I am trying to uncomment "/usr/lib/sa/sa1" or "/usr/lib/sa/sa2" entries from one of the configuration file.
Below is the regex:
"^[0-9].*/usr/lib/sa/sa(1|2)"
Eg:
#0 * * * * /usr/lib/sa/sa1 1200 3 &
#5 23 * * * /usr/lib/sa/sa2 -s 0:00 -e 23:01 -i 3600 -ubcwyaqvm &
Output should be:
0 * * * * /usr/lib/sa/sa1 1200 3 &
5 23 * * * /usr/lib/sa/sa2 -s 0:00 -e 23:01 -i 3600 -ubcwyaqvm &
Tried the below "sed" command but no luck:
sed -e '/^#.*\/usr\/lib\/sa\/sa\(1\|2\)/s/^#//g' adm
Can you please let me know where am I going wrong here.
Thanks in advance!
You can use
sed '/^#.*\/usr\/lib\/sa\/sa[12]/s/^#//'
Here, no grouping construct is used, no alternation operator is used either. It is possible as the 1 and 2 are single char alternatives, and bracket expressions are supported in all versions of sed.
Also, see this online demo.
Related
I have a cron job with below details
The timing section is unknown, it might be hourly or daily.
cat zyx.txt
SHELL=/bin/bash
PATH=/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin
0 * * * * root /usr/local/bin/log-cleaner
I have to update it on the fly using sed, so that it should run every minute .
Here is what I have tried
sed -in 's/\(.*\) \(.*\) \(.*\)/\* \* \* \* \*/1' zyx.txt
But it is removing the user and path section
cat zyx.txt
SHELL=/bin/bash
PATH=/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin
* * * * *
Any help / suggestion using sed is much appreciated
How can I remove useless ".0" strings in a txt file of numbers?
If I have this file:
43.65 24.0 889.5 5.0
32.14 32.0 900.0 6.0
38.27 43.0 899.4 5.0
I want to get:
43.65 24 889.5 5
32.14 32 900 6
38.27 43 899.4 5
I tried: sed 's|\.0 | |g' but that obviously does not work with new lines and EOF.
Any suggestion without getting into python, etc?
This might work for you (GNU sed):
sed 's/\.0\b//g' file
Or
sed 's/\.0\>//g' file
Remove any period followed by a zero followed by a word boundary.
You can use
sed -E 's,([0-9])\.0($| ),\1\2,g' file
Details:
-E - enables POSIX ERE syntax
([0-9])\.0($| ) - finds and captures into Group 1 a digit, then matches .0, and then matches and captures into Group 2 a space or end of string
\1\2 - replaces with Group 1 + Group 2 concatenated values
g - replaces all occurrences.
See the online demo:
s='43.65 24.0 889.5 5.0
32.14 32.0 900.0 6.0
38.27 43.0 899.4 5.0'
sed -E 's,([0-9])\.0($| ),\1\2,g' <<< "$s"
Output:
43.65 24 889.5 5
32.14 32 900 6
38.27 43 899.4 5
So i have simple shell commands to ping websites to retrieve data about said websites.
For example one of my pinging.sh looks like this:
ping -R -c 120 blar.org.cn >> pingdata.txt
ping -R -c 120 another.net >> pingdata.txt
But then my crontabs look like this:
7 * * * ./pinging.sh >> pingdata.log
The pingdata.log doesn't output. Is it best to do it through the crontab or through the script? I assumed the crontab would be better because it would cover the entire script rather than having to write it out multiple times.
You need to indicate the full path of your script in the cronjob, together with the binary running it.
For example:
7 * * * * /bin/sh /home/you/pinging.sh >> /home/you/pingdata.log
Note also you are just adding 4 parameters to the cronjob, whereas you need at least 5:
+---------------- minute (0 - 59)
| +------------- hour (0 - 23)
| | +---------- day of month (1 - 31)
| | | +------- month (1 - 12)
| | | | +---- day of week (0 - 6) (Sunday=0 or 7)
| | | | |
* * * * * command to be executed
You can test your cron syntax with Crontab guru (---> http://crontab.guru/).
First, the executable must be provided as full path in cron.
Example:
7 * * * * /bin/bash /path/to/pinging.sh
Second, create a wrapper script for pinging.sh >> pingdata.log and add that to crontab.
Third, your crontab entry is wrong. There must be 5 fields whereas your's have 4 (maybe that's a typo ?)
I've been trying to pull a field from a row in a file although each row may have plus or minus 2 or 3 fields per row. They aren't always equal in the number of fields per row.
