Not able to append command line variable in shell script - shell

Here is my code:
export ALLOW_RPM_UPGRADE=True
path='/opt/rpm/latest/'
echo $1
file=$1
echo $file
dest=${path}${file}
echo $dest
cp $source $dest
Problem:
The three echo statements are printing the same value.
The third one is not appending path to $dest variable.

This is because I had created shell script on windows machine. Before executing it on linux we need to covert it through DOS2Unix utility. OR Just create shell script on Linux and save.
After this it is working fine.
Thanks

Related

Usage of Perl custom script from shell

I'm trying to pass a variable to a custom Perl script.
$file="my_wa/fin"
if [[ "$file" == *"fin" ]]; then
script.pl
fi
In my script I try to use $file as my path to open and read a file
my $input_filename = "$ENV{file}";
I understand that I need to export it somehow and receive it in my Perl code,
but it doesn't recognize $file. - it works perfectly if I just write my_wa/fin.
You did not export the variable in the shell before calling the perl program.
Paste your shell code into http://www.shellcheck.net/ -- it will tell you your other shell error.
you can pass $file as an argument
$file="my_wa/fin"
if [[ "$file" == *"fin" ]]; then
script.pl \"$file\"
fi
Then access the data
my $input_filename = $ARGV[0];
Yes, you will need to export the environment variable file in your shell before the perl script execution. Also you may want to make sure to include full path to the file.
export file=/full/path/to/my/file/my_wa/fin
Or making use of pwd if it is relative to current working directory:
export file="$(pwd)/my_wa/fin"

shell script to simulate the mv command in linux

I am a beginner in shell script and have to code a script to simulate the functionality of mv command of linux without actually using the mv command in my script. Please help me I will be very thankful !
If you want to simulate the mv command you need to copy the file that you want to "move" to the destination folder then you have to delete the old one.
Here is a little script:
#!/bin/bash
from=$1
to=$2
echo "Copy from $1 to $2"
cp $1 $2
echo "Removing the old file $1"
rm $1
echo "Done."`
Hope that will help you.

Shell Script to load multiple FTP files

I am trying to upload multiple files from one folder to a ftp site and wrote this script:
#!/bin/bash
for i in '/dir/*'
do
if [-f /dir/$i]; then
HOST='x.x.x.x'
USER='username'
PASSWD='password'
DIR=archives
File=$i
ftp -n $HOST << END_SCRIPT
quote USER $USER
quote PASS $PASSWD
ascii
put $FILE
quit
END_SCRIPT
fi
It is giving me following error when I try to execute:
username#host:~/Documents/Python$ ./script.sh
./script.sh: line 22: syntax error: unexpected end of file
I can't seem to get this to work. Any help is much appreciated.
Thanks,
Mayank
It's complaining because your for loop does not have a done marker to indicate the end of the loop. You also need more spaces in your if:
if [ -f "$i" ]; then
Recall that [ is actually a command, and it won't be recognized if it doesn't appear as such.
And... if you single quote your glob (at the for) like that, it won't be expanded. No quotes there, but double quotes when using $i. You probably also don't want to include the /dir/ part when you use $i as it's included in your glob.
If I'm not mistaken, ncftp can take wildcard arguments:
ncftpput -u username -p password x.x.x.x archives /dir/*
If you don't already have it installed, it's likely available in the standard repo for your OS.
First, the literal, fixing-your-script answer:
#!/bin/bash
# no reason to set variables that don't change inside the loop
host='x.x.x.x'
user='username'
password='password'
dir=archives
for i in /dir/*; do # no quotes if you want the wildcard to be expanded!
if [ -f "$i" ]; then # need double quotes and whitespace here!
file=$i
ftp -n "$host" <<END_SCRIPT
quote USER $user
quote PASS $password
ascii
put $file $dir/$file
quit
END_SCRIPT
fi
done
Next, the easy way:
lftp -e 'mput -a *.i' -u "$user,$password" "ftp://$host/"
(yes, lftp expands the wildcard internally, rather than expecting this to be done by the outer shell).
First of all my apologies in not making myself clear in the question. My actual task was to copy a file from local folder to a SFTP site and then move the file to an archive folder. Since the SFTP is hosted by a vendor I cannot use the key sharing (vendor limitation. Also, SCP will require password entering if used in a shell script so I have to use SSHPASS. SSHPASS is in the Ubuntu repo however for CentOS it needs to be installed from here
Current thread and How to run the sftp command with a password from Bash script? did gave me better understanding on how to write the script and I will share my solution here:
#!/bin/bash
#!/usr/bin
for i in /dir/*; do
if [ -f "$i" ]; then
file=$i
export SSHPASS=password
sshpass -e sftp -oBatchMode=no -b - user#ftp.com << !
cd foldername/foldername
put $file
bye
!
mv $file /somedir/test
fi
done
Thanks everyone for all the responses!
--Mayank

read the contents of a directory using shell script

I'm trying to get the contents of a directory using shell script.
My script is:
for entry in `ls`; do
echo $entry
done
However, my current directory contains many files with whitespaces in their names. In that case, this script fails.
What is the correct way to loop over the contents of a directory in shell scripting?
PS: I use bash.
for entry in *
do
echo "$entry"
done
don't parse directory contents using ls in a for loop. you will encounter white space problems. use shell expansion instead
for file in *
do
if [ -f "$file" ];then
echo "$file"
fi
done

Shell scripting debug help - Iterating through files in a directory

#!/bin/sh
files = 'ls /myDir/myDir2/myDir3/'
for file in $files do
echo $file
java myProg $file /another/directory/
done
What i'm trying to do is iterate through every file name under /myDir/myDir2/myDir3/, then use that file name as the first argument in calling a java program (second argument is "/another/directory")
When I run this script: . myScript.sh
I get this error:
-bash: files: command not found
What did I do wrong in my script? Thanks!
Per Neeaj's answer, strip off the whitespace from files =.
Better yet, use:
#!/bin/sh -f
dir=/myDir/MyDir2/MyDir3
for path in $dir/*; do
file=$(basename $path)
echo "$file"
java myProg "$file" arg2 arg3
done
Bash is perfectly capable of expanding the * wildcard itself, without spawning a copy of ls to do the job for it!
EDIT: changed to call basename rather than echo to meet OP's (previously unstated) requirement that the path echoed be relative and not absolute. If the cwd doesn't matter, then even better I'd go for:
#!/bin/sh -f
cd /myDir/MyDir2/MyDir3
for file in *; do
echo "$file"
java myProg "$file" arg2 arg3
done
and avoid the calls to basename altogether.
strip off the whitespace in and after files = as files=RHS of assignment
Remove the space surrounding the '=' : change
files = 'ls /myDir/myDir2/myDir3/'
into:
files='ls /myDir/myDir2/myDir3/'
and move the 'do' statement to its own line:
for file in $files
do
....
quote your variables and no need to use ls.
#!/bin/sh
for file in /myDir/myDir2/*
do
java myProg "$file" /another/directory/
done

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