Bash - Remove upper and lower lines from file using variables [duplicate] - bash

This question already has answers here:
Difference between single and double quotes in Bash
(7 answers)
Closed 1 year ago.
I have to build a bash script that removes upper and lower lines from a file.
The script ask the person for a word, searches the word in the file and removes 4 upper lines, 9 lower lines and the line that contains the word.
Bellow are the commands that works:
vi -e -c 'g/word/.-4,-d' -c 'wq' fileName
sed -i '/word/,+9d' fileName
The problem is that I want to ask the user for a word that I will use as a variable to do all that.
The bellow code doesn't work!
#!/bin/bash
read -p "Insert the word:" word
vi -e -c 'g/$word/.-4,-d' -c 'wq' fileName
sed -i '/$word/,+9d' fileName
What should I do to solve it?

something like this will work, assumes the search word appears only once in the file
$ awk -v b=4 -v a=9 -v word="$word" 'NR==FNR{if($0~word) n=NR; next}
FNR<n-b || FNR>n+a' file{,}
double scans the file, first to find the line number of the match, second round prints out the lines based on before/after context set.
Also doesn't handle if the search word not found. If you want to print everything add !n || to the last condition

You can do it one shot with vi or ed, no need to use sed. Variables inside single quotes won't be interpolated, you need double quotes for that. See Difference between single and double quotes in Bash for details.
vi -e -c 'g/'"$word"'/.-4,+9d' -c 'wq' fileName
printf '/%s/\n-4,+9d\nwq\n' "$word" | ed -s fileName > /dev/null
I don't know ed enough to prevent the line matching the given word from being printed, hence the use of /dev/null

Related

How to insert dynamic matching pattern in bash? [duplicate]

This question already has answers here:
Difference between single and double quotes in Bash
(7 answers)
Closed 3 years ago.
I was looking at this link Check if a Bash array contains a value which says how to check for existence of an item in a list as follows:
if printf '%s\n' ${myarray[#]} | grep -q -P '^mypattern$'; then
# ...
fi
However, I want mypattern value to be passed as a variable as follows:
mynewpattern="xyz"
then I was expecting the following to work
if printf '%s\n' ${myarray[#]} | grep -q -P '^"$mynewpattern"$'; then
# ...
fi
But it is not picking the new pattern of xyz. What is the appropriate syntax to insert the new pattern?
I have just started learning bash.
The single quotes are wrong; you want double quotes instead of single.
However, grep -P is also slightly wrong here; it's not properly portable, and your pattern doesn't use any of the syntax which -P enables; also, you should quote your array properly.
if printf '%s\n' "${myarray[#]}" |
grep -q "^$mypattern\$"
then
...
Text between single quotes is passed through verbatim. If you want the shell to perform variable interpolation, use double quotes (and then you need to escape any literal backslash, dollar sign, or backtick).
Could you please try using like grep -q -P "^$var$"(in your script)
Here is an example script for same scenario for an Input_file(since no samples for array elements were provided so explaining it with an sample/example script here).
##Shell variable
var="bla"
##A sample Input_file
cat << EOF > Input_file
blabla test test
123abcd123test
bla
EOF
##Following is the code to check.
if grep -q -P "^$var$" Input_file
then
echo "match found."
fi
Above will only match lines which are starting with variable val's value.

