I want to use a script that checks whether a list of directories exists or not and at the same time it should print some custom message that I am sending.
For example:
I have a script that validates if directory exists or not:
**check.sh**
for i in $*
if [ -d "$i" ]; then
echo Found <msg-i> directory.
else
echo <msg-i> directory not found.
Now I want to call this script like this:
./check.sh $DIR1 msg1 $Dir2 msg2 $Dir3 msg3
So if DIR1 doesn't exist then I want to display message as "msg1 directory not found", similarly for DIR2 I want to show "msg2 directory not found". Here msg1 and msg2 are something I want to pass as string. How to achieve this? I am using bash shell.
Try this:
while [ -n "$1" ]
do
dir="$1"
msg="$2"
if [ -d "$dir" ]; then
echo "$msg dir FOUND"
else
echo "$msg dir NOT FOUND"
fi
shift 2
done
shift <n> command simply shifts left positional parameters passed to the script of n positions.
For example if you call a script with:
./myscript 1 2 3 4
$1 is "1" and $2 is "2"
but if you shift 2 then $1 is "3" and $2 is "4".
In this way the loop consumes 2 parameters per cycle until $1 parameter is an empty string ( -n "$1").
while condition can be written more elegantly as:
while (( $# ))
obtaining the same result.
You can also check for the second parameter (while [ -n "$2" ]) but this changes the behavior when user provides an odd number of parameters:
in the first case last directory will be checked but you'll have a strange message because $msg il empty
il the second case you'll not have strange messages, but last directory will silently not be checked
Better test parameters at the beginning:
if (( $# % 2 ))
then
echo "Provide an even number of parameters"
exit 1
fi
Chepner Says:
The while condition can simply be (( $# )) (test if the number of positional parameters is non-zero).
Chaitanya Says:
Hi Chepner, thanks for providing alternate solution, can you please tell me how the while condition should actually look like in order to use $# , I tried different ways but it is not working for me.
Here's a quick sample:
while (( $# ))
do
dir=$1
msg=$2
shift 2
[...]
done
The while (( $# )) will be true as long as there are any command line arguments. Doing the shift twice removes arguments from the list. When no more arguments, the while loop ends.
#Zac has the correct answer.
One tip for the message: use a printf format string:
./check.sh dir1 "can't locate %s directory"
and in the script:
if [[ ! -d "$dir" ]]; then
printf "$msg" "$dir"
Related
I was doing this little script in which the first argument must be a path to an existing directory and the second any other thing.
Each object in the path indicated in the first argument must be renamed so that the new
name is the original that was added as a prefix to the character string passed as the second argument. Example, for the string "hello", the object OBJECT1 is renamed hello.OBJECT1 and so on
Additionally, if an object with the new name is already present, a message is shown by a standard error output and the operation is not carried out continuing with the next object.
I have the following done:
#! /bin/bash
if [ "$#" != 2 ]; then
exit 1
else
echo "$2"
if [ -d "$1" ]; then
echo "directory"
for i in $(ls "$1")
do
for j in $(ls "$1")
do
echo "$i"
if [ "$j" = "$2"."$i" ]; then
exit 1
else
mv -v "$i" "$2"."$i"
echo "$2"."$i"
fi
done
done
else
echo "no"
fi
fi
I am having problems if I run the script from another file other than the one I want to do it, for example if I am in /home/pp and I want the changes to be made in /home/pp/rr, since that is the only way It does in the current.
I tried to change the ls to catch the whole route with
ls -R | sed "s;^;pwd;" but the route catches me badly.
Using find you can't because it puts me in front of the path and doesn't leave the file
Then another question, to verify that that object that is going to create new is not inside, when doing it with two for I get bash errors for all files and not just for coincidences
I'm starting with this scripting, so it has to be a very simple solution thing
An obvious answer to your question would be to put a cd "$2 in the script to make it work. However, there are some opportunities in this script for improvement.
