Create parameters for command Linux shell script - linux

Hello I am trying to create parameters for my shell script but I am having trouble.
lets say for example the file is called test.
When I call ./test -parameter1 input_file.txt
I get an error saying 'no such file or directory'.
Here is an example of my code.
#!/bin/sh
if [ "$1" == "parameter" ]
then
while read line
do
#echo "$line"
done <$1
else
echo "Not working"
fi
My over all goal of this is to read in a file of numbers line by line which I have working, then to calculate the average values of the rows or by columns. Which is why I am trying to create parameters so the user will have to specify ./test -rows input_file.txt or ./test -columns input_file.txt

You are using the string -parameter as the input file name. Perhaps you want:
#!/bin/sh
if [ "$1" = "-parameter" ]
then
while read line
do
#echo "$line"
done <$2 # Use $2 instead of $1 here. Or use shift
else
echo "Not working" >&2
fi

Related

How can I remove the extension of specific files in a directory?

I want to remove the extension of specific files with a given extension.
So for instance, in a directory foobar, we have foo.txt, bar.txt foobar.jpg.
Additionally, the extension that I've put in to be removed is txt
After calling the program, my output should be foo bar foobar.jpg
Here is my code so far:
#!/bin/bash
echo "Enter an extension"
read extension
echo "Enter a directory"
read directory
for file in "$directory"/*; do //
if [[ $file == *.txt ]]
then
echo "${file%.*}"
else
echo "$file"
fi
done
However when I run this on a given directory, nothing shows up.
I'm assuming that there is a problem with how I referred to the directory ( in the line where I placed a //) and I've tried to research on how to solve it but to no avail.
What am I doing wrong?
If files do exist in a valid directory you've entered then they should show up — with one exception. If you are using ~/ (shorthand home directory) then it will be treated as plain text in your for loop. The read variable should be substituted into another variable so the for loop can treat it as a directory (absolute paths should work normally as well).
#!/bin/bash
echo "Enter an extension"
read -r extension
echo "Enter a directory"
read -r directory
dir="${directory/#\~/$HOME}"
for file in "$dir"/*; do
if [[ $file == *."$extension" ]]
then
echo "${file%.*}"
else
echo "$file"
fi
done
You can simplify your for-loop:
for file in "$directory"/*; do
echo "${f%.$extension}";
done
The % instructions removes only matching characters. If nothing matches, the original string (here f) is returned.
When you write bash scripts it's more common to pass arguments to your script via command line arguments rather than by reading it from standard input via read program.
Passing arguments via command line:
#!/bin/bash
# $# - a bash variable which holds a number of arguments passed
# to script via command line arguments
# $0 holds the name of the script
if [[ $# -ne 2 ]]; then # checks if exactly 2 arguments were passed to script
echo "Usage: $0 EXTENSION DIRECTORY"
exit -1;
fi
echo $1; # first argument passed to script
echo $2; # second arugment passed to script
This approach is more efficient because a subprocess is spawn for read command to run and there is no subprocess spawn for reading command line arguments.
There is no need to manually loop through directory, you can use find command to find all files with given extension within given directory.
find /path/to/my/dir -name '*.txt'
find $DIRECTORY -name "*.$EXTENSION"
# note that single quotes in this context would prevent $EXTENSION
# variable to be resolved, so double quotes are used " "
# find searches for files inside $DIRECTORY and searches for files
# matching pattern '*.$EXTENSION'
Note that to avoid bash filename expansion sometimes it is required to wrap actual pattern in single quotes ' ' or double quotes " ". See Bash Filename Expansion
So now your script can look like this:
#!/bin/bash
if [[ $# -ne 2 ]]; then
echo "Usage: $0 EXTENSION DIRECTORY"
exit -1;
fi
$EXTENSION = $1 # for better readability
$DIRECTORY = $2
for file in `find $DIRECTORY -name "*.$EXTENSION"`; do
mv $file ${file%.$EXTENSION}
done
Construct ${file%.$EXTENSION} is called Shell Parameter Expansion it searches for occurrence of .$EXTENSION inside file variable and deletes it.
Notice that in the script it is easy to pass extension as directory and vice versa.
We can check if second argument is in fact directory, we can use following construction:
if ! [[ -d $DIRECTORY ]]; then
echo $DIRECTORY is not a dir
exit -1
fi
This way we can exit from the script earlier with more readable error.
To sum up entire script could look like this:
#!/bin/bash
if [[ $# -ne 2 ]]; then
echo "Usage: $0 EXTENSION DIRECTORY"
exit -1;
fi
EXTENSION=$1 # for better readability
DIRECTORY=$2
if ! [[ -d $DIRECTORY ]]; then
echo $DIRECTORY is not a directory.
exit -1
fi
for file in `find $DIRECTORY -name "*.$EXTENSION"`; do
mv $file ${file%.$EXTENSION}
done
Example usage:
$ ./my-script.sh txt /path/to/directory/with/files

