I have been using the following code to get the current directory, in each file I am sourcing. In this way I can source the code path relatively from where I am located. However, this does not look that is working when I have 2 sources as below because $curr is overwritten in the first sourced child file.
#!/bin/bash
set +e
curr=$(dirname -- "$(readlink -f -- "$0")")
echo $curr
source $curr/bash/index.sh
echo $curr
source $curr/layout.sh
Is any alternative or option to what I am trying to do here?
Related
I have some csv files in location A like this
abc1.csv,
abc2.csv,
abc3.csv
I have a cron job which runs every 30 mins and in each execution I want to copy only 1 file(which shouldn't be repeated) and placed it in location B
I had though of 2 ways of doing this
1)I will pick the first file in the list of files and copy it to location B and will delete it once copied.Problem with this is I am not sure when the file will get copied completely and if i delete before its completed copied it can be an issue
2)I will have a temp folder.So i will copy the file from location A to location B and also keep it in temp location.In next iteration, when I pick the file from list of files I will compare its existence in the temp file location.If it exists I will move to next file .I think this will be more time consuming etc.
Please suggest if there is any other better way
you can use this bash script for your use case:
source="/path/to/.csv/directory"
dest="/path/to/destination/directory"
cd $source
for file in *.csv
do
if [ ! -f $dest/"$file" ]
then
cp -v $file $dest
break
fi
done
You can ensure you move the already copied file with:
cp abc1.csv destination/ && mv abc1.csv.done
(here you can make your logic to find only *.csv files, and not take into account *.done files.. that have been already processed by your script... or use any suffix you want..
if the cp does not succeed, nothing after that will get executed, so the file will not be moved.
You can also replace mv with rm to delete it:
cp abc1.csv destination/ && rm -f abc1.csv
Further more, you can add to the above commands error messages in case you want to be informed if the cp failed:
cp abc1.csv destination/ && mv abc1.csv.done || echo "copy of file abc1.csv failed"
And get informed via CRON/email output
Finally I took some idea from both the opted solution.Here is the final script
source="/path/to/.csv/directory"
dest="/path/to/destination/directory"
cd $source
for file in *.csv
do
if [ ! -f $dest/"$file" ]
then
cp -v $file $dest || echo "copy of file $file failed"
rm -f $file
break
fi
done
/tmp/target_dir IN_MODIFY,IN_CREATE,IN_MOVED_TO /tmp/script.sh $#
contents of script.sh
echo $1 > /tmp/script.log
on executing
cp -r some_dir /tmp/target_dir
contents of /tmp/script.log
/tmp/target_dir
on executing
cp some_file /tmp/target_dir/some_dir
contents of /tmp/script.log
/tmp/target_dir/some_dir
here instead of watched directory directory on which work is done is echoed.
As far as i understand from here $# is used to display watched system path.
As far as i understand from here $# is used to display watched system path.
Stated there is: When monitoring a directory, the events marked with an asterisk (*) above can occur for files in the directory, in which case the name field in the returned event data identifies the name of the file within the directory.
instead of watched directory directory on which work is done is echoed.
The directory on which work is done is exactly the file within the directory as documented. (Copying some_file into /tmp/target_dir/some_dir has modified /tmp/target_dir/some_dir.)
For a school project, I have a shell script that is supposed to copy the files in two directories (without looking at subdirectories) into a third directory. I'm testing out the -u command so that if two files have the same name, only the newer one will get copied over (that's also a spec). My shell script looks like this (excluding #! and error checking):
cd $1 #first directory
for file in `ls`; do
if [ -f $file ]; then
cp "$file" ../$3 # $3 is the third directory
fi
done
cd ../$2
for file in `ls`; do
if [ -f $file ]; then
cp -u "$file" ../$3
fi
done
My current shell script will copy files that don't exist in directory 3 already, and it won't overwrite a newer file with an older file with the same name. However, my shell script doesn't overwrite an older file with a newer file of the same name in directory 3. I don't think there's anything wrong with the -u command. Can you help find the bug in my code? Thanks!
You are missing the -u option in the first loop:
cp "$file" ../$3 # $3 is the third directory
should instead read:
cp-u"$file" ../$3 # $3 is the third directory
I'm having problems moving files into a folder after I create it in a shell script.
