I have a script that has uses cut to pick some information out of a search that contains absolute paths. It looks something like:
PLACE=$(grep foo flatfile.txt | cut -d '/' -f 1-6)
The output looks like this:
machine1:/path/to/where/foo/is
machine2:/another/path/to/find/foo
I need it to look like this:
machine1:/path/to/where/foo/is/
machine2:/another/path/to/find/foo/
This needs to be printed to the console at the end of the script with echo "$PLACE" or something like that. The output will always be at least 2 lines, but usually more.
I tried about everything I can think of with echo, but it either shows no output at all or gives the output:
grep: '/' is a directory
I am running bash 3.00 on Solaris, if that helps any. I would really like to K.I.S.S. this by just having something tacked onto the end of the cut command, and not having to monkey with sed or awk. But, if that is the only way, so be it.
try this :
PLACE=$(grep foo flatfile.txt | cut -d '/' -f 1-6 | xargs -I "%" echo %/)
Try the following. Note that $PLACE is not quoted. This should allow word-splitting to happen and each word (which corresponds to one line of your output, assuming no embedded spaces) is then printed by printf with a following / and newline. Haven't been able to test this on Bash 3 on Solaris though
printf "%s/\n" $PLACE
You can use one sed command, instead of multiple other commands:
$ PLACE=$(sed -n '/foo/s:\(\([^/]\+/\)\{6\}\)\(.*\)\+:\1:p' flatfile.txt)
$ echo "$PLACE"
machine1:/path/to/where/foo/is/
machine2:/another/path/to/find/foo/
$
... and then I'm reminded of why sed sometimes makes me shudder :-/
Related
I have a shell script that contains the following line
PROC_ID=$(cat myfile.sh | grep running)
which, after you echo out the value would be 1234 or something like that.
What I want to do is find and replace instances of this line with a literal value
I want to replace it with PROC_ID=1234 instead of having the function call.
I've tried doing this in another shell script using sed but I can't get it to work
STR_TO_USE="PROC_ID=${1}"
STR_TO_REP='PROC_ID=$(cat myfile.sh | grep running)'
sed -i "s/$STR_TO_REP/$STR_TO_USE/g" sample.sh
but it complains stating sed: 1: "sample.sh": unterminated substitute pattern
How can I achieve this?
EDIT:
sample.sh should contain beforehand
#!/bin/bash
....
PROC_ID=$(cat myfile.sh | grep running)
echo $PROC_ID
....
After, it should contain
#!/bin/bash
....
PROC_ID=1234
echo $PROC_ID
....
The script I'm using as described above will be taking the in an arg from the command line, hence STR_TO_USE="PROC_ID=${1}"
Simply:
sed /^PROC_ID=/s/=.*/=1234/
Translation:
At line begining by PROC_ID=
replace = to end of line by =1234.
or more accurate
sed '/^[ \o11]*PROC_ID=.*myfile.*running/s/=.*/=1234/'
could be enough
([ \o11]* mean some spaces and or tabs could even prepand)
Well, first, I want to point out something obvious. This: $(cat myfile.sh | grep running) will at the very least NOT only contain the string 1234 but will certainly also contain the string running. But since you aren't asking for help with that, I'll leave it alone.
All you need in your above sed, is first to backslash the $.
STR_TO_REP='PROC_ID=\$(cat myfile.sh | grep running)'
This allows the sed command to be terminated.
We use some package called Autosys and there are some specific commands of this package. I have a list of variables which i like to pass in one of the Autosys commands as variables one by one.
For example one such variable is var1, using this var1 i would like to launch a command something like this
autosys_showJobHistory.sh var1
Now when I launch the below written command, it gives me the desired output.
echo "var1" | while read line; do autosys_showJobHistory.sh $line | grep 1[1..6]:[0..9][0..9] | grep 24.12.2012 | tail -1 ; done
But if i put the var1 in a file say Test.txt and launch the same command using cat, it gives me nothing. I have the impression that command autosys_showJobHistory.sh does not work in that case.
cat Test.txt | while read line; do autosys_showJobHistory.sh $line | grep 1[1..6]:[0..9][0..9] | grep 24.12.2012 | tail -1 ; done
What I am doing wrong in the second command ?
Wrote all of below, and then noticed your grep statement.
Recall that ksh doesn't support .. as an indicator for 'expand this range of values'. (I assume that's your intent). It's also made ambiguous by your lack of quoting arguments to grep. If you were using syntax that the shell would convert, then you wouldn't really know what reg-exp is being sent to grep. Always better to quote argments, unless you know for sure that you need the unquoted values. Try rewriting as
grep '1[1-6]:[0-9][0-9]' | grep '24.12.2012'
Also, are you deliberately using the 'match any char' operator '.' OR do you want to only match a period char? If you want to only match a period, then you need to escape it like \..
Finally, if any of your files you're processing have been created on a windows machine and then transfered to Unix/Linux, very likely that the line endings (Ctrl-MCtrl-J) (\r\n) are causing you problems. Cleanup your PC based files (or anything that was sent via ftp) with dos2unix file [file2 ...].
