I have a problem that has been bugging me for a few hours now. I have created a parameter --file-dir using getopt, which assigns a directory for the program to use. Following the parameter, the user has the choice to choose whatever directory they please. To keep the program stable, I check to see whether that directory even exists. The following code is what I have currently and it always returns "Directory does not exist. Terminating." even when I search for my /home directory.
-a|--file-dir) FILE_DIR=$2 ;
if [ ! -d "$FILE_DIR" ]; then
echo "Directory does not exist. Terminating." ;
exit 1;
else
echo "Directory exists." ;
fi ;
shift;;
Any input is much appreciated. The getopt's work fine with echo tests and such but fail when checking for directories.
It would be a good idea to check if you're really having the right argument for it:
-a|--file-dir) FILE_DIR=$2 ;
if [ ! -d "$FILE_DIR" ]; then
echo "Directory \"$FILE_DIR\" does not exist. Terminating." ;
exit 1;
else
echo "Directory exists." ;
fi ;
shift;;
If not certainly the problem is not in the checker but somewhere in your argument-parsing loop.
I had an issue with the same behavior: checking for a directory in the command line worked as expected, but always failed when done in a script.
I was running this script under git bash for Windows:
while read -r i; do
[ ! -d "$i" ] && echo "No $i"
done < "$1"
Windows' line endings (\r\n) can cause issues when splitting lines. Each test actually checks for directory\r instead of directory. Therefore, I needed to run the read command with the correct delimiter:
while IFS=$'\r\n' read -r i; do
It is possible that OP also had a similar issue, where non-printable characters got in the way.
Related
I currently have a directory that is supposed to have 500 files. Each file is of the name form List.1.rds, ... List.500.rds. The way I can see which ones are missing is by the following code in bash:
for((i=1; i<=500; i++)); do name="List.${i}.rds"; [[ ! -e "$name" ]] && echo "missing $name"; done
If a file is missing, it returns the missing file name. However, I would like to go one step further and stop the entire script should any file be missing. Is there a way to do this? thanks.
It can be as simple as setting a flag when a file is missing:
miss=0
for ((i=1;i<=500;i++)); do
file=List.$i.rds
if [[ ! -e $file ]]; then
echo "Missing $file"
miss=1
fi
done
# exit if "miss" flag is 1
((miss)) && exit 1
This a small bash program that is tasked with looking through a directory and counting how many files are in the directory. It's to ignore other directories and only count the files.
Below is my bash code, which seems to fail to count the files specifically in the directory, I say this because if I remove the if statement and just increment the counter the for loop continues to iterate and prints 4 in the counter (this is including directories though). With the if statement it prints this to the console.
folder1 has files
Looking at other questions I think the expression in my if statement is right and I am getting no compilation errors for syntax or another problems.
So I just simply dumbfounded as to why it is not counting the files.
#!/bin/bash
folder=$1
if [ $1 = empty ]; then
folder=empty
counter=0
echo $folder has $counter files
exit
fi
for d in $(ls $folder); do
if [[ -f $d ]]; then
let 'counter++'
fi
done
echo $folder has $counter files
Thank you.
Your entire script could be very well simplified as below with enhancements made. Never use output of ls programmatically. It should be used only in the command-line. The -z construct allows to you assert if the parameter following it is empty or non-empty.
For looping over files, use the default glob expansion provided by the shell. Note the && is a short-hand to do a action when the left-side of the operand returned a true condition, in a way short-hand equivalent of if <condition>; then do <action>; fi
#!/usr/bin/env bash
[ -z "$1" ] && { printf 'invalid argument passed\n' >&2 ; exit 1 ; }
shopt -s nullglob
for file in "$1"/*; do
[ -f "$file" ] && ((count++))
done
printf 'folder %s had %d files\n' "$1" "$count"
I have checked everywhere and tried many different "Solutions" on checking to see if the directory exists. Here's my code:
#!/bin/bash
echo -e "Where is the directory/file located?"
read $DIRECTORY
if [ -d "$DIRECTORY" ]; then
echo "Exists!"
else
echo "Does not exist!"
fi
What I am trying to do is have the user input a directory and for the script to check if it exists or not and return a result. This will ultimately tar/untar a directory. Regardless of whether the directory exists or not, it returns the answer "Does not exist!". (The input i'm trying is ~/Desktop, and from what I know that is 100% correct. Any concise answers are much appreciated :).
Your script can be refactored to this:
#!/bin/bash
read -p 'Where is the directory/file located?' dir
[[ -d "$dir" ]] && echo 'Exists!' || echo 'Does not exist!'
Basically use read var instead of read $var
Better not to use all caps variable names in BASH/shell
Use single quotes while using ! in BASH since it denotes a history event
I'm trying to compile a script that will read user input, and check if the file after the y/n statement. Then it will make files executable. I think the problem with my script is conditional ordering but check it out yourself:
target=/home/user/bin/
cd $target
read -p "This will make the command executable. Are you sure? (y/n)" CONT
if [ "$CONT" == "y" ];
then
chmod +x $1
echo "File $1 is now executable."
else
if [ "$(ls -A /home/user/bin/)" ];
then
echo "File not found."
else
echo "Terminating..."
fi
fi
As I said, I need the script to scan for the file after the y/n statement is printed. The script works fine how it is but still gives the "file is now executable" even if the argument file doesn't exist (but just gives the standard system "cannot find file" message after the echo'd text).