My script looks like:
#!/bin/bash
echo -e "Processing\033[36m" $1 "\033[0mwith the German script";
if [ ! -d ${1%.dat} ]; then
echo -e "making directory\033[33m" ${1%.dat} "\033[0msince it didn't exist...";
mkdir ${1%.dat};
fi
...processing occurs here... (irrelevant to issue)
if [ -d ${1%.dat} ]; then
mv useragents_$1 /${1%.dat}/useragents_$1;
mv summary_$1 /${1%.dat}/summary_$1;
more /${1%.dat}/useragents_$1;
else
echo -e "\033[31mERROR: cannot move files to folder.\033[0m";
fi
As you can see I create the folder if it doesn't exist in the top section and then if it exists I move the files into that folder in the bottom section, the problem is that it doesn't create the folder in time to move the files in (I'm assuming) so when it reaches the lower code, I only get the ERROR.
I tried using, sleep 5, but it only slows down the script and has no effect on the ERROR.
I would really appreciate some advice.
Errors below:
mv: cannot move `useragents_100_stns2_stns6.dat' to `/100_stns2_stns6/useragents_100_stns2_stns6.dat': No such file or directory
mv: cannot move `summary_100_stns2_stns6.dat' to `/100_stns2_stns6/summary_100_stns2_stns6.dat': No such file or directory
/100_stns2_stns6/useragents_100_stns2_stns6.dat: No such file or directory
Pass 1
Your check:
if [ ! -d ${1%.dat} ]; then
should be:
if [ -d ${1%.dat} ]; then
You created the directory; if it is a directory, move stuff into it.
Typo in question
Pass 2
You create:
mkdir ${1%.dat}
You try to move files:
mv useragents_$1 /${1%.dat}/useragents_$1;
Note the leading slash in the move compared to the create. Make those consistent.
Are you sure of this part ? It uses a root directory.
/${1%.dat}/summary_$1;
You probably want to do this instead:
${1%.dat}/summary_$1;
It allows you to move the file into the directory IN your current directory.
I'm converting some Windows batch files to Unix scripts using sh. I have problems because some behavior is dependent on the %~dp0 macro available in batch files.
Is there any sh equivalent to this? Any way to obtain the directory where the executing script lives?
The problem (for you) with $0 is that it is set to whatever command line was use to invoke the script, not the location of the script itself. This can make it difficult to get the full path of the directory containing the script which is what you get from %~dp0 in a Windows batch file.
For example, consider the following script, dollar.sh:
#!/bin/bash
echo $0
If you'd run it you'll get the following output:
# ./dollar.sh
./dollar.sh
# /tmp/dollar.sh
/tmp/dollar.sh
So to get the fully qualified directory name of a script I do the following:
cd `dirname $0`
SCRIPTDIR=`pwd`
cd -
This works as follows:
cd to the directory of the script, using either the relative or absolute path from the command line.
Gets the absolute path of this directory and stores it in SCRIPTDIR.
Goes back to the previous working directory using "cd -".
Yes, you can! It's in the arguments. :)
look at
${0}
combining that with
{$var%Pattern}
Remove from $var the shortest part of $Pattern that matches the back end of $var.
what you want is just
${0%/*}
I recommend the Advanced Bash Scripting Guide
(that is also where the above information is from).
Especiall the part on Converting DOS Batch Files to Shell Scripts
might be useful for you. :)
If I have misunderstood you, you may have to combine that with the output of "pwd". Since it only contains the path the script was called with!
Try the following script:
#!/bin/bash
called_path=${0%/*}
stripped=${called_path#[^/]*}
real_path=`pwd`$stripped
echo "called path: $called_path"
echo "stripped: $stripped"
echo "pwd: `pwd`"
echo "real path: $real_path
This needs some work though.
I recommend using Dave Webb's approach unless that is impossible.
In bash under linux you can get the full path to the command with:
readlink /proc/$$/fd/255
and to get the directory:
dir=$(dirname $(readlink /proc/$$/fd/255))
It's ugly, but I have yet to find another way.
I was trying to find the path for a script that was being sourced from another script. And that was my problem, when sourcing the text just gets copied into the calling script, so $0 always returns information about the calling script.
I found a workaround, that only works in bash, $BASH_SOURCE always has the info about the script in which it is referred to. Even if the script is sourced it is correctly resolved to the original (sourced) script.