Using value inside a variable without expanding

I am trying to find and replace a specific text content using the sed command and to run it via a shell script.
Below is the sample script that I am using:
fp=/asd/filename.txt
fd="sed -i -E 's ($2).* $2:$3 g' ${fp}"
eval $fd
and executing the same by passing the arguments:
./test.sh update asd asdfgh
But if the argument string contains $ , it breaks the commands and it is replacing with wrong values, like
./test.sh update asd $apr1$HnIF6bOt$9m3NzAwr.aG1Yp.t.bpIS1.
How can I make sure that the values inside the variables are not expanded because of the $?
Updated
sh file test.sh
set -xv
fp="/asd/filename.txt"
sed -iE "s/(${2//'$'/'\$'}).*/${2//'$'/'\$'}:${3//'$'/'\$'}/g" "$fp"
text file filename.txt
hello:world
Outputs
1)
./test.sh update hello WORLD
sed -iE "s/(${2//'$'/'\$'}).*/${2//'$'/'\$'}:${3//'$'/'\$'}/g" "$fp"
++ sed -iE 's/(hello).*/hello:WORLD/g' /asd/filename.txt
2)
./test.sh update hello '$apr1$hosgaxyv$D0KXp5dCyZ2BUYCS9BmHu1'
sed -iE "s/(${2//'$'/'\$'}).*/${2//'$'/'\$'}:${3//'$'/'\$'}/g" "$fp"
++ sed -iE 's/(hello).*/hello:'\''$'\''apr1'\''$'\''hosgaxyv'\''$'\''D0KXp5dCyZ2BUYCS9BmHu1/g' /asd/filename.txt
In both the case , its not replacing the content
You don't need eval here at all:
fp=/asd/filename.txt
sed -i -E "s/(${2//'$'/'\$'}).*/\1:${3//'$'/'\$'}/g" "$fp"
The whole sed command is in double quotes so variables can expand.
I've replaced the blank as the s separator with / (doesn't really matter in the example).
I've used \1 to reference the first capture group instead of repeating the variable in the substitution.
Most importantly, I've used ${2//'$'/'\$'} instead of $2 (and similar for $3). This escapes every $ sign as \$; this is required because of the double quoting, or the $ get eaten by the shell before sed gets to see them.
When you call your script, you must escape any $ in the input, or the shell tries to expand them as variable names:
./test.sh update asd '$apr1$HnIF6bOt$9m3NzAwr.aG1Yp.t.bpIS1.'
Put the command-line arguments that are filenames in single quotes:
./test.sh update 'asd' '$apr1$HnIF6bOt$9m3NzAwr.aG1Yp.t.bpIS1'
must protect all the script arguments with quotes if having space and special shell char, and escape it if it's a dollar $, and -Ei instead of -iE even better drop it first for test, may add it later if being really sure
I admit i won't understant your regex so let's just get in the gist of solution, no need eval;
fp=/asd/filename.txt
sed -Ei "s/($2).*/$2:$3/g" $fp
./test.sh update asd '\$apr1\$HnIF6bOt\$9m3NzAwr.aG1Yp.t.bpIS1.'

