using sed to set a variable works on command line, but not bash script - linux

I have looked quite a bit for answers but I am not finding any suggestions that have worked so far.
on command line, this works:
$ myvar=$( cat -n /usr/share/dict/cracklib-small | grep $myrand | sed -e "s/$myrand//" )
$ echo $myvar
$ commonness
however, inside a bash script the same exact lines just echoes out a blank line
notes - $myrand is a number, like 10340 generated with $RANDOM
cat prints out a dictionary with line numbers
grep grabs the line with $myrand in it ; e.g. 10340 commonness
sed is intended to remove the $myrand part of the line and replace it with nothing. here is my sample script
#!/bin/bash
# prints out a random word
myrand=$RANDOM
export myrand
myword=$( cat -n /path/to/dict/cracklib-small | grep myrand | sed -e "s/$myrand//g" <<<"$myword" )
echo $myword

Your command line code is running:
grep $myrand
Your script is running:
grep myrand
These are not the same thing; the latter is looking for a word that contains "myrand" within it, not a random number.
By the way -- I'd suggest a different way to get a random line. If you have GNU coreutils, the shuf tool is built-to-purpose:
myword=$(shuf -n 1 /path/to/dict/cracklib-small)

#!/bin/bash
# prints out a random word
myrand=$RANDOM
export myrand
myword=$( cat -n /path/to/dict/cracklib-small | grep myrand | sed -e "s/$myrand//g" <<<"$myword" )
echo $myword
where is the $ sign in grep myrand ?
you must put in some work before posting it here.

Related

Bash script and manually running commands on the command line

I have the following simple bash script which takes input from stdin and prints the third line given as input.
#!/bin/bash
var=$(cat)
echo $var | head -n 3 | tail -n 1
The problem with this script is that it prints all the lines but here is the funny part, when I type the commands individually on the command line I am getting the expected result i.e. the third line. Why this anomaly? Am I doing something wrong here?
The aim of head -n 3 | tail -n 1 is to keep the third line into variable
It will be more efficient to use read builtin
read
read
read var
echo "${var}"
Or to keep heading white-spaces
IFS= read
and not join lines ending with \ or not give special meaning to \
read -r
You don't need $(cat) in your script. If script is reading data from stdin then just have this single line in your script:
head -n 3 | tail -n 1
And run it as:
bash myscript.sh < file.txt
This will print 3rd line from file.txt
PS: You can replace head + tail with this faster sed to print 3rd line from input:
sed '3q;d'
The shell is splitting the var variable so echo get multiple parameters. You need to quote your variable to prevent this to happen:
#!/bin/bash
var=$(cat)
echo "$var" | head -n 3 | tail -n 1
This should do the trick, as far as I understand your question:
#!/bin/bash
var=$(cat)
echo "$var" | head -n 3 | tail -n 1
var=$(cat) will not allow you to escape out of stdin mode. you need to specify the EOF for the script to understand to stop reading from stdin.
read -d '' var << EOF
echo "$var" | head -n 3 | tail -n 1

UNIX shell script to run a list of grep commands from a file and getting result in a single delimited file

I am beginner in unix programming and a way to automate my work
I want to run a list a grep commands and get the output of all the grep command in a in a single delimited file .
i am using the following bash script. But it's not working .
Mockup sh file:
!/bin/sh
grep -l abcd123
grep -l abcd124
grep -l abcd125
and while running i used the following command
$ ./Mockup.sh > output.txt
Is it the right command?
How can I get both the grep command and output in the output file?
how can i delimit the output after each command and result?
How can I get both the grep command and output in the output file
You can use bash -v (verbose) to print each command before execution on stderr and it's output will be as usual be available on stdout:
bash -v ./Mockup.sh > output.txt 2>&1
cat output.txt
Working Demo
A suitable shell script could be
#!/bin/sh
grep -l 'abcd123\|abcd124\|abcd125' "$#"
provided that the filenames you pass on the invocation of the script are "well behaved", that is no whitespace in them. (Edit Using the "$#" expansion takes care of generic whitespace in the filenames, tx to triplee for his/her comment)
This kind of invocation (with alternative matching strings, as per the \| syntax) has the added advantage that you have exactly one occurrence of a filename in your final list, because grep -l prints once the filename as soon as it finds the first occurrence of one of the three strings in a file.
Addendum about "$#"
% ff () { for i in "$#" ; do printf "[%s]\n" "$i" ; done ; }
% # NB "a s d" below is indeed "a SPACE s TAB d"
% ff "a s d" " ert " '345
345'
[a s d]
[ ert ]
[345
345]
%
cat myscript.sh
########################
#!/bin/bash
echo "Trying to find the file contenting the below string, relace your string with below string"
grep "string" /path/to/folder/* -R -l
########################
save above file and run it as below
sh myscript.sh > output.txt
once the command prmpt get return you can check the output.txt for require output.
Another approach, less efficient, that tries to address the OP question
How can I get both the grep command and output in the output file?
% cat Mockup
#!/bin/sh
grep -o -e string1 -e string2 -e string3 "$#" 2> /dev/null | sort -t: -k2 | uniq
Output: (mocked up as well)
% sh Mockup file{01..99}
file01:string1
file17:string1
file44:string1
file33:string2
file44:string2
file48:string2
%
looking at the output from POV of a consumer, one foresees problems with search strings and/or file names containing colons... oh well, that's another Q maybe

