How to use grep with quotes inside a shell script? - bash

I am trying to use the following pattern inside a shell script eg. I want to use the following:
grep ^"#" $fn | grep -v NM$ > $op
but inside bash script. The problem is normally bash considers everything after the "#" as a comment. If i use
grep ^"\#" $fn
I think it changes the meaning. I am a newbie.
Any help will be appreciated.

There are multiple possibilities:
grep '^#' $fn
or
grep "^#" $fn
or
grep ^\# $fn

Here is a little example which I saved as foo.sh
echo "#"
echo '#' # this is a comment
echo '\#'
when run as sh foo.sh this is what is printed out
#
#
\#

Related

Some tips to improve a bash script for count fastq files

Hi guys I got this bash one line that i wish to make a script
for i in 'ls *.fastq.gz'; do echo $(zcat ${i} | wc -l)/4|bc; done
I would like to make it as a script to read from a data dir and print out the result with the name of the file.
I tried to put the dir in front of the 'data/*.fastq.gz' but got am error No such dir exist...
I would like some like this:
name1.fastq.gz 1898516
name2.fastq.gz 2467421
namen.fastq.gz 1234532
I am not experienced in bash.
Could you guys give a help?
Thanks
Take the dir as an argument, but default to the current dir if it's not set.
dir="${1-.}"
Then put it in the glob: "$dir"/*.fastq.gz
As well:
Quote variables and command expansions.
Don't parse ls.
Don't trust echo with arbitrary data (filenames). Use printf instead.
Use an end-of-options flag -- when giving filenames to commands.
I prefer to not have any inline command expansions, but that's just personal preference
Putting it together:
#!/bin/bash
dir="${1-.}"
for file in "$dir"/*.fastq.gz; do
printf '%s ' "$file"
lines="$(zcat -- "$file" | wc -l)"
bc <<< "$lines/4" # Using a here-string (Bash feature)
done
There is no need to escape to bc for integer math (divide by 4), or to use 'ls' to enumerate the files. The original version will do with minor changes:
#!/bin/bash
dir="${1-.}"
for i in "$dir"/*.fastq.gz; do
lines=$(zcat "${i}" | wc -l)
printf '%s %d\n' "$i" "$((lines/4))"
done

Shell script for replacing characters?

I'm trying to write a shell script that takes in a file(ex. file_1_2.txt) and replaces any "_" with "."(ex. file.1.2.txt). This is what I have but its giving me a blank output when I run it.
read $var
x= `echo $var | sed 's/\./_/g'`
echo $x
I'm trying to store the changed filename in the variable "x" and then output x to the console.
I am calling this script by writing
./script2.sh < file_1_2.txt
There is two problems. First, your code has some bugs:
read var
x=`echo $var | sed 's/_/\./g'`
echo $x
will work. You had an extra $ in read var, a space too much (as mentioned before) and you mixed up the replacement pattern in sed (it was doing the reverse of what you wanted).
Also if you want to replace the _ by . in the filename you should do
echo "file_1_2.txt" | ./script2.sh
If you use < this will read the content of `file_1_2.txt" into your script.
Another solution, with bash only:
$ x=file_1_2.txt; echo "${x//_/.}"
file.1.2.txt
(See “Parameter expansion” section in bash manual page for details)
And you can also do this with rename:
$ touch file_1_2.txt
$ ls file*
file_1_2.txt
$ rename 'y/_/\./' file_1_2.txt
$ ls file*
file.1.2.txt
Threre is not need for sed as bash supports variable replacement:
$ cat ./script2
#!/bin/bash
ofile=$1
nfile=${ofile//_/./}
echo mv "$ofile" "$nfile"
$ ./script2 file_1_2.txt
mv "file_1_2.txt" "file.1.2.txt"
Then just remove echo if you are satisfied with the result.