How can I keep file name to variable from this code

readLBL.sh
#!/bin/bash
while IFS='' read -r line || [[ -n "$line" ]]; do
echo "Text read from file: $line"
done < "$1"
When I run this shell script in Terminal and I have to insert file name for run it
Example :
./readLBL.sh science.txt
output :
58050364;Tom Jones
58050365;Marry Jane
how can i keep "science.txt" into some variable like a = "science.txt"
Use $1 as this is the argument you are already using to read the file in the first instance.
#!/bin/bash
while IFS='' read -r line || [[ -n "$line" ]]; do
echo "Text read from file $1: $line"
done < "$1"
Running this gives the following:
$ ./readLBL.sh science.txt
Text read from file science.txt: 58050364;Tom Jones
Text read from file science.txt: 58050365;Marry Jane
a="science.txt"
from the Linux command line, will set $a equal to science.txt
If you want this permanent, you could perhaps add it to the bottom of one of your linux profiles. For example add it to the bottom of ~/.bashrc

Run commands from a text file through a bash script

I am attempting to write a script that will read through a text file, and then execute every line that begins with the word "run" or "chk" as a command. This is what I have thus far:
#!/bin/bash
counter=1
for i in $#
do
while read -r line
do
if [[ ${line:0:4} == "run " ]]
then
echo "Now running line $counter"
${line:4:${#line}}
elif [[ ${line:0:4} == "chk " ]]
then
echo "Now checking line $counter"
${line:4:${#line}}
elif [[ ${line:0:2} == "# " ]]
then
echo "Line $counter is a comment"
else
echo "Line $counter: '$line' is an invalid line"
fi
counter=$((counter+1))
done<$i
done
However, when I feed it a text file with, for example the commands
run echo > temp.txt
It does not actually create a file called temp.txt, it just echoes "> temp.txt" back to the stdout. It also does a similar thing when I attempt to do something like
run program arguments > filename.txt
It does not put the output of the program in a file as I want, but it rather tries to treat the '>' as a file name.
I know this is a super specific and probably obvious thing, but I am very new to bash and all shell scripting.
Thanks
You need to use eval to do all the normal shell parsing of the variable:
eval "${line:4}"
You also don't need :${#line}. If you leave out the length, it defaults to the rest of the string.

Copy multiple files with bash script from command line arguments?

I want to create a script that allows me to enter multiple filenames from the command line, and have the script copy those files to another directory. This is what I am trying but I keep getting an error of
line 10: binary operator expected
#!/bin/bash
DIRECTORY=/.test_files
FILE=$*
if [ -e $DIRECTORY/$FILE ]; then
echo "File already exists"
else
cp $FILE $DIRECTORY
fi
So if the script was named copfiles.sh, I am writing...
./copyfiles.sh doc1.txt doc2.txt
It will move the files, but if they already exist, it won't read the error message.
Also I get the "line 10: binary operator expected" error regardless of it the files are there or not. Can anyone tell me what I am doing wrong?
As a possible problem, if you had a filename with a space or had multiple arguments $* would have spaces in it so [ -e $DIR/$FILE ] will expand to have lots of words, like [ -e /.test_files/First word and more ] and -e expects just 1 word after it. Try putting it in quotes like
if [ -e "$DIRECTORY/$FILE" ]
Of course, you may only want to store $1 in $FILE to get just the first argument.
To test all the arguments you want to loop over the arguments and test each with something like
for FILE in "$#"; do
if [ -e "$DIRECTORY/$FILE" ]; then
echo "$FILE already exists"
else
cp "$FILE" $DIRECTORY
fi
done
Using quotes around $# to preserve spaces in the original arguments as well

How to create a shell script that can scan a file for a specific word?

one of the questions that I have been given to do for my Computer Science GCSE was:
Write a shell script that takes a string input from a user, asks for a file name and reports whether that string is present in the file.
However way I try to do it, I cannot create a shell script.
I don't need you to tell me the whole number, however, I have no idea where to start. I input the variable and the file name, however, I have no idea how to search for the chosen word in the chosen file. Any ideas?
Using grep can get this working, for example
viewEntry()
{
echo "Entering view entry"
echo -n "Enter Name: "
read input
if grep -q "$input" datafile
then
echo ""
echo -n "Information -> "
grep -w "$input" datafile
echo ""
else
echo "/!\Name Not Found/!\\"
fi
echo "Exiting view entry"
echo ""
}
dataFile is the file you would be reading from. Then making use of -q and -w arguments of grep, you should be able to navigate your chosen file.
This site does a great job explaining grep and your exact problem: http://www.cyberciti.biz/faq/howto-use-grep-command-in-linux-unix/
The following shell-script is a very quick approach to do what you suggested:
#!/bin/sh # Tell your shell with what program this script should be exectued
echo "Please enter the filename: "
read filename # read user input into variable filename
count=`grep -c $1 $filename` # store result of grep into variable count
if [ $count -gt 0 ] # check if count is greater than 0
then
echo "String is present:" $1
else
echo "String not found:" $1
fi
You should look at some tutorials to get the basics of shell-scripting. Your task isn't very complex, so after some reading you should be able understand what the script does and modify it according your needs.

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