Find line number in a text file - without opening the file

In a very large file I need to find the position (line number) of a string, then extract the 2 lines above and below that string.
To do this right now - I launch vi, find the string, note it's line number, exit vi, then use sed to extract the lines surrounding that string.
Is there a way to streamline this process... ideally without having to run vi at all.
Maybe using grep like this:
grep -n -2 your_searched_for_string your_large_text_file
Will give you almost what you expect
-n : tells grep to print the line number
-2 : print 2 additional lines (and the wanted string, of course)
You can do
grep -C 2 yourSearch yourFile
To send it in a file, do
grep -C 2 yourSearch yourFile > result.txt
Use grep -n string file to find the line number without opening the file.
you can use cat -n to display the line numbers and then use awk to get the line number after a grep in order to extract line number:
cat -n FILE | grep WORD | awk '{print $1;}'
although grep already does what you mention if you give -C 2 (above/below 2 lines):
grep -C 2 WORD FILE
You can do it with grep -A and -B options, like this:
grep -B 2 -A 2 "searchstring" | sed 3d
grep will find the line and show two lines of context before and after, later remove the third one with sed.
If you want to automate this, simple you can do a Shell Script. You may try the following:
#!/bin/bash
VAL="your_search_keyword"
NUM1=`grep -n "$VAL" file.txt | cut -f1 -d ':'`
echo $NUM1 #show the line number of the matched keyword
MYNUMUP=$["NUM1"-1] #get above keyword
MYNUMDOWN=$["NUM1"+1] #get below keyword
sed -n "$MYNUMUP"p file.txt #display above keyword
sed -n "$MYNUMDOWN"p file.txt #display below keyword
The plus point of the script is you can change the keyword in VAL variable as you like and execute to get the needed output.

BASH: can grep on command line, but not in script

Did this million times already, but this time it's not working
I try to grep "$TITLE" from a file. on command line it's working, "$TITLE" variable is not empty, but when i run the script it finds nothing
*title contains more than one word
echo "$TITLE"
cat PAGE.$TEMP.2 | grep "$TITLE"
what i've tried:
echo "cat PAGE.$TEMP.2 | grep $TITLE"
to see if title is not empty and file name is actually there
Are you sure that $TITLE does not have leading or trailing whitespace which is not in the file? Your fix with the string would strip out whitespace before execution, so it would not see it.
For example, with a file containing 'Line one':
/home/user1> TITLE=' one '
/home/user1> grep "$TITLE" text.txt
/home/user1> cat text.txt | grep $TITLE
Line one
Try echo "<$TITLE>", or echo "$TITLE"|od -xc which sould enable you to spot errant chars.
This command
echo "cat PAGE.$TEMP.2 | grep $TITLE"
echoes a string that starts with 'cat'. It does not run a command. You would want
echo "$( cat PAGE.$TEMP.2 | grep $TITLE )"
although that is identical in functionality to the simpler
cat PAGE.$TEMP.2 | grep $TITLE
And as pointed out by others, there is no need to pipe a single file using cat; grep can read from files just fine:
grep "$TITLE" "PAGE.$TEMP.2"
(Your default behavior should be to quote parameter expansions, unless you can show it is incorrect to do so.)
Works for me:
~> cat test.dat
abc
cda
xyz
~> export GRP=cda
~> cat test.dat | grep $GRP
cda
Edit:
Also the proper way to use grep is:
~> grep $GRP test.dat