If the above doesn't help, You'll have to "divide and conquer" to debug your problem.
When I did the following tests, I got the expected output
$ echo "var1" | while read line ; do print "line=${line}" ; done
line=var1
$ vi Test.txt
$ cat Test.txt
var1
$ cat Test.txt | while read line ; do print "line=${line}" ; done
line=var1
Unrelated to your question, but certain to cause comment is your use of the cat commnad in this context, which will bring you the UUOC award. That can be rewritten as
while read line ; do print "line=${line}" ; done < Test.txt
But to solve your problem, now turn on the shell debugging/trace options, either by changing the top line of the script (the shebang line) like
#!/bin/ksh -vx
Or by using a matched pair to track the status on just these lines, i.e.
set -vx
while read line; do
print -u2 -- "#dbg: Line=${line}XX"
autosys_showJobHistory.sh $line \
| grep 1[1..6]:[0..9][0..9] \
| grep 24.12.2012 \
| tail -1
done < Test.txt
set +vx
I've added an extra debug step, the print -u2 -- .... (u2=stderror, -- closes option processing for print)
Now you can make sure no extra space or tab chars are creeping in, by looking at that output.
They shouldn't matter, as you have left your $line unquoted. As part of your testing, I'd recommend quoting it like "${line}".
Then I'd comment out the tail and the grep lines. You want to see what step is causing this to break, right? So does the autosys_script by itself still produce the intermediate output you're expecting? Then does autosys + 1 grep produce out as expected, +2 greps, + tail? You should be able to easily see where you're loosing your output.
IHTH
I am writing shell script for embedded Linux in a small industrial box. I have a variable containing the text pid: 1234 and I want to strip first X characters from the line, so only 1234 stays. I have more variables I need to "clean", so I need to cut away X first characters and ${string:5} doesn't work for some reason in my system.
The only thing the box seems to have is sed.
I am trying to make the following to work:
result=$(echo "$pid" | sed 's/^.\{4\}//g')
Any ideas?
The following should work:
var="pid: 1234"
var=${var:5}
Are you sure bash is the shell executing your script?
Even the POSIX-compliant
var=${var#?????}
would be preferable to using an external process, although this requires you to hard-code the 5 in the form of a fixed-length pattern.
Here's a concise method to cut the first X characters using cut(1). This example removes the first 4 characters by cutting a substring starting with 5th character.
echo "$pid" | cut -c 5-
Use the -r option ("use extended regular expressions in the script") to sed in order to use the {n} syntax:
$ echo 'pid: 1234'| sed -r 's/^.{5}//'
1234
Cut first two characters from string:
$ string="1234567890"; echo "${string:2}"
34567890
pipe it through awk '{print substr($0,42)}' where 42 is one more than the number of characters to drop. For example:
$ echo abcde| awk '{print substr($0,2)}'
bcde
$
Chances are, you'll have cut as well. If so:
[me#home]$ echo "pid: 1234" | cut -d" " -f2
1234
Well, there have been solutions here with sed, awk, cut and using bash syntax. I just want to throw in another POSIX conform variant:
$ echo "pid: 1234" | tail -c +6
1234
-c tells tail at which byte offset to start, counting from the end of the input data, yet if the the number starts with a + sign, it is from the beginning of the input data to the end.
Another way, using cut instead of sed.
result=`echo $pid | cut -c 5-`
I found the answer in pure sed supplied by this question (admittedly, posted after this question was posted). This does exactly what you asked, solely in sed:
result=\`echo "$pid" | sed '/./ { s/pid:\ //g; }'\``
The dot in sed '/./) is whatever you want to match. Your question is exactly what I was attempting to, except in my case I wanted to match a specific line in a file and then uncomment it. In my case it was:
# Uncomment a line (edit the file in-place):
sed -i '/#\ COMMENTED_LINE_TO_MATCH/ { s/#\ //g; }' /path/to/target/file
The -i after sed is to edit the file in place (remove this switch if you want to test your matching expression prior to editing the file).
(I posted this because I wanted to do this entirely with sed as this question asked and none of the previous answered solved that problem.)
Rather than removing n characters from the start, perhaps you could just extract the digits directly. Like so...
$ echo "pid: 1234" | grep -Po "\d+"
This may be a more robust solution, and seems more intuitive.
This will do the job too:
echo "$pid"|awk '{print $2}'
Sometimes I receive a CSV file which has a carriage return inside a cell. This is not an acceptable format to a program that will use it as input.
In order to detect if an input line is split, I determined that a bad line would not have the expected number of commas in it. Is there a bash or other common unix command line tool that would allow me to count the commas in the line? If necessary, I can write a Python or Perl program to do it, but if possible, I'd like to add a line or two to an existing bash script to cause it to fail if the comma count is wrong. Any ideas?
Strip everything but the commas, and then count number of characters left:
$ echo foo,bar,baz | tr -cd , | wc -c
2
To count the number of times a comma appears, you can use something like awk:
string=(line of input from CSV file)
echo "$string" | awk -F "," '{print NF-1}'
But this really isn't sufficient to determine whether a field has carriage returns in it. Fields can have commas inside as long as they're surrounded by quotes.