#! /bin/bash
if [ "$#" != 2 ]; then
You might put an error message here, for example, echo "Usage: $0 dir prefix" or even a more elaborate help text.
exit 1
else
echo $2
Please quote, as in echo "$2".
if [ -d $1 ]; then
Here, the quotes are important. Suppose that your directory name has a space in it; then this if would fail with bash: [: a: binary operator expected. So, put quotes around the $1: if [ -d "$1" ]; then
echo "directory"
This is where you could insert the cd "$1".
for i in $(ls $1)
do
It is almost always a bad idea to parse the output of ls. Once again, this for-loop will fail if a file name has a space in it. A possible improvement would be for i in "$1"/* ; do.
for j in $(ls $1)
do
echo $i
if [ $j = $2.$i ]; then
exit 1
else
The logic of this section seems to be: if a file with the prefix exists, then exit instead of overwriting. It is always a good idea to tell why the script fails; an echo before the exit 1 will be helpful.
The question is why you use the second loop? a simple if [ -f "$2.$i ] ; then would do the same, but without the second loop. And it will therefore be faster.
mv -v $i $2.$i
echo $2.$i
Once again: use quotes!
fi
done
done
else
echo "no"
fi
fi
So, with all the remarks, you should be able to improve your script. As tripleee said in his comment, running shellcheck would have provided you with most of the comment above. But he also mentioned basename, which would be useful here.
With all that, this is how I would do it. Some changes you will probably only appreciate in a few months time when you need some changes to the script and try to remember what the logic was that you had in the past.
#!/bin/bash
if [ "$#" != 2 ]; then
echo "Usage: $0 directory prefix" >&2
echo "Put a prefix to all the files in a directory." >&2
exit 1
else
directory="$1"
prefix="$2"
if [ -d "$directory" ]; then
for f in "$directory"/* ; do
base=$(basename "$f")
if [ -f "Sdirectory/$prefix.$base" ] ; then
echo "This would overwrite $prefix.$base; exiting" >&2
exit 1
else
mv -v "$directory/$base" "$directory/$prefix.$base"
fi
done
else
echo "$directory is not a directory" >&2
fi
fi
While iterating through the arguments, how do you determine if there is a next argument?
The way I tried to approach this was to check if the next argument is not empty but I ran into some problems.
Here in this example I print the value of the current argument and if there is an argument that comes after that then print some message.
My approach:
use $i+1 where $i+1 will give you the value of the next index.
#!/bin/sh
for i in "$#"
do
echo $i
if ! [ ${i+1}="" ]; then
echo "test"
fi
done
sh test 1 2 3 4 5
but that didn't work. I also tried expr i + 1, but that didn't work as well.
If anyone could give me a hint on how to approach this problem that would be really appreciated.
#!/bin/sh
while [ $# -gt 0 ] ; do
echo $1
if [ -n "${2+x}" ]; then
echo another arg follows
fi
shift
done
$ ./test.sh 1 2 3
1
another arg follows
2
another arg follows
3
The trick here is that we use shift for consuming the argument list instead of iterating over it. The next argument is always $1, which we know exists because we only execute the loop if $# (the count of the positional arguments, not including $0) is positive. To check whether the argument after that, $2, exist, we can use the ${PARAM+WORD} expansion, which produces nothing if PARAM doesn't exist, otherwise produces WORD.
Of course, shift destroys the argument list. If you don't want that, move things into a function. The following example shows how we can process the same argument list twice by passing a copy into a function in which shift locally eats it:
#!/bin/sh
func() {
while [ $# -gt 0 ] ; do
echo $1
if [ -n "${2+x}" ]; then
echo another arg follows
fi
shift
done
}
func "$#"
func "$#"
$ ./test.sh 1 2 3
1
another arg follows
2
another arg follows
3
1
another arg follows
2
another arg follows
3
You can use a counter and check for $#:
n=1
for i in "$#"; do
echo "$i"
if [ $n -eq $# ]; then
echo "test"
fi
n=$(expr $n + 1)
done
I have to write a shell script that creates a file containing the name of each text files from a folder (given as parameter) and it's subfolders that contain words longer than n characters (read n from keyboard).