shell script grep to grep a string

The output is blank fr the below script. What is it missing? I am trying to grep a string
#!/bin/ksh
file=$abc_def_APP_13.4.5.2
if grep -q abc_def_APP $file; then
echo "File Found"
else
echo "File not Found"
fi
In bash, use the <<< redirection from a string (a 'Here string'):
if grep -q abc_def_APP <<< $file
In other shells, you may need to use:
if echo $file | grep -q abc_def_APP
I put my then on the next line; if you want your then on the same line, then add ; then after what I wrote.
Note that this assignment:
file=$abc_def_APP_13.4.5.2
is pretty odd; it takes the value of an environment variable ${abc_def_APP_13} and adds .4.5.2 to the end (it must be an env var since we can see the start of the script). You probably intended to write:
file=abc_def_APP_13.4.5.2
In general, you should enclose references to variables holding file names in double quotes to avoid problems with spaces etc in the file names. It is not critical here, but good practices are good practices:
if grep -q abc_def_APP <<< "$file"
if echo "$file" | grep -q abc_def_APP
Yuck! Use the shell's string matching
if [[ "$file" == *abc_def_APP* ]]; then ...

Bash script for downloading files with Curl

I'm trying to knock together a little batch script for downloading files, that takes a URL as its first parameter and a local filename as its second parameter. In testing I've learned that its tripping up on spaces in the output filename, so I've tried using sed to escape them, but its not working.
#!/bin/bash
clear
echo Downloading $1
echo
filename=`sed -e "s/ /\\\ /g" $2`
echo $filename
echo eval curl -# -C - -o $filename $1
but I get the message
sed: outfile.txt: No such file or directory
which suggests its trying to load the output file as input to sed instead of treating the output filename as a string literal.
What would be the correct syntax here?
quoting the arguments correctly, rather than transforming them might be a better approach
It's quite normal to expect to have to quote spaces in arguments to shell scripts
e.g.
#!/bin/bash
clear
echo Downloading $1
echo `curl -# -C - -o "${2}" "${1}"`
called like so
./myscript http://www.foo.com "my file"
alternatively, escape the spaces with a '\' as you call them
./myscript http://www.example.com my\ other\ filename\ with\ spaces
I agree with cms. Quoting the input arguments correctly is much better style - what will you do with the next problem character?
The following is much better.
curl -# -C - -o "$2" $1
However, I hate people not answering the asked question, so here's an answer :-)
#!/bin/bash
clear
echo Downloading $1
echo
filename=`echo $2 | sed -e "s/ /\\\ /g"`
echo $filename
echo eval curl -# -C - -o $filename $1
curl -# -C - -o "$2" $1
if $2 is a text input then try
echo $2 | sed 's: :\\\ :g '
I generally avoid backslashes in sed delimiters are they are quite confusing.

How can I use bash to parse out only a section of a variable with different delimiters?

I have a loop in a bash file to show me all of the files in a directory, each as its own variable. I need to take that variable (filename) and parse out only a section of it.
Example:
92378478234978ehbWHATIWANT#98712398712398723
Now, assuming "ehb" and the pound symbol never change, how can I just capture WHATIWANT into its own variable?
So far I have:
#!/bin/bash
for FILENAME in `dir -d *` ; do
done
You can use sed to edit out the parts you don't want.
want=$(echo "$FILENAME" | sed -e 's/.*ehb\(.*\)#.*/\1/')
Or you can use Bash's parameter expansion to strip out the tail and head.
want=${FILENAME%#*}; want=${want#*ehb}
One possibility:
for i in '92378478234978ehbWHATIWANT#98712398712398723' ; do
j=$(echo $i | sed -e 's/^.*ehb//' -e 's/#.*$//')
echo $j
done
produces:
WHATIWANT
using only the bash shell, no need external tools
$ string=92378478234978ehbWHATIWANT#98712398712398723
$ echo ${string#*ehb}
WHATIWANT#98712398712398723
$ string=${string#*ehb}
$ echo ${string%#*}
WHATIWANT

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