I have a for loop written in a script like so:
for((i=0;i<${#hours[#]})); do
dates=("$(last | egrep -v "reboot|wtmp|^$" | sort | tr -s " " | sed "$i q;d" | cut -f5-7 -d' ')")
done
If I simply execute the command assigned to dates in the terminal, replacing $i (inside the sed command) by a number (0,1,2...), it returns me exactly what I want, which is, for instance Nov 15 23:15.
However, when inside the for loop, I seem to have a problem with the sed command not incrementing $i. What am I doing wrong?
Your problem is that you never change i. You should probably update it in the third part of your for statement:
for((i=0;i<${#hours[#]};++i)); do
# ^^^ here
Related
I have a variable called $dirs storing directories in a dir tree:
root/animals/rats/mice
root/animals/cats
And I have another variable called $remove for example that holds the names of the directories I want to remove from the dirs variable:
rats
crabs
I am using a for loop to do that:
for d in $remove; do
dirs=$(echo "$dirs" | sed "/\b$d\b/d")
done
After that loop is done, what I should be left with is:
root/animals/cats
because the loop found rats.
I have tested this approach on 3 systems but it only works as expected on 2.
Is there a more universal approach that would work on all shells?
You are looking for something like
echo "${dirs}" | grep -Ev "rats|crabs"
When you can't store the exclusion list in the format with |, try to change it on the fly:"
echo "${dirs}" | grep -Ev $(echo "${remove}" | tr -s "\n" "|" | sed 's/|$//')
You can use the excludeFile technique without a temp file with
echo "${dirs}" | grep -vf <(echo "${remove}")
I am not sure which of there solutions will be best supported.
Hi I'm using GNU awk version 3.1.5 and I've specified 2 variables in a KSH script
PKNAME= ls -lt /var/db/pkg | tr -s " " | cut -d" " -f9
PKDATE= ls -lt /var/db/pkg/$PKNAME/ | tr -s " " | cut -d" " -f6-8
I'm trying to prove that I'm getting the correct output, by running a test using
echo bar
awk -F, -v pkname="$PKNAME" -v pkdate="$PKDATE" 'BEGIN{print pkname, pkdate, "foo"; exit}'
echo baz
The output from this results in 2 blank spaces and foo, like so
bar
foo
baz
I have tried, double quoting the variables, single quotes and back ticks. Also tried double quotes with back ticks.
Any ideas why the variables are not being executed? I'm fairly new to awk and appreciate any help! Thanks
I suppose it is possible that it is not possible to run a sub shell comand within an awk statement. Is this true?
This has nothing to do with awk. The problem is in the way you're assigning your variables. Your lines should be like this:
PKNAME=$(ls -lt /var/db/pkg | tr -s " " | cut -d" " -f9)
There can be no spaces around either side of an assignment in the shell.
At the moment, you're running the command ls -lt ... with a variable PKNAME temporarily assigned to an empty string. In subsequent commands the variable remains unset.
Your awk command should remain unchanged, i.e. the shell variables should be passed like -v pkname="$PKNAME". As an aside, it's generally considered bad practice to use uppercase variable names, as these should be reserved for internal use by the shell.
I need to extract PID, UID and command fields from 'ps' and I have tried it like this:
ps -L u n | cut -f 1,2,13
For some reason, this behaves as there is no cut command whatsoever. It just returns normal ps output. Then, I tried
ps -L u n | tr -s " " | cut -d " " -f 1,2,13 and this returns total nonsense. Then, I tried playing with it and with this:
ps -L u n | tr -s " " | cut -d " " -f 2,3,14
and this somehow returns what I need (almost, and I don't understand why that almost works), except that it cuts out the command field in the middle of it. How can I get what I need?
ps is printing out space separators, but cut without -d uses the tab character. The tr -s squeezes the spaces together to get more of the separation that you want, but remember that there is the initial set of spaces (squeezed to one) hence why you need to add 1 to each field. Also, there are spaces in the commands for each word. This should work:
ps -L u n | tr -s " " | cut -d " " -f 2,3,14-
Is there any particular reason for using cut?
I guess this will do what you want:
ps -eopid,uid,cmd
You can use awk to clean up your command, like so:
ps -L u n | awk '{ print $1,$2,$13 }'
The question is what to do once you have a list. I find cut kludgy, so instead of cut I pass the list to a while read loop.
"While read" recognizes non-blank values on a line, so in this example, "a" is the first value, "b" is the second and "c" is the rest of the line. I am only interested in the first 2 values, process owner and process ID; and I basically abuse the case statement rather than use an "if". (Even though grep filtered, I don't want to kill processes where the owner name might be embedded elsewhere in the line)
ps -ef | grep owner | grep -v grep | while read a b c;
do
case $a in
"owner")
kill -9 $b
;;
esac;
done
I have a configuration file that contains lines like "hallo;welt;" and i want to do a grep on this file.