sed - inserting line with /c\ that has a variable that contains spaces

I have just recently got back into learning bash. Currently working on a project of mine and when using sed I've run into an issue, I've tried looking around the web for help but haven't had any joy. I suspect as I may not be using the correct terminology so I can't find what I'm looking for. ANYHOW.
So in my script I'm trying to assign the output of date to a variable. Here's the line from my script.
origdate=$(date)
When I call it the output looks like this:
Wed Oct 5 19:40:45 BST 2016
Part of my script then generates a file and writes information to it, part of which I am trying to use sed to find lines and replace parts of it. This is the first I've been playing around with sed, I've used it successfully so far for my needs. However I'm getting stuck when I try this:
sed -i '/origdate=empty/c\'$origdate'' $sd/pingcheck-email-$job.txt
When I run the script and it gets to this line, this is the error I'm getting:
sed: can't read Oct: No such file or directory
sed: can't read 5: No such file or directory
sed: can't read 19:52:56: No such file or directory
sed: can't read BST: No such file or directory
sed: can't read 2016: No such file or directory
I suspect it's something to do with the spaces in the date (variable), my question is: how can I work around this? Can I get sed to 'ignore' the spaces? or should I just use cut to cut the field for the date, and set that to a variable and the same thing again to set the time to another variable?
Even if someone could kindly point me in the right direction that'd be great!
Thanks in advance!
double quote the variable
sed -i '/origdate=empty/c\'"$origdate"'' $sd/pingcheck-email-$job.txt
or alternatively, the whole script
sed -i "/origdate=empty/c\$origdate" $sd/pingcheck-email-$job.txt
The problem is not with sed but rather with how bash word splits on your date given your command.
Bash
In bash, word splitting is performed on the command line so that text is broken up into a list of arguments. To illustrate, I'm going to run a simple script that outputs the first argument only.
bash -c 'echo $1' ignored_0 foo bar
Think of bash -c 'echo $1' ignored_0 as the command (sed in your case) and foo bar as the arguments. In this case, foo bar is split into two arguments, foo and bar.
To pass foo bar in as the first parameter, you need to have the text in either single or double quotes. See the GNU manual on quoting.
bash -c 'echo $1' ignored_0 'foo bar'
bash -c 'echo $1' ignored_0 "foo bar"
Parameter expansion does not occur when the variable is inside a single quote.
var="foo bar"
bash -c 'echo $1' ignored_0 '$var'
bash -c 'echo $1' ignored_0 "$var"
NOTE: In the command `bash -c 'echo $1', I do not want $1 to expand before being passed as an argument to bash because that's part of the code I want to execute.
Parameter expansion occurs when variables are outside of quotes, but word splitting will apply after the parameter is expanded. From the bash man page in the Word Splitting section:
The shell scans the results of parameter expansion, command
substitution, and arithmetic expansion that did not occur within
double quotes for word splitting.
From the GNU bash manual on Word Splitting:
The shell scans the results of parameter expansion, command
substitution, and arithmetic expansion that did not occur within
double quotes for word splitting.
var="foo bar"
bash -c 'echo $1' ignored_0 $var
The last step in Shell Expansions in Quote Removal where unquoted quote characters are removed before being passed to commands. The following command shows that ''"" has no effect on the arguments passed.
bash -c 'echo $1' ignored_0 foo''""
Application
In your example, the trailing '' after $origdate is extraneous. The important part is that $origdate is not quoted so word splitting applies to the expanded variable.
When -e is not passed to the sed command, sed expects the expression to be in one argument, or word from bash. When you run your command, your expression is /origdate=empty/c\Wed and the rest of the date is considered to be files for the expression to be applied to.
The simple fix is to put double quotes around the string for which you want to prevent word splitting. I've modified the command so that anyone can run this example without having the files on their system.
In this example, the \ must be escaped so that it is not considered an escape character for $.
echo "origdate=empty" | sed "/origdate=empty/c\\$origdate"
You can also change the type of quotes you are using without affecting word splitting like so.
echo "origdate=empty" | sed '/origdate=empty/c\'"$origdate"
You need escape by double slash
\ / \%

proper syntax for the s command along to the addressing in sed

I want to issue this command from the bash script
sed -e $beginning,$s/pattern/$variable/ file
but any possible combination of quotes gives me an error, only one that works:
sed -e "$beginning,$"'s/pattern/$variable/' file
also not good, because it do not dereferences the variable.
Does my approach can be implemented with sed?
Feel free to switch the quotes up. The shell can keep things straight.
sed -e "$beginning"',$s/pattern/'"$variable"'/' file
You can try this:
$ sed -e "$beginning,$ s/pattern/$variable/" file
Example
file.txt:
one
two
three
Try:
$ beginning=1
$ variable=ONE
$ sed -e "$beginning,$ s/one/$variable/" file.txt
Output:
ONE
two
three
There are two types of quotes:
Single quotes preserve their contents (> is the prompt):
> var=blah
> echo '$var'
$var
Double quotes allow for parameter expansion:
> var=blah
> echo "$var"
blah
And two types of $ sign:
One to tell the shell that what follows is the name of a parameter to be expanded
One that stands for "last line" in sed.
You have to combine these so
The shell doesn't think sed's $ has anything to do with a parameter
The shell parameters still get expanded (can't be within single quotes)
The whole sed command is quoted.
One possibility would be
sed "$beginning,\$s/pattern/$variable/" file
The whole command is in double quotes, i.e., parameters get expanded ($beginning and $variable). To make sure the shell doesn't try to expand $s, which doesn't exist, the "end of line" $ is escaped so the shell doesn't try anything funny.
Other options are
Double quoting everything but adding a space between $ and s (see Ren's answer)
Mixing quoting types as needed (see Ignacio's answer)
Methods that don't work
sed '$beginning,$s/pattern/$variable/' file
Everything in single quotes: the shell parameters are not expanded (doesn't follow rule 2 above). $beginning is not a valid address, and pattern would be literally replaced by $variable.
sed "$beginning,$s/pattern/$variable/" file
Everything in double qoutes: the parameters are expanded, including $s, which isn't supposed to (doesn't follow rule 1 above).
the following form worked for me from within script
sed $beg,$ -e s/pattern/$variable/ file
the same form will also work if executed from the shell