Here is a snippet:
A orarpp 45286124 1 1 0 20 60 Nov 25 9-16:42:32 01:04:58 11176 117056 0 - oracleXXX (LOCAL=NO)
A orarpp 45351560 1 1 3 20 61 Nov 30 5-03:54:42 02:24:48 4804 110684 0 - ora_w002_XXX
A orarpp 45548236 1 1 22 20 71 Nov 26 8-19:36:28 00:56:18 10628 116508 0 - oracleXXX (LOCAL=NO)
A orarpp 45679190 1 1 0 20 60 Nov 28 6-23:42:20 00:37:59 10232 116112 0 - oracleXXX (LOCAL=NO)
A orarpp 45744808 1 1 0 20 60 10:52:19 23:08:12 00:04:58 11740 117620 0 - oracleXXX (LOCAL=NO)
A root 45810380 1 1 0 -- 39 Nov 25 9-19:54:34 00:00:00 448 448 0 - garbage
In the case of the first line, I'm interested in 9-16:42:32 and the similar fields for each row.
I've tried to pull it by using ':' as the field separator and then filter from there however, what I am trying to accomplish is to do something if the number before the dash (in the example it's 9) is greater than one.
cat file.txt | grep oracle | awk -F: '{print substr($1, length($1)-5)}'
This is because the number of fields on either side of the actual field I need can be different from line to line.
Definitely not the most efficient but I've been trying to do this with an awk one liner.
Hints or a direction would be appreciated to get me moving again. I am not opposed to doing in a better way than awk.
Thanks.
Maybe cut is the right tool for this job? For example, with your snippet:
$ cut -c 62-71 file.txt
9-16:42:32
5-03:54:42
8-19:36:28
6-23:42:20
23:08:12
9-19:54:34
The arguments tell cut to snip columns (-c) 62 through 71.
For additional processing, you can pipe it to awk.
You can also accomplish the whole thing in awk by accepting entire lines and then using substr to extract the columns you want. For example, this awk command produces the same output as the cut command above:
awk '{ print substr($0, 62, 10) }' file.txt
Whether you create a pipeline or do the processing entirely in awk is at least in part a matter of personal taste / style.
Would this do?
awk -F: '/oracle/ {print substr($0,62,10)}' file.txt
9-16:42:32
8-19:36:28
6-23:42:20
23:08:12
This search for oracle and then print 10 characters starting from position 62
You can grab those identifiers with one of
grep -o '[[:digit:]]\+-[[:digit:]]\{2\}:[[:digit:]]\{2\}:[[:digit:]]\{2\}'
grep -oP '\d+-\d\d:\d\d:\d\d' # GNU grep
It sounds like you want to do something with the lines, not just find the ids. Please elaborate.
Using GNU awk:
gawk --re-interval '
/oracle/ && \
match($0, /([[:digit:]]+)-([[:digit:]]{2}:){2}[[:digit:]]{2}/, a) && \
a[1]>1 {
# do something with the matching line
print
}
' file
Q1: Sed specify the whole line and if the line is nothing but the string then delete
I have a file that contains several of the following numbers:
1 1
3 1
12 1
1 12
25 24
23 24
I want to delete numbers that are the same in each line. For that I have either been using:
sed '/1 1/d' < old.file > new.file
OR
sed -n '/1 1/!p' < old.file > new.file
Here is the main problem. If I search for pattern '1 1' that means I get rid of '1 12' as well. So for I want the pattern to specify the whole line and if it does, to delete it.
Q2: Automation of question 1
I am also trying to automate this problem. The range of numbers in the first column and the second column could be from 1 to 25.
So far this is what I got:
for ((i=1;i<26;i++)); do
sed "/'$i' '$i'/d" < oldfile > newfile; mv newfile oldfile;
done
This does nothing to the oldfile in the end. :(
This would be more readable with awk:
awk '$1 == $2 {next} {print}' oldfile > newfile
Update based on comment:
If the requirement is to remove lines where the two values are within 1 of each other:
awk '{d = $1-$2; if (-1 <= d && d <= 1) next; else print}' oldfile
Unfortunately, awk does not have abs() (at least nawk and gawk don't)
Just put the first number in a group (\([0-9]*\)) and then look for it with a backreference (\1). Since the line to delete should contain only the group, repeated, use the ^ to mark the beginning of line and the $ to mark the end of line. For example, for the following file:
$ cat input
1 1
3 1
12 1
1 12
12 12
12 13
13 13
25 24
23 24
...the result is:
$ sed '/^\([0-9]*\) \1$/d' input
3 1
12 1
1 12
12 13
25 24
23 24
You can also do it with grep:
grep -E -v "([0-9])*\s\1" testfile
Look for multiple digits in a row and remember them, followed by a single whitespace, followed by whatever digits you remembered.