xargs with multiple arguments

I have a source input, input.txt
a.txt
b.txt
c.txt
I want to feed these input into a program as the following:
my-program --file=a.txt --file=b.txt --file=c.txt
So I try to use xargs, but with no luck.
cat input.txt | xargs -i echo "my-program --file"{}
It gives
my-program --file=a.txt
my-program --file=b.txt
my-program --file=c.txt
But I want
my-program --file=a.txt --file=b.txt --file=c.txt
Any idea?
Don't listen to all of them. :) Just look at this example:
echo argument1 argument2 argument3 | xargs -l bash -c 'echo this is first:$0 second:$1 third:$2'
Output will be:
this is first:argument1 second:argument2 third:argument3
None of the solutions given so far deals correctly with file names containing space. Some even fail if the file names contain ' or ". If your input files are generated by users, you should be prepared for surprising file names.
GNU Parallel deals nicely with these file names and gives you (at least) 3 different solutions. If your program takes 3 and only 3 arguments then this will work:
(echo a1.txt; echo b1.txt; echo c1.txt;
echo a2.txt; echo b2.txt; echo c2.txt;) |
parallel -N 3 my-program --file={1} --file={2} --file={3}
Or:
(echo a1.txt; echo b1.txt; echo c1.txt;
echo a2.txt; echo b2.txt; echo c2.txt;) |
parallel -X -N 3 my-program --file={}
If, however, your program takes as many arguments as will fit on the command line:
(echo a1.txt; echo b1.txt; echo c1.txt;
echo d1.txt; echo e1.txt; echo f1.txt;) |
parallel -X my-program --file={}
Watch the intro video to learn more: http://www.youtube.com/watch?v=OpaiGYxkSuQ
How about:
echo $'a.txt\nb.txt\nc.txt' | xargs -n 3 sh -c '
echo my-program --file="$1" --file="$2" --file="$3"
' argv0
It's simpler if you use two xargs invocations: 1st to transform each line into --file=..., 2nd to actually do the xargs thing ->
$ cat input.txt | xargs -I# echo --file=# | xargs echo my-program
my-program --file=a.txt --file=b.txt --file=c.txt
You can use sed to prefix --file= to each line and then call xargs:
sed -e 's/^/--file=/' input.txt | xargs my-program
Here is a solution using sed for three arguments, but is limited in that it applies the same transform to each argument:
cat input.txt | sed 's/^/--file=/g' | xargs -n3 my-program
Here's a method that will work for two args, but allows more flexibility:
cat input.txt | xargs -n 2 | xargs -I{} sh -c 'V="{}"; my-program -file=${V% *} -file=${V#* }'
I stumbled on a similar problem and found a solution which I think is nicer and cleaner than those presented so far.
The syntax for xargs that I have ended with would be (for your example):
xargs -I X echo --file=X
with a full command line being:
my-program $(cat input.txt | xargs -I X echo --file=X)
which will work as if
my-program --file=a.txt --file=b.txt --file=c.txt
was done (providing input.txt contains data from your example).
Actually, in my case I needed to first find the files and also needed them sorted so my command line looks like this:
my-program $(find base/path -name "some*pattern" -print0 | sort -z | xargs -0 -I X echo --files=X)
Few details that might not be clear (they were not for me):
some*pattern must be quoted since otherwise shell would expand it before passing to find.
-print0, then -z and finally -0 use null-separation to ensure proper handling of files with spaces or other wired names.
Note however that I didn't test it deeply yet. Though it seems to be working.
xargs doesn't work that way. Try:
myprogram $(sed -e 's/^/--file=/' input.txt)
It's because echo prints a newline. Try something like
echo my-program `xargs --arg-file input.txt -i echo -n " --file "{}`
I was looking for a solution for this exact problem and came to the conclution of coding a script in the midle.
to transform the standard output for the next example use the -n '\n' delimeter
example:
user#mybox:~$ echo "file1.txt file2.txt" | xargs -n1 ScriptInTheMiddle.sh
inside the ScriptInTheMidle.sh:
!#/bin/bash
var1=`echo $1 | cut -d ' ' -f1 `
var2=`echo $1 | cut -d ' ' -f2 `
myprogram "--file1="$var1 "--file2="$var2
For this solution to work you need to have a space between those arguments file1.txt and file2.txt, or whatever delimeter you choose, one more thing, inside the script make sure you check -f1 and -f2 as they mean "take the first word and take the second word" depending on the first delimeter's position found (delimeters could be ' ' ';' '.' whatever you wish between single quotes .
Add as many parameters as you wish.
Problem solved using xargs, cut , and some bash scripting.
Cheers!
if you wanna pass by I have some useful tips http://hongouru.blogspot.com
Actually, it's relatively easy:
... | sed 's/^/--prefix=/g' | xargs echo | xargs -I PARAMS your_cmd PARAMS
The sed 's/^/--prefix=/g' is optional, in case you need to prefix each param with some --prefix=.
The xargs echo turns the list of param lines (one param in each line) into a list of params in a single line and the xargs -I PARAMS your_cmd PARAMS allows you to run a command, placing the params where ever you want.
So cat input.txt | sed 's/^/--file=/g' | xargs echo | xargs -I PARAMS my-program PARAMS does what you need (assuming all lines within input.txt are simple and qualify as a single param value each).
There is another nice way of doing this, if you do not know the number of files upront:
my-program $(find . -name '*.txt' -printf "--file=%p ")
Nobody has mentioned echoing out from a loop yet, so I'll put that in for completeness sake (it would be my second approach, the sed one being the first):
for line in $(< input.txt) ; do echo --file=$line ; done | xargs echo my-program
Old but this is a better answer:
cat input.txt | gsed "s/\(.*\)/\-\-file=\1/g" | tr '\n' ' ' | xargs my_program
# i like clean one liners
gsed is just gnu sed to ensure syntax matches version brew install gsed or just sed if your on gnu linux already...
test it:
cat input.txt | gsed "s/\(.*\)/\-\-file=\1/g" | tr '\n' ' ' | xargs echo my_program

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