i need to use variable in instead of direct date.
cat file | awk -F, '{ if ($1>"2012-08-20 11:30" && $1<"2012-08-22 16:00") print }'
thanks in advance
Based on your shown code, could you please try following and let me know if this helps you.(In lack of samples I haven't tested it)
awk -v date1="2012-08-20 11:30" -v date2="2012-08-22 16:00" -F, '($1>date1 && $1<date2)' Input_file
In case your variables are coming from shell to awk then following could help you on same, you could change date subtraction order as per your need too:
date1="2012-08-20 11:30"
date2="2012-08-22 16:00"
awk -v date_1="$date1" -v date_2="$date2" -F, '($1>date_1 && $1<date_2)' Input_file
I'm trying to get the numbers that's in between two colons and other numbers in a file.
Example:
1234:12345678:1234
1234:12345678:1234
1234:12345678:1234
I want the output to show all of the 12345678's and nothing else.
Like this:
12345678
12345678
12345678
I achieved this using:
egrep -o "[0-9]{8}" file
Problem is that I need a different solution than egrep -o (awk or sed?)
I searched and tried a couple of things but without succes.
Any help would be appreciated!
If the "number" is always 2nd column, you could do with awk:
awk -F: '{print $2}' file
For the awk solution,
awk -F: '{print $2}' file
Or simply use cut to do that,
cut -d: -f2 file
would this work?
awk -F':' '{print $2}' test > test results.txt
I have a file which contains text as follows:
Directory /home/user/ "test_user"
bunch of code
another bunch of code
How can I get from this file only the /home/user/ part?
I've managed to use awk -F '"' 'NR==1{print $1}' file.txt to get rid of rest of the file and I'm gettig output like this:
Directory /home/user/
How can I change this command to get only /home/user/ part? I'd like to make it as simple as possible. Unfortunately, I can't modify this file to add/change the content.
this should work the fastest, noticeable if your file is large
awk '{print $2; exit}' file
it will print the second field of the first line and stop processing the rest of the file.
With awk it should be:
awk 'NR==1{print $2}' file.txt
Setting the field delimiter to " was wrong Since it splits the line into these fields:
$1 = 'Directory /home/user/'
$2 = 'test_user'
$3 = '' (empty)
The default record separator, which is [[:space:]]+, splits like this:
$1 = 'Directory'
$2 = '/home/user/'
$3 = '"test_user"'
As an alternate, you can use head and cut:
$ head -n 1 file | cut -d' ' -f2
Not sure why you are using the -F" as that changes the delimiter. If you remove that, then $2 will get you what you want.
awk 'NR==1{print $2}' file.txt
You can also use awk to execute the print when the line contains /home/user instead of counting records:
awk '/\home\/user\//{print $2}' file.txt
In this case, if the line were buried in the file, or if you had multiple instances, you would get the name for every occurrence wherever it was.
Adding some grep
grep Directory file.txt|awk '{print $2}'
I have a file which contains data like below.
appid=TestApp
version=1.0.1
We want to parse the file and capture the value assigned to appid field.
I have tried with awk command as below
awk '/appid=/{print $1}' filename.txt
However it outputs the whole line
appid=TestApp
but we required only
TestApp
Please let me know how I can achieve this using awk/grep/sed shell commands.
You need to change the field separator:
awk -F'=' '$1 ~ /appid/ {print $2}' filename.txt
or with an exact match
awk -F'=' '$1 == "appid" {print $2}' filename.txt
outputs
TestApp
There's about 20 different ways to do this but it's usually a good idea when you have name = value statements in a file to simply build an array of those assignments and then just print whatever you care about using it's name, e.g.:
$ cat file
appid=TestApp
version=1.0.1
$
$ awk -F= '{a[$1]=$2} END{print a["appid"]}' file
TestApp
$ awk -F= '{a[$1]=$2} END{print a["version"]}' file
1.0.1
$ awk -F= '{a[$1]=$2} END{for (i in a) print i,"=",a[i]}' file
appid = TestApp
version = 1.0.1
If you are in the shell already then simply sourcing the file will let you get what you want.
. filename.txt
echo $appid
I have data in the text file like val1,val2 with multiple lines
and I want to change it to 1,val1,val2,0,0,1
I tried with print statement in awk(solaris) to add constants by it didn't work.
What is the correct way to do it ?
(From the comments) This is what I tried
awk -F, '{print "%s","1,"$1","$2"0,0,1"}' test.txt
Based on the command you posted, a little change makes it:
$ awk -F, 'BEGIN{OFS=FS} {print 1,$1,$2,0,0,1}' file
1,val1,val2,0,0,1
OR using printf (I prefer print):
$ awk -F, '{printf "1,%s,%s,0,0,1", $1, $2}' file
1,val1,val2,0,0,1
To prepend every line with the constant 1 and append with 0,0,1 simply do:
$ awk '{print 1,$0,0,0,1}' OFS=, file
1,val1,val2,0,0,1
A idiomatic way would be:
$ awk '$0="1,"$0",0,0,1"' file
1,val1,val2,0,0,1
Using sed:
sed 's/.*/1,&,0,0,1/' inputfile
Example:
$ echo val1,val2 | sed 's/.*/1,&,0,0,1/'
1,val1,val2,0,0,1