Whenever i try something like grep "$1;$2" my.config or echo "$1;$2 of even line="$1;$2" my script fails with something like:
: command not found95: line 155: =hallo...
How can i tell bash to ignore ; while evaluating "..." blocks?
EDIT: an example of my code.
# find entry
$line=$(grep "$1;$2;" $PERMISSIONSFILE)
# splitt line
reads=$(echo $line | cut -d';' -f3)
writes=$(echo $line | cut -d';' -f4)
admins=$(echo $line | cut -d';' -f5)
# do some stuff on the permissions
# replace old line with new line
nline="$1;$2;$reads;$writes;$admins"
sed -i "s/$line/$nline/g" $TEMPPERM
my script should be called like this: sh script "table" "a.b.*.>"
EDIT: another, simpler example
$test=$(grep "$1;$2;" temp.authorization.config)
the temp file:
table;pattern;read;write;stuff
the call sh test.sh table pattern results in: : command not foundtable;pattern;read;write;stuff
Don't use $ on the left side of an assignment in bash -- if you do it'll substitute the current value of the variable rather than assigning to it. That is, use:
test=$(grep "$1;$2;" temp.authorization.config)
instead of:
$test=$(grep "$1;$2;" temp.authorization.config)
Edit: also, variable expansions should be in double-quotes unless there's a good reason otherwise. For example, use:
reads=$(echo "$line" | cut -d';' -f3)
instead of:
reads=$(echo $line | cut -d';' -f3)
This doesn't matter for semicolons, but does matter for spaces, wildcards, and a few other things.
A ; inside quotes has no meaning at all for bash. However, if $1 contains a doublequote itself, then you'll end up with
grep "something";$2"
which'll be parsed by bash as two separate commands:
grep "something" ; other"
^---command 1----^ ^----command 2---^
Show please show exactly what your script is doing around the spot the error is occurring, and what data you're feeding into it.
Counter-example:
$ cat file.txt
hello;welt;
hello;world;
hell;welt;
$ cat xx.sh
grep "$1;$2" file.txt
$ bash -x xx.sh hello welt
+ grep 'hello;welt' file.txt
hello;welt;
$
You have not yet classified your problem accurately.
If you try to assign the result of grep to a variable (like I do) your example breaks.
Please show what you mean. Using the same data file as before and doing an assignment, this is the output I get:
$ cat xx.sh
grep "$1;$2" file.txt
output=$(grep "$1;$2" file.txt)
echo "$output"
$ bash -x xx.sh hello welt
+ grep 'hello;welt' file.txt
hello;welt;
++ grep 'hello;welt' file.txt
+ output='hello;welt;'
+ echo 'hello;welt;'
hello;welt;
$
Seems to work for me. It also demonstrates why the question needs an explicit, complete, executable, minimal example so that we can see what the questioner is doing that is different from what people answering the question think is happening.
I see you've provided some sample code:
# find entry
$line=$(grep "$1;$2;" $PERMISSIONSFILE)
# splitt line
reads=$(echo $line | cut -d';' -f3)
writes=$(echo $line | cut -d';' -f4)
admins=$(echo $line | cut -d';' -f5)
The line $line=$(grep ...) is wrong. You should omit the $ before line. Although it is syntactically correct, it means 'assign to the variable whose name is stored in $line the result of the grep command'. That is unlikely to be what you had in mind. It is, occasionally, useful. However, those occasions are few and far between, and only for people who know what they're doing and who can document accurately what they're doing.
For safety if nothing else, I would also enclose the $line values in double quotes in the echo lines. It may not strictly be necessary, but it is simple protective programming.
The changes lead to:
# find entry
line=$(grep "$1;$2;" $PERMISSIONSFILE)
# split line
reads=$( echo "$line" | cut -d';' -f3)
writes=$(echo "$line" | cut -d';' -f4)
admins=$(echo "$line" | cut -d';' -f5)
The rest of your script was fine.
It seems like you are trying to read a semicolon-delimited file, identify a line starting with 'table;pattern;' where table is a string you specify and pettern is a regular expression grep will understand. Once the line is identified you wish to replaced the 3rd, 4th and 5th fields with different data and write the updated line back to the file.
Does this sound correct?