Using egrep and regular expression together

I want to search the below text file for words that ends in _letter, and get the whole portion upto "::". There is no space between any letter
blahblah:/blahblah::abc_letter:/blahblah/blahblah
blahblah:/blahblah::cd_123_letter:/blahblah/blahblah
blahblah:::/blahblah::24_cde_letter:/blahblah/blahblah
blahblah::/blahblah::45a6_letter:/blahblah/blahblah
blahblah:/blahblah::fgh_letter:/blahblah/blahblah
blahblah:/blahblah::789_letter:/blahblah/blahblah
I tried
egrep -o '*_letter'
and
egrep -o "*_letter"
But it only returns the word _letter
then I want to feed the input to the parametre of a shell script for loop. So the script will look like following
for i in [grep command]
mkdir $i
end
It will create the following directories
abc_letter/
cd_123_letter/
24_cde_letter/
45a6_letter/
fgh_letter/
789_letter/
ps: The result between :: and _letter doesn't contain any special character, only alphanumeric character
also my system doesn't have perl
Assuming no spaces or new-lines:
for i in $(sed 's/^.*:\([^/]*_letter\):.*$/\1/g' infile); do
mkdir $i
done
To extract after : to _letter strings from a file.txt and use them in your for loop, you can use the following egrep and revise your: script.sh, like this:
#!/bin/bash
for i in $(egrep -o "[^:]+_letter" file.txt); do
mkdir -p $i
done
Then you run ./script.sh, and later you check with ls, you see:
$ ls -1
24_cde_letter
45a6_letter
789_letter
abc_letter
cd_123_letter
fgh_letter
file.txt
script.sh
Explanation
Your original egrep -o '*_letter' probably just confused bash filename expansion with regular expression,
In bash, *something uses star globbing character to match * = anything here + something.
However in regular expression star * means the preceding character zero or more times. Since * is at the beginning of what you wrote, there is nothing before it, so it does not match anything there.
The only thing egrep can match is _letter, and since we are using the -o option it only displays the match, on an individual line, and thus why you originally only saw a line of _letter matches
Our new changes:
egrep pattern starts with [^ ... ], a negation, matches the opposite of what characters you put within. We put : within.
The + says to match the preceding one or more times.
So combined, it says look for anything-but-:, and do this one or more times.
Thus of course it matches anything after :, and keeps matching, until the next part of the pattern
The next part of the pattern is just _letter
egrep -o so only matched text will be shown, one per line
So in this way, from lines such as:
blahblah:/blahblah::abc_letter:/blahblah/blahblah
It successfully extracts:
abc_letter
Then, changes to your bash script:
Bash command substitution $() to have the results of the egrep command sent to the for-loop
for i value...; do ... done syntax
mkdir -p just a convenience in case you are re-testing, it will not error if directory was already made.
So altogether it helps to extract the pattern you wanted and generate directories with those names.

Resources