The correct answer is this one:
How do I determine the location of my script? I want to read some config files from the same place.
It is important to realize that in the general case, this problem has no solution. Any approach you might have heard of, and any approach that will be detailed below, has flaws and will only work in specific cases. First and foremost, try to avoid the problem entirely by not depending on the location of your script!
Before we dive into solutions, let's clear up some misunderstandings. It is important to understand that:
Your script does not actually have a location! Wherever the bytes end up coming from, there is no "one canonical path" for it. Never.
$0 is NOT the answer to your problem. If you think it is, you can either stop reading and write more bugs, or you can accept this and read on.
...
Try this:
${0%/*}
This should work for bash shell:
dir=$(dirname $(readlink -m $BASH_SOURCE))
Test script:
#!/bin/bash
echo $(dirname $(readlink -m $BASH_SOURCE))
Run test:
$ ./somedir/test.sh
/tmp/somedir
$ source ./somedir/test.sh
/tmp/somedir
$ bash ./somedir/test.sh
/tmp/somedir
$ . ./somedir/test.sh
/tmp/somedir
This is a script can get the shell file real path when executed or sourced.
Tested in bash, zsh, ksh, dash.
BTW: you shall clean the verbose code by yourself.
#!/usr/bin/env bash
echo "---------------- GET SELF PATH ----------------"
echo "NOW \$(pwd) >>> $(pwd)"
ORIGINAL_PWD_GETSELFPATHVAR=$(pwd)
echo "NOW \$0 >>> $0"
echo "NOW \$_ >>> $_"
echo "NOW \${0##*/} >>> ${0##*/}"
if test -n "$BASH"; then
echo "RUNNING IN BASH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${BASH_SOURCE[0]}
elif test -n "$ZSH_NAME"; then
echo "RUNNING IN ZSH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${(%):-%x}
elif test -n "$KSH_VERSION"; then
echo "RUNNING IN KSH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${.sh.file}
else
echo "RUNNING IN DASH OR OTHERS ELSE..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=$(lsof -p $$ -Fn0 | tr -d '\0' | grep "${0##*/}" | tail -1 | sed 's/^[^\/]*//g')
fi
echo "EXECUTING FILE PATH: $SH_FILE_RUN_PATH_GETSELFPATHVAR"
cd "$(dirname "$SH_FILE_RUN_PATH_GETSELFPATHVAR")" || return 1
SH_FILE_RUN_BASENAME_GETSELFPATHVAR=$(basename "$SH_FILE_RUN_PATH_GETSELFPATHVAR")
# Iterate down a (possible) chain of symlinks as lsof of macOS doesn't have -f option.
while [ -L "$SH_FILE_RUN_BASENAME_GETSELFPATHVAR" ]; do
SH_FILE_REAL_PATH_GETSELFPATHVAR=$(readlink "$SH_FILE_RUN_BASENAME_GETSELFPATHVAR")
cd "$(dirname "$SH_FILE_REAL_PATH_GETSELFPATHVAR")" || return 1
SH_FILE_RUN_BASENAME_GETSELFPATHVAR=$(basename "$SH_FILE_REAL_PATH_GETSELFPATHVAR")
done
# Compute the canonicalized name by finding the physical path
# for the directory we're in and appending the target file.
SH_SELF_PATH_DIR_RESULT=$(pwd -P)
SH_FILE_REAL_PATH_GETSELFPATHVAR=$SH_SELF_PATH_DIR_RESULT/$SH_FILE_RUN_BASENAME_GETSELFPATHVAR
echo "EXECUTING REAL PATH: $SH_FILE_REAL_PATH_GETSELFPATHVAR"
echo "EXECUTING FILE DIR: $SH_SELF_PATH_DIR_RESULT"
cd "$ORIGINAL_PWD_GETSELFPATHVAR" || return 1
unset ORIGINAL_PWD_GETSELFPATHVAR
unset SH_FILE_RUN_PATH_GETSELFPATHVAR
unset SH_FILE_RUN_BASENAME_GETSELFPATHVAR
unset SH_FILE_REAL_PATH_GETSELFPATHVAR
echo "---------------- GET SELF PATH ----------------"
# USE $SH_SELF_PATH_DIR_RESULT BEBLOW
I have tried $0 before, namely:
dirname $0
and it just returns "." even when the script is being sourced by another script:
. ../somedir/somescript.sh