I'm trying to go the path and remove the files starting collectorlist*, so I write the shell script like below., however -e option always give output not found. kindly suggest me.
Script
if [ -d "/abc" ] ; then
for c in `ls -1 "/abc/"`; do
if [ -d "/abc/${c}" ] ; then
if [ -d "/abc/${c}/dynatrace/agent/linux-x86-64/agent/conf/" ] ; then
if [ -e "/bxp/${c}/dynatrace/agent/linux-x86-64/agent/conf/collectorlist* " ] && echo "Found" || echo "Not found" ; then
`ls -ltr "sat.txt"` > /users/home/abc/test.txt
fi
fi
fi
done
fi
There are some problems with your code.
There's a trailing space after the asterisk and before the closing double quote.
You are using an asterisk within double quotes; within double quotes wildcards like the asterisk are not evaluated but taken verbatim.
You are using backticks together with redirection. This will execute the stuff within the backticks, replace the backtick-expression by the output of it and execute the resulting line. So your ls command needs to give out the name of a command which then will be executed. I guess you don't want that. Just remove the backticks, you probably just want to execute ls and redirect its output into a file.
You try to find out if any file matching a pattern exists and use [ -e pattern* ] for this. The pattern* evaluates to one or more file names or (of none exists) to the pattern* itself (unless you tweak your shell to behave differently). The problem is the case of the many values. These will make a problem within your [ ... ] expression.
Result:
if [ -d "/abc" ] ; then
for c in $(ls -1 "/abc/"); do
if [ -d "/abc/${c}/dynatrace/agent/linux-x86-64/agent/conf/" ] ; then
if ls "/bxp/${c}/dynatrace/agent/linux-x86-64/agent/conf"/collectorlist* &> /dev/null; then
echo "Found"
ls -ltr "sat.txt" > /users/home/abc/test.txt
else
echo "Not Found"
fi
fi
done
fi
Related
I was doing this little script in which the first argument must be a path to an existing directory and the second any other thing.
Each object in the path indicated in the first argument must be renamed so that the new
name is the original that was added as a prefix to the character string passed as the second argument. Example, for the string "hello", the object OBJECT1 is renamed hello.OBJECT1 and so on
Additionally, if an object with the new name is already present, a message is shown by a standard error output and the operation is not carried out continuing with the next object.
I have the following done:
#! /bin/bash
if [ "$#" != 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
I am trying to create a "Lambda" style WHERE script.
I want lambdaWHERE to take piped input and pass it through if condition after given as arguments is met. Like xargs I use {} to represent what comes down the pipe.
I call command like:
ls -d EqAAL* | lambdaWHERE.sh -f {}/INFO_ACTIVETICK
I want the folder names passed through if they contain a file called INFO_ACTIVETICK
Here is the script:
#!/bin/sh
#set -x
ARGS=$*
while read i
do
CMD=`echo $ARGS | sed 's/{}/'$i'/g'`
if [[ $CMD ]]
then
echo $i
fi
done
But when I run it a mysterious "-n" appears...
$ ls -d EqAAL* | /q/lambdaWHERE.sh -f {}/INFO_ACTIVETICK
+ ARGS='-f {}/INFO_ACTIVETICK'
+ read i
++ echo -f '{}/INFO_ACTIVETICK'
++ sed 's/{}/EqAAL-1m/g'
+ CMD='-f EqAAL-1m/INFO_ACTIVETICK'
+ [[ -n -f EqAAL-1m/INFO_ACTIVETICK ]]
+ echo EqAAL-1m
EqAAL-1m
+ read i
How can I get the bit in the [[ ]] correct?
You were quite close. you only need to switch to the standard POSIX [ $CMD ] and it will work.
The main difference between using [[ $CMD ]] and [ $CMD ] is that the first has fewer surprises and you need not quote variables. That also means that a variable is though of as one token and cannot have a whole expression in it like you are trying. [ $CMD ] however works the same way as the original shell where [ was just a command an thus need explicit quotations in order to interpret something with spaces as one argument.
There is a relevant question about the differences between [[ ...]] and [ ..]
I want to see if files exist in a certain folder and if they exist run certain event else skip this. Bash see files in my empty folder.
I tried a few different ways and all show that a file exits.
$ [ -f ] && echo "Found" || echo "Not found"
Found
$ [ -r ] && echo "Found" || echo "Not found"
Found
$ [ -e ] && echo "Found" || echo "Not found"
Found
Whereas it does not:
$ ls -lrt
total 0
What am I missing here?
My actual code is:
get_last_parsed_file_time ()
{
if [ -s "$DATA_DIR$EPG_XML_FILES" ]
then
NEWEST_FILE=$( ls -tr | tail -1 )
LAST_PARSED_TIME=$( stat -c %Y $NEWEST_FILE )
else
NO_FILE=1
fi
}
[ -f filename ] checks if filename is a file that exists.
[ str ] checks whether str is non-empty.
In your case, str is -f, but it still just checks whether string "-f" is nonempty, which it obviously isn't (it's two characters, a dash and an "f").
