I want to know if there's any way to use the 'find' command in bash for testing. in other words, how do I go about making a conditional statement that tests whether a file was found with 'find' or not?
To expound more on what I mean, I've included the following code. The purpose of which should:
find a directory inputted by user
find a file in that directory, filename inputted by user
find a word in that file, which is also inputted by user
#!/bin/bash
#Task9.sh
cd /
echo "please enter a directory"
read direc
path= find home -depth -name "$direc"
if [ -z $path ]; then
echo "not a valid directory"
exit 1
else
cd $path
counter=3
while [ $counter -gt 0 ]; do
echo "enter a filename"
read FileName
FilePath= find $direc -depth -name "$FileName"
if [ -z $FilePath ]; then
break
else
let counter--
fi
done
if [ -z $FilePath ]; then
cd $FilePath
echo "input a word"
read SearchWord
echo "Found!"
grep $SearchWord $FileName
else
echo "No such file"
exit 2
fi
fi
exit 0
I'm very new to Bash, so I apologize for any easy mistakes I have overlooked. I'm not used to parsing the language yet. Thank you to anyone who helps me, it is greatly appreciated
you're using the find command plus the shell's -z $foo operator to figure out whether or not the user has specified the path to an existing file. There might be a way just to ask bash directly if the file exists or not, using a file-based operator, instead of running an unnecessary command and then using a string-based operator on the result (hint: there is). Some of the links on this page might help you:
http://tldp.org/LDP/abs/html/tests.html
good luck!
Related
New to shell scripting and I want to test to see if the variables I created are valid directories and if not send the user into a while loop to enter the directory and only allow exit when a valid directory is entered.
So far this is what my script looks like:
~/bin/bash
source_dir="$1"
dest_dir="$2"
mkdir /#HOME/$source_dir
mkdir /#HOME$dest_dir
if [ -d "$source_dir" ]
then
echo "$source_dir is a valid directory"
fi
while [[ ! -d "$source_dir" ]]
do
echo "Please enter a valid directory"
read source_dir
done
Is there any way to combine these into a single statement?
The while code will never execute if the directory is valid. Therefore just move the echo "$source_dir is a valid directory" after the loop:
#!/bin/bash
source_dir="$1"
dest_dir="$2"
mkdir "$HOME/$source_dir"
mkdir "$HOME/$dest_dir"
until [[ -d "$source_dir" ]]
do
read -p "Please enter a valid directory" source_dir
done
echo "$source_dir is a valid directory"
Notes:
a few code typos were fixed, e.g. /#HOME$dest_dir should be "$HOME/$dest_dir".
any while ! can be shortened to until.
The above code lacks a few things:
It tries create a new dir, then if that fails, has the user enter an already existing directory. It might be better to let the user create a new directory, but only if it doesn't already exist.
It would be better to check if $dest_dir exists.
Here's a more thorough approach using a shell function:
#!/bin/bash
untilmkdir ()
{
d="$1";
until mkdir "$d" ; do
read -p "Please enter a valid directory: " d
[ -d "$d" ] && break
done;
echo "$d is a valid directory" 1>&2
echo "$d"
}
source_dir=$(untilmkdir "$HOME/$1")
dest_dir=$(untilmkdir "$HOME/$2")
Notes:
The prompts in untilmkdir are printed to stderr.
The name of whatever directory untilmkdir creates is printed to stdout.
Having untilmkdir print to both stderr and stdout allows storing the successfully created name to a variable.
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.
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.
I'm writing a reasonably lengthy script (or what I would consider lengthy - you could probably do it in a few hours). I basically have a file (named .restore.info) which contains files of names. In part of the script, I want to test "If cannot find filename in .restore.info, then says “Error: restored file does not exist”. Apologies if this doesn't make sense for you guys (for me, it does in the grand scheme of things). So if type this in the command line:
sh MYSCRIPT filename
It will search for the string filename in the .restore.info file, and if it cant find anything, it should produce an error message.
Basically, I need the top line of this coded translated into a UNIX bash statement and something that actually makes sense!:
If grep $1 .restore.info returns an exist status of 1; then
echo “Filename does not exist!”
fi
Thanks in advance! Please ask me if you need me to clarify anything more clearly as I know I'm not the best explainer, and I'll get back to you in less than a minute! (+rep and best answer of course will be given!)
You probably only care if grep exits with a non-zero exit status:
if ! grep -q "$1" .restore.info; then
echo "Filename does not exist!"
fi
but if you really do care about a specific exit status (1, in this case):
if ! grep -q "$1" .restore.info && [[ $? -eq 1 ]]; then
echo "Filename does not exist!"
fi
Use grep -q
grep -q "filename" .restore.info && echo "found match"
or
! grep -q "filename" .restore.info && echo "not found"
grep -l 'filename' .restore.info
if [ $? = 0 ];then
echo "found it"
else
echo "not found"
fi