I wrote the following code so far :
#!/bin/bash
Verifies if the first given parameter is a folder:
if [ ! -d $1 ]
then echo $1 is not a directory\!
exit 1
fi
Reading n
echo -n "Give the number n: "
read n
echo "You entered: $n"
Destination where to write the name of the files:
destinatie="destinatie"
the actual part that i think it makes me problems:
nr=0;
#while read line;
#do
for fisier in `find $1 -type f`
do
counter=0
for word in $(<$fisier);
do
file=`basename "$fisier"`
length=`expr length $word`
echo "$length"
if [ $length -gt $n ];
then counter=$(($counter+1))
fi
done
if [ $counter -gt $nr ];
then echo "$file" >> $destinatie
fi
done
break
done
exit
The script works but it does a few more steps that i don't need.It seems like it reads some files more than 1 time. If anyone can help me please?
Does this help?
egrep -lr "\w{$n,}" $1/* >$destinatie
Some explanation:
\w means: a character that words consist of
{$n,} means: number of consecutive characters is at least $n
Option -l lists files and does not print the grepped text and -r performs a recursive scan on your directory in $1
Edit:
a bit more complete version around the egrep command:
#!/bin/bash
die() { echo "$#" 1>&2 ; exit 1; }
[ -z "$1" ] && die "which directory to scan?"
dir="$1"
[ -d "$dir" ] || die "$dir isn't a directory"
echo -n "Give the number n: "
read n
echo "You entered: $n"
[ $n -le 0 ] && die "the number should be > 0"
destinatie="destinatie"
egrep -lr "\w{$n,}" "$dir"/* | while read f; do basename "$f"; done >$destinatie
This code has syntax errors, probably leftovers from your commented-out while loop: It would be best to remove the last 3 lines: done causes the error, break and exit are unnecessary as there is nothing to break out from and the program always terminates at its end.
The program appears to output files multiple times because you just append to $destinatie. You could simply delete that file when you start:
rm "$destinatie"
You echo the numbers to stdout (echo "$length") and the file names to $destinatie (echo "$file" >> $destinatie). I do not know if that is intentional.
I found the problem.The problem was the directory in which i was searching.Because i worked on the files from the direcotry and modified them , it seems that there remained some files which were not displayed in file explorer but the script would find them.i created another directory and i gived it as parameter and it works. Thank you for your answers
.
I am making a bash script that you have to give 2 files or more as arguments.
I want to test if the given files exist. I'm using a while loop because I don't know how many files are given. The problem is that the if statement sees the $t as a number and not as the positional parameter $number. Does somebody have a solution?
t=1
max=$#
while [ $t -le $max ]; do
if [ ! -f $t ]; then
echo "findmagic.sh: $t is not a regular file"
echo "Usage: findmagic.sh file file ...."
exit
fi
t=`expr $t + 1`
done
You can do it with the bash Special parameter # in this way:
script_name=${0##*/}
for t in "$#"; do
if [ ! -f "$t" ]; then
echo "$script_name: $t is not a regular file"
echo "Usage: $script_name file file ...."
exit 1
fi
done
With "$#" you are expanding the positional parameters, starting from one as separate words (your arguments).
Besides, remember to provide a meaningful exit status (e.g. exit 1 instead of exit alone). If not provided, the exit status is that of the last command executed (echo in your case, which succes, so you're exiting with 0).
And for last, instead of write the script name (findmagic.sh in your case), you can set a variable at the beginning in your script:
script_name=${0##*/}
and then use $script_name when necessary. In this way you don't need to update your script if it changes its name.