Your script is mostly correct, you just need to check if the file exists first. Also, it's not the best practice to use cd in shell scripts and not needed here.
So re-writing it
#!/bin/bash
target="/home/user/bin/$1"
if [[ ! -f $target ]]; then
echo "File not found."
else
read -p "This will make the command executable. Are you sure? (y/n) " CONT
if [[ $CONT == "y" ]]; then
chmod +x "$target"
echo "File $1 is now executable."
else
echo "Terminating..."
fi
fi
To get an understanding:
Your script will take one argument (a name of a file).
You ask if you want to make that file executable.
If the answer is 'yes', you make the file executable.
Otherwise, you don't.
You want to verify that the file exists too?
I'm trying to understand your logic. What does this:
if [ "$(ls -A /home/user/bin/)" ];
suppose to do. The [ ... ] syntax is a test. And, it has to be one of the valid tests you see here. For example, There's a test:
-e file: True if file exists.
That mean, I can see if your file is under /home/user/bin:
target="/home/user/bin"
if [ -e "$target/$file" ] # The "-e" test for existence
then
echo "Hey! $file exists in the $target directory. I can make it executable."
else
echo "Sorry, $file is not in the $target directory. Can't touch it."
fi
Your $(ls -A /home/user/bin/) will produce a file listing. It's not a valid test like -e unless it just so happens that the first file in your listing is something like -e or -d.
Try to clarify what you want to do. I think this is something more along the lines you want:
#! /bin/bash
target="/home/user/bin"
if [ -z "$1" ] # Did the user give you a parameter
then
echo "No file name given"
exit 2
fi
# File given, see if it exists in $target directory
if [ ! -e "$target/$1" ]
then
echo "File '$target/$1' does not exist."
exit 2
fi
# File was given and exists in the $target directory
read -p"Do you want $target/$1 to be executable? (y/n)" continue
if [ "y" = "$continue" ]
then
chmod +x "$target/$1"
fi
Note how I'm using the testing, and if the testing fails, I simply exit the program. This way, I don't have to keep embedding if/then statements in if/then statements.
Ok so I have written a .sh file in Linux Ubuntu and it works perfectly. However on a Mac it always returns that the file was not found even when it is in the same directory. Can anyone help me out?
.sh file:
if [ ! -f file-3*.jar ]; then
echo "[INFO] jar could not be found."
exit
fi
Just thought I'd add, this isn't for more than one file, it's for a file that is renamed to multiple endings.
In a comment to #Paul R's answer, you said "The shell script is also in the same directory as the jar file. So they can just double click it after assigning SH files to open with terminal by default." I suspect that's the problem -- when you run a shell script by double-clicking it, it runs with the working directory set to the user's home directory, not the directory where the script's located. You can work around this by having the script cd to the directory it's in:
cd "$(dirname "$BASH_SOURCE")"
EDIT: $BASH_SOURCE is, of course, a bash extension not available in other shells. If your script can't count on running in bash, use this instead:
case "$0" in
*/*)
cd "$(dirname "$0")" ;;
*)
me="$(which "$0")"
if [ -n "$me" ]; then
cd "$(dirname "$me")"
else
echo "Can't locate script directory" >&2
exit 1
fi ;;
esac
BTW, the construct [ ! -f file-3*.jar ] makes me nervous, since it'll fail bizarrely if there's ever more than one matching file. (I know, that's not supposed to happen; but things that aren't supposed to happen have any annoying tendency to happen anyway.) I'd use this instead:
matchfiles=(file-3*.jar)
if [ ! -f "${matchfiles[0]}" ]; then
...
Again, if you can't count on bash extensions, here's an alternative that should work in any POSIX shell:
if [ ! -f "$(echo file-3*.jar)" ]; then
Note that this will fail (i.e. act as though the file didn't exist) if there's more than one match.
I think the problem lies elsewhere, as the script works as expected on Mac OS X here:
$ if [ ! -f file-3*.jar ]; then echo "[INFO] jar could not be found."; fi
[INFO] jar could not be found.
$ touch file-302.jar
$ if [ ! -f file-3*.jar ]; then echo "[INFO] jar could not be found."; fi
$
Perhaps your script is being run under the wrong shell, or in the wrong working directory ?
It's not that it doesn't work for you, it doesn't work for your users? The default shell for OS X has changed over the years (see this post) - but it looks like your comment says you have the #! in place.
Are you sure that your users have the JAR file in the right place? Perhaps it's not the script being wrong as much as it's telling you the correct answer - the required file is missing from where the script is being run.
This isn't so much an answer, as a strategy: consider some serious logging. Echo messages such as "[INFO] jar could not be found." both to the screen and to a log file, then add extra logging, such as the values of $PWD, $SHELL and $0 to the log. Then, when your customers/co-workers try to run the script and fail, they can email the log to you.
I would probably use something like
screenlog() {
echo "$*"
echo "$*" >> $LOGFILE
}
log() {
echo "$*" >> $LOGFILE
}
Define $LOGFILE at the top of your script. Then pepper your script with statements like screenlog "[INFO] jar could not be found." or log "\$PWD: $PWD".