What worked for me better than the other solutions was this. If test.txt has:
foo,bar,baz
baz,foo,foobar,bar
Then cat test.txt | xargs -I % sh -c 'echo % | tr -cd , | wc -c' produces
2
3
This works very well for streaming sources, or tailing logs, etc.
In pure Bash:
while IFS=, read -ra array
do
echo "$((${#array[#]} - 1))"
done < inputfile
or
while read -r line
do
count=${line//[^,]}
echo "${#count}"
done < inputfile
Try Perl:
$ perl -ne 'print 0+#{[/,/g]},"\n"'
a
0
a,a
1
a,a,a,a,a
4
Depending on what you are trying to do with the CSV data, it may be helpful to use a wrapper script like csvquote to temporarily replace the problematic newlines (and commas) inside quoted fields, then restore them. For instance:
csvquote inputfile.csv | wc -l
and
csvquote inputfile.csv | cut -d, -f1 | csvquote -u
may be the sort of thing you're looking for. See [https://github.com/dbro/csvquote][1] for the code and more information
An example Python command you could run (since it's going to be installed on most modern shells) is:
python -c "import pathlib; print({l.count(',') for l in pathlib.Path('my_file.csv').read_text().splitlines()})"
This counts the number of commas per line, then makes a set from them (so if your lines all have the same number of commas in, you'll get a set with just that number in).
Just remove all of the carriage returns:
tr -d "\r" old_file > new_file
[Editorial insertion: Possible duplicate of the same poster's earlier question?]
Hi, I need to extract from the file:
first
second
third
using the grep command, the following line:
second
third
How should the grep command look like?
Instead of grep, you can use pcregrep which supports multiline patterns
pcregrep -M 'second\nthird' file
-M allows the pattern to match more than one line.
Your question abstract "bash grep newline", implies that you would want to match on the second\nthird sequence of characters - i.e. something containing newline within it.
Since the grep works on "lines" and these two are different lines, you would not be able to match it this way.
So, I'd split it into several tasks:
you match the line that contains "second" and output the line that has matched and the subsequent line:
grep -A 1 "second" testfile
you translate every other newline into the sequence that is guaranteed not to occur in the input. I think the simplest way to do that would be using perl:
perl -npe '$x=1-$x; s/\n/##UnUsedSequence##/ if $x;'
you do a grep on these lines, this time searching for string ##UnUsedSequence##third:
grep "##UnUsedSequence##third"
you unwrap the unused sequences back into the newlines, sed might be the simplest:
sed -e 's/##UnUsedSequence##/\n'
So the resulting pipe command to do what you want would look like:
grep -A 1 "second" testfile | perl -npe '$x=1-$x; s/\n/##UnUsedSequence##/ if $x;' | grep "##UnUsedSequence##third" | sed -e 's/##UnUsedSequence##/\n/'
Not the most elegant by far, but should work. I'm curious to know of better approaches, though - there should be some.
I don't think grep is the way to go on this.
If you just want to strip the first line from any file (to generalize your question), I would use sed instead.
sed '1d' INPUT_FILE_NAME
This will send the contents of the file to standard output with the first line deleted.
Then you can redirect the standard output to another file to capture the results.
sed '1d' INPUT_FILE_NAME > OUTPUT_FILE_NAME
That should do it.
If you have to use grep and just don't want to display the line with first on it, then try this:
grep -v first INPUT_FILE_NAME
By passing the -v switch, you are telling grep to show you everything but the expression that you are passing. In effect show me everything but the line(s) with first in them.
However, the downside is that a file with multiple first's in it will not show those other lines either and may not be the behavior that you are expecting.
To shunt the results into a new file, try this:
grep -v first INPUT_FILE_NAME > OUTPUT_FILE_NAME
Hope this helps.
I don't really understand what do you want to match. I would not use grep, but one of the following:
tail -2 file # to get last two lines
head -n +2 file # to get all but first line
sed -e '2,3p;d' file # to get lines from second to third
(not sure how standard it is, it works in GNU tools for sure)
So you just don't want the line containing "first"? -v inverts the grep results.
$ echo -e "first\nsecond\nthird\n" | grep -v first
second
third
Line? Or lines?
Try
grep -E -e '(second|third)' filename
Edit: grep is line oriented. you're going to have to use either Perl, sed or awk to perform the pattern match across lines.
BTW -E tell grep that the regexp is extended RE.
grep -A1 "second" | grep -B1 "third" works nicely, and if you have multiple matches it will even get rid of the original -- match delimiter
grep -E '(second|third)' /path/to/file
egrep -w 'second|third' /path/to/file
you could use
$ grep -1 third filename
this will print a string with match and one string before and after. Since "third" is in the last string you get last two strings.
I like notnoop's answer, but building on AndrewY's answer (which is better for those without pcregrep, but way too complicated), you can just do:
RESULT=`grep -A1 -s -m1 '^\s*second\s*$' file | grep -s -B1 -m1 '^\s*third\s*$'`
grep -v '^first' filename
Where the -v flag inverts the match.