This is especially puzzling in the case of [ -f $filename ]. When $filename is empty and unquoted, the command will become [ -f ] and will be true. Always quote your variables.
that other guy's answer explains why [ -f ] doesn't work.
More generally, it is important to note that you cannot pass globs (filename patterns such as * to a bash file-test operator such as -f. The file-test operators expect a single, literal filename or file path (either as a string literal or as a string stored in a variable).
[[ -f '/path/to/file' ]] # correct
[[ -f /path/to/* ]] # WRONG
Here's a snippet that tests whether a folder is empty, i.e. whether it contains ANY items - whether files, hidden files, or subfolders:
folder='/path/to/folder'
[[ -n $(ls -A "$folder" | head -n 1) ]] && echo "NON-empty" || echo "EMPTY"
To only consider non-hidden items, remove -A from the ls command.
To quietly ignore the case where the input folder doesn't exist or is not accessible, append 2> /dev/null to the ls command.
#!/bin/bash
traverse() {
local x=$1
if [ -d $x ]
then
lst=(`ls $x`)
for((i=${#lst[#]}; --i;)); do
echo "${lst[i]}"
done
else echo "not a directory"
fi
}
traverse
I want to pass a parameter such as "/path/to/this/directory/" when executing program but only works if I'm running the program in the same directory as my bash script file and any other parameter I pass is completely ignored.
the script is supposed to take a parameter and check if it's a directory and if it's a directory then list all the files/folders in descending order. If not display error message.
What is wrong with code, thanks!
This happens because $1 in the function refers to traverse's parameters, not your script's parameters.
To run your function once with each argument, use
for arg in "$#" # "$#" is also the default, so you can drop the 'in ..'
do
traverse "$arg"
done
If you in the future want to pass all the script's parameters to a function, use
myfunc "$#"
This is just the problem at hand, though. Other problems include not quoting your variables and using command expansion of ls, lst=(`ls $x`), instead of globs, lst=( "$x"/* )
You don't need to call ls for that. You can use this code:
traverse() {
local x="$1"
if [ -d "$x" ]; then
arr=( "$x/"* )
for ((i=${#arr[#]}; i>0; i--)); do
echo "${arr[$i]}"
done
else
echo "not a directory"
fi
}
"That other guy" has the right answer. The reason it always looks at the current directory:
you invoke traverse with no arguments
the $1 in the traverse function is empty, therefore $x is empty
the test is therefore [ -d ], and when [ is given 1 argument, it returns success if the argument is not empty. Your if command always executes the "true" block and ls $x is just ls when x is empty
Use [[ ... ]] with bash: it is smarter about empty arguments. Otherwise, quote your variables:
$ x=; [ -d $x ] && echo always true || echo not a directory
always true
$ x=; [[ -d $x ]] && echo always true || echo not a directory
not a directory
$ x=; [ -d "$x" ] && echo always true || echo not a directory
not a directory
I am trying to run the below logic
if [-f $file] && [$variable -lt 1] ; then
some logic
else
print "This is wrong"
fi
It fails with the following error
MyScipt.ksh[10]: [-f: not found
Where 10th line is the if condition , I have put in .
I have also tried
if [-f $file && $variable -lt 1] ; then
which gives the same error.
I know this is a syntax mistake somehwere , but I am not sure , what is the correct syntax when I am using multiple conditions with && in a if block
[ is not an operator, it's the name of a program (or a builtin, sometimes). Use type [ to check. Regardless, you need to put a space after it so that the command line parser knows what to do:
if [ -f $file ]
The && operator might not do what you want in this case, either. You should probably read the bash(1) documentation. In this specific case, it seems like what you want is:
if [ -f $file -a $variable -lt 1 ]
Or in more modern bash syntax:
if [[ -f $file && $variable -lt 1 ]]
The [ syntax is secretly a program!
$ type [
[ is a shell builtin
$ ls -l $(which [)
-rwxr-xr-x 1 root root 35264 Nov 19 16:25 /usr/bin/[
Because of the way the shell parses (technically "lexes") your command line, it sees this:
if - keyword
[-f - the program [-f
$file] - A string argument to the [-f program, made by the value of $file and ]. If $file was "asdf", then this would be asdf]
And so forth, down your command. What you need to do is include spaces, which the shell uses to separate the different parts (tokens) of your command:
if [ -f "$file" ]; then
Now [ stands on its own, and can be recognized as a command/program. Also, ] stands on its own as an argument to [, otherwise [ will complain. A couple more notes about this:
You don't need to put a space before or after ;, because that is a special separator that the shell recognizes.
You should always "double quote" $variables because they get expanded before the shell does the lexing. This means that if an unquoted variable contains a space, the shell will see the value as separate tokens, instead of one string.
Using && in an if-test like that isn't the usual way to do it. [ (also known as test) understands -a to mean "and," so this does what you intended:
if [ -f "$file" -a "$variable" -lt 1 ]; then
Use -a in an if block to represent AND.
Note the space preceding the -f option.
if [ -f $file -a $variable -lt 1] ; then
some logic
else
print "This is wrong"
fi