If so, try this code
#!/bin/bash
in_table="$1"
in_pattern="$2"
file="$3"
while IFS=';' read -r -d$'\n' tuple pattern reads writes admins ; do
line=$(cut -d: -f1<<<"$tuple")
table=$(cut -d: -f2<<<"$tuple")
# do some stuff with the variables
# e.g., update the values
reads=1
writes=2
admins=12345
# replace the old line with the new line
sed -i'' -n $line'{i\
'"$table;$pattern;$reads;$writes;$admins"'
;d;}' "$file"
done < <(grep -n '^'"${in_table}"';'"${in_pattern}"';' "${file}")
I chose to update by line number here to avoid problems of unknown characters in the left hand of the substitution.
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2`
outputs something like this
"Abc Inc";
What I want to do is I want to remove the trailing ";" as well. How can i do that? I am a beginner to bash. Any thoughts or suggestions would be helpful.
This will remove the last character contained in your COMPANY_NAME var regardless if it is or not a semicolon:
echo "$COMPANY_NAME" | rev | cut -c 2- | rev
I'd use sed 's/;$//'. eg:
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2 | sed 's/;$//'`
foo="hello world"
echo ${foo%?}
hello worl
I'd use head --bytes -1, or head -c-1 for short.
COMPANY_NAME=`cat file.txt | grep "company_name" | cut -d '=' -f 2 | head --bytes -1`
head outputs only the beginning of a stream or file. Typically it counts lines, but it can be made to count characters/bytes instead. head --bytes 10 will output the first ten characters, but head --bytes -10 will output everything except the last ten.
NB: you may have issues if the final character is multi-byte, but a semi-colon isn't
I'd recommend this solution over sed or cut because
It's exactly what head was designed to do, thus less command-line options and an easier-to-read command
It saves you having to think about regular expressions, which are cool/powerful but often overkill
It saves your machine having to think about regular expressions, so will be imperceptibly faster
I believe the cleanest way to strip a single character from a string with bash is:
echo ${COMPANY_NAME:: -1}
but I haven't been able to embed the grep piece within the curly braces, so your particular task becomes a two-liner:
COMPANY_NAME=$(grep "company_name" file.txt); COMPANY_NAME=${COMPANY_NAME:: -1}
This will strip any character, semicolon or not, but can get rid of the semicolon specifically, too.
To remove ALL semicolons, wherever they may fall:
echo ${COMPANY_NAME/;/}
To remove only a semicolon at the end:
echo ${COMPANY_NAME%;}
Or, to remove multiple semicolons from the end:
echo ${COMPANY_NAME%%;}
For great detail and more on this approach, The Linux Documentation Project covers a lot of ground at http://tldp.org/LDP/abs/html/string-manipulation.html
Using sed, if you don't know what the last character actually is:
$ grep company_name file.txt | cut -d '=' -f2 | sed 's/.$//'
"Abc Inc"
Don't abuse cats. Did you know that grep can read files, too?
The canonical approach would be this:
grep "company_name" file.txt | cut -d '=' -f 2 | sed -e 's/;$//'
the smarter approach would use a single perl or awk statement, which can do filter and different transformations at once. For example something like this:
COMPANY_NAME=$( perl -ne '/company_name=(.*);/ && print $1' file.txt )
don't have to chain so many tools. Just one awk command does the job
COMPANY_NAME=$(awk -F"=" '/company_name/{gsub(/;$/,"",$2) ;print $2}' file.txt)
In Bash using only one external utility:
IFS='= ' read -r discard COMPANY_NAME <<< $(grep "company_name" file.txt)
COMPANY_NAME=${COMPANY_NAME/%?}
Assuming the quotation marks are actually part of the output, couldn't you just use the -o switch to return everything between the quote marks?
COMPANY_NAME="\"ABC Inc\";" | echo $COMPANY_NAME | grep -o "\"*.*\""
you can strip the beginnings and ends of a string by N characters using this bash construct, as someone said already
$ fred=abcdefg.rpm
$ echo ${fred:1:-4}
bcdefg
HOWEVER, this is not supported in older versions of bash.. as I discovered just now writing a script for a Red hat EL6 install process. This is the sole reason for posting here.
A hacky way to achieve this is to use sed with extended regex like this:
$ fred=abcdefg.rpm
$ echo $fred | sed -re 's/^.(.*)....$/\1/g'
bcdefg
Some refinements to answer above. To remove more than one char you add multiple question marks. For example, to remove last two chars from variable $SRC_IP_MSG, you can use:
SRC_IP_MSG=${SRC_IP_MSG%??}
cat file.txt | grep "company_name" | cut -d '=' -f 2 | cut -d ';' -f 1
I am not finding that sed 's/;$//' works. It doesn't trim anything, though I'm wondering whether it's because the character I'm trying to trim off happens to be a "$". What does work for me is sed 's/.\{1\}$//'.