I wrote a bash script that uploads a file on my home server. It gets activated from a folder action script using applescript. The setup is the folder on my desktop is called place_on_server. Its supposed to have an internal file structure exactly like the folder I want to write to: /var/www/media/
usage goes something like this:
if directory etc added to place_on_server: ./upload DIR etc
if directory of directory: etc/movies ./upload DIR etc movies //and so on
if file to place_on_server: ./upload F file.txt
if file in file in place_on_server ./upload F etc file.txt //and so on
for creating a directory its supposed to execute a command like:
ssh root#192.168.1.1<<EOF
cd /var/www/media/wherever
mkdir newdirectory
EOF
and for file placement:
rsync -rsh='ssh -p22' file root#192.168.1.1:/var/www/media/wherever
script:
#!/bin/bash
addr=$(ifconfig -a | ./test)
if ($# -le "1")
then
exit
elif ($1 -eq "DIR")
then
f1="ssh -b root#$addr<<EOF"
list = "cd /var/www/media\n"
if($# -eq "2")
then
list=list+"mkdir $2\nEOF\n"
else
num=2
i=$(($num))
while($num < $#)
do
i=$(($num))
list=list+"mkdir $i\n"
list=list+"cd $i\n"
$num=$num+1
done
fi
echo $list
elif ($1 -eq "F")
then
#list = "cd /var/www/media\n"
f2="rsync -rsh=\'ssh -p22\' "
f3 = "root#$addr:/var/www/media"
if($# -eq "2")
then
f2=f2+$2+" "+f3
else
num=3
i=$(($num))
while($num < $#)
do
i=$(($num))
f2=f2+"/"+$i
$num=$num+1
done
i=$(($num))
f2=f2+$i+" "+$f3
fi
echo $f2
fi
exit
output:
(prompt)$ ./upload2 F SO test.txt
./upload2: line 3: 3: command not found
./upload2: line 6: F: command not found
./upload2: line 25: F: command not found
So as you can see I'm having issues handling input. Its been awhile since I've done bash. And it was never extensive to begin with. Looking for a solution to my problem but also suggestions. Thanks in advance.
For comparisons, use [[ .. ]]. ( .. ) is for running commands in subshells
Don't use -eq for string comparisons, use =.
Don't use < for numerical comparisons, use -lt
To append values, f2="$f2$i $f3"
To add line feeds, use $'\n' outside of double quotes, or a literal linefeed inside of them.
You always need "$" on variables in strings to reference them, otherwise you get the literal string.
You can't use spaces around the = in assignments
You can't use $ before the variable name in assignments
To do arithmetics, use $((..)): result=$((var1+var2))
For indirect reference, such as getting $4 for n=4, use ${!n}
To prevent word splitting removing your line feeds, double quote variables such as in echo "$line"
Consider writing smaller programs and checking that they work before building out.
Here is how I would have written your script (slightly lacking in parameter checking):
#!/bin/bash
addr=$(ifconfig -a | ./test)
if [[ $1 = "DIR" ]]
then
shift
( IFS=/; echo ssh "root#$addr" mkdir -p "/var/www/media/$*"; )
elif [[ $1 = "F" ]]
then
shift
last=$#
file=${!last}
( IFS=/; echo rsync "$file" "root#$addr:/var/www/media/$*" )
else
echo "Unknown command '$1'"
fi
$* gives you all parameters separated by the first character in $IFS, and I used that to build the paths. Here's the output:
$ ./scriptname DIR a b c d
ssh root#somehost mkdir -p /var/www/media/a/b/c/d
$ ./scriptname F a b c d somefile.txt
rsync somefile.txt root#somehost:/var/www/media/a/b/c/d/somefile.txt
Remove the echos to actually execute.
The main problem with your script are the conditional statements, such as
if ($# -le "1")
Despite what this would do in other languages, in Bash this is essentially saying, execute the command line $# -le "1" in a subshell, and use its exit status as condition.
in your case, that expands to 3 -le "1", but the command 3 does not exist, which causes the error message
./upload2: line 3: 3: command not found
The closest valid syntax would be
if [ $# -le 1 ]
That is the main problem, there are other problems detailed and addressed in that other guy's post.
One last thing, when you're assigning value to a variable, e.g.
f3 = "root#$addr:/var/www/media"
don't leave space around the =. The statement above would be interpreted as "run command f3 with = and "root#$addr:/var/www/media" as arguments".