire people named Eve.
(Try it out if you don't understand.)
grep -oP '^[^:]*Jenny[^:]*:\K[^,:]+' file
Or to do a whole-word match:
grep -oP '^[^:]*\bJenny\b[^:]*:\K[^,:]+' file
Or when you are confident that "Jenny" is the full name:
grep -oP '^Jenny:\K[^,:]+' file
Output:
Mon
Explanation:
The stuff up until \K speaks for itself: it selects the line(s) with the desired name.
[^,:]+ captures the day of week (in this case Mon).
\K cuts off everything preceding Mon.
-o cuts off anything following Mon.
file1
ABY37499|ANK37528|DEL37508|SRILANKA|195203230000|445500759
ARJU7499|CHA38008|DEL37508|SRILANKA|195203230000|445500759
IB1704174|ANK37528|DEL37508|SRILANKA|195203230000|445500759
IB1704174|CHA38008|DEL37508|SRILANKA|195203230000|445500759
ABY37500|ANK37529|DEL37509|BRAZIL|195203240000|445500757
ARJU7500|CHA38009|DEL37509|BRAZIL|195203240000|445500757
IB1704175|ANK37529|DEL37509|BRAZIL|195203240000|445500757
i want to convert the fifth column date to another format script below
#!/bin/sh
dt="%Y-%m-%d %H:%M"
awk -F '|' '{print $5}' file1 | sed 's/.\{8\}/& /g'> f1.txt
aa=`(date -f f1.txt +"$dt")`
echo "$aa"
awk -F '|' '$5=$aa' file1
echo "$aa" got desired output but i cannot assign $aa to $5 please help me.
Thanks
I corrected my answer after the commento of Etan Reisner
from AWK man:
The input is read in units called records, and processed by the rules
of your program one record at a time. By default, each record is one
line. Each record is automatically split into chunks called fields.
This makes it more convenient for programs to work on the parts of a
record.
Fields are stored in variables $1, $2, ...
And
The contents of a field, as seen by awk, can be changed within an awk
program; this changes what awk perceives as the current input record.
see the man page
thus, this expression:
awk -F '|' '$5=$aa' file1
does not have the effect of substitute the fifth column of file1.
You have to write the modified output to a second file.
May be this could help you in sed
echo 195203240000 | sed -n -e "s_\(....\)\(..\)\(..\)\(..\)\(..\)_\1-\2-\3 \4:\5_p"
1952-03-24 00:00
This awk script should do what you want.
It isn't exactly pretty but it works assuming the input format is consistent.
awk '{$5=sprintf("%s-%s-%s %s:%s\n",
substr($5,1,4), substr($5,5,2), substr($5,7,2),
substr($5,9,2), substr($5,11,2))} 7' file1 > file1.new
It assigns the new value for the field to $5 and then uses 7 (as a truth-y value) to get the default awk {print} action to print the modified line.
I have an input file that looks like this:
"LEVEL1","cn=APP_GROUP_ABC,ou=dept,dc=net","uid=A123456,ou=person,dc=net"
"LEVEL1","cn=APP_GROUP_DEF,ou=dept,dc=net","uid=A123456,ou=person,dc=net"
"LEVEL1","cn=APP_GROUP_ABC,ou=dept,dc=net","uid=A567890,ou=person,dc=net"
I want to read each line, parse and then output like this:
A123456,ABC
A123456,DEF
A567890,ABC
In other words, retrieve the user id from "uid=" and then the identifier from "cn=APP_GROUP_". Repeat for each input record, writing to a new output file.
Note that the column positions aren't fixed, so can't rely on positions, guessing I have to search for the "uid=" string and somehow use the position maybe?
Any help much appreciated.
You can do this easily with sed:
sed 's/.*cn=APP_GROUP_\([^,]*\).*uid=\([^,]*\).*/\2,\1/'
The regex captures the two desired strings, and outputs them in reverse order with a comma between them. You might need to change the context of the captures, depending on the precise nature of your data, because the uid= will match the last uid= in the line, if there are more than one.
You can use awk to split in columns, split by ',' and then split by =, and grab the result. You can do it easily as awk -F, '{ print $5}' | awk -F= '{print $2}'
Take a look at this line looking at the example you provided:
cat file | awk -F, '{ print $5}' | awk -F= '{print $2}'
A123456
A123456
A567890
This is an example of a filename.csv file
"Sort Order","Common Name","Formal Name","Type","Sub Type"
"1","Afghanistan","Islamic State of Afghanistan","Independent State"
"2","Albania","Republic of Albania","Independent State"
"3","Algeria","People's Democratic Republic of Algeria","Independent State"
"4","Andorra","Principality of Andorra","Independent State"
"5","Angola","Republic of Angola","Independent State"
So what is the grep command to search for angola in common name and print it like this:
"5","Angola","Republic of Angola","Independent State"
I know that we can use:
grep '"Algeria"' filename.csv
However, what if I am being more specific. Let's say Algeria might exist in other column however, we only need to print the one in Common Name column.
I tried
grep '"Common Name | Algeria"' filename.csv
Seems not to work.
You could try the below grep command to print the lines which contains the string Angola in the second column.
grep -E '^"[^"]*","Angola"' file
This could be easily done through awk,
awk -F, '$2=="\"Angola\""' file
try awk
awk -F"," '$2~/Algeria/' file
Use cut command,
grep Algeria filename.csv|cut -f2 -d','|awk -F '[""]' '{print $2}'
Output:
Algeria
Here is the explanation of all command I have added with your grep and discarded some unnecessary quotes you used in it.
cut command:
-f It will extract 2nd column depending on delimiter
-d It will choose delimiter i.e.: which is here ,
awk command:
It will extract value between two "" (block quotes) and print it.