when i execute this script that i had made
#!/bin/bash
for object in $(ls -l)
do
echo $object
done
it's displayed as below
when i execute my script i want to have a result like this
i was trying a lot of things but its not working
please i need your help
thank you in advance
You want to read the output of ls -l line by line; to do this, replace the line
for object in $(ls -l)
by
ls -l|while read object
and change
echo $object
to
echo "$object"
(to preserve spaces).
If you insist on using for, you can do:
readarray -t < <(ls -l)
for object in "${MAPFILE[#]}"; do echo "$object"; done
Related
I have a text file which contains multiple paths like below
$ cat directory.txt
/aaaa/bbbbb/ccccc/
/aaaa/bbbbb/eeeee/
/aaaa/bbbbb/ddddd/
I need to change directory to each path in text file and need to get count of files under that paths.Below is the code i used, But it is not working.
i=cat /aaaa/bbbbb/directory.txt
while read $i ;do
cd $i
ls |wc -l
done < /aaaa/bbbbb/count.txt
Actually you're almost there. The line i=... is not needed, read $i should be read i, and you simply need to call ls with the path instead of cd it first.
#!/bin/bash
while read i; do
ls "$i" | wc -l
done < "/xxx/yyy/count.txt"
Thanks every one i tried this code it is working fine
!/bin/bash
for i in cat /nrt/home/directory.txt;
do
cd $i
ls | wc -l
done > /nrt/home/count.txt
I want to list the files recursively in the HOME directory. I'm trying to write my own script , so I should not use the command find or ls. My script is:
#!/bin/bash
minSize=102400;
printFiles() {
for x in "$1/"*; do
if [ -d "$x" ]; then
printFiles "$x";
else
size=$(wc -c "$x");
if [[ "$size" -gt "$minSize" ]]; then
echo "$size";
fi
fi
done
}
printFiles "/~";
So, the problem here is that when I run this script, the terminal throws Line 11: division by 0 and /home/gandalf/Videos/*: No such file or directory. I have not divided by any number, why I'm getting this error?. And the second one?
Alternatively, I can't use find or ls because I have to display the files one by one asking to the user if he want to see the next file or not. This is possible using the command find or ls or only can be done writing my own function?
Thanks.
size=$(wc -c "$x");
That's the line that is failing. When you run that wc command manually you should be able to see why:
$ wc -c /tmp/out
5 /tmp/out
The output contains not only the file size but also the file name. So you can't use $size with the -gt comparator on the next line. One way to fix that is to change the wc line to use cut (or awk, or sed, etc) to keep just the file size.
size=$(wc -c "$x" | cut -f1 -d " ")
A simpler alternative suggested by #mklement0:
size=$(wc -c < "$x")
I searched SO but could not find any relevant post with this specific problem. I would like to know how to call a shell script which is stored in a variable of another shell script.
In the below script I am trying to read service name & corresponding shellscript, check if the service is running, if not, start the service using the shell script associated with that service name. tried multiple options shared in various forums(like 'eval' etc) with no luck. please help to provide your suggestions on this.
checker.sh
#!/bin/sh
while read service
do
servicename=`echo $service | cut -d: -f1`
servicestartcommand=`echo $service | rev | cut -d: -f1 | rev`
if (( $(ps -ef | grep -v grep | grep $servicename | wc -l) > 0 ))
then
echo "$servicename Running"
else
echo "!!$servicename!! Not Running, calling $servicestartcommand"
eval "$servicestartcommand"
fi
done < names.txt
Names.txt
WebSphere:\opt\software\WebSphere\startServer.sh
WebLogic:\opt\software\WebLogic\startWeblogic.sh
Your script can be refactored into this:
#!/bin/bash
while IFS=: read -r servicename servicestartcommand; do
if ps cax | grep -q "$servicename"; then
echo "$servicename Running"
else
echo "!!$servicename!! Not Running, calling $servicestartcommand"
$servicestartcommand
fi
done < names.txt
No need to use wc -l after grep's output as you can use grep -q
No need to use read full line and then use cut, rev etc later. You can use IFS=: and read the line into 2 separate variables
No need to use eval in the end
It is much simpler than you expect. Instead of:
eval "$servicestartcommand"
eval should only be used in extreme circumstances. All you need is
$servicestartcommand
Note: no quotes.
As an example, try this on the command-line:
cmd='ls -l'
$cmd
That should work. But:
"$cmd"
will fail. It will look for a program with a space in its name called 'ls -l'.
May be I don't get the idea, but why not use system variables?
export FOO=bar
echo $FOO
bar
I am working on a project for one of my professors and he asked me to sort a couple hundred .fits images based on their header files (specifically what star they are images of) I think that grep would be the best way to do this however I can't seam to figure out how to use grep based on the header.
I am entering:
ls | imhead *.fits | grep -E -r "PG\ 1104+243" *
to just list them out for now, once they are listed I know how to copy them into a directory.
I am new to using grep so I am unsure as to where my error lies? any help would be greatly appreciated! Thanks!
Assuming that imghead will extract the headers of the .fits as txt, you can use a simple shell script to do it:
script.sh
#!/bin/bash
grep "$1" "$2" > /dev/null 2>&1 && echo "$2"
Note that the + is a special character if you use extended regular expression, meaning if you pass the -E as in the question. A simple grep without any options should do the trick here.
Use find to exec the script on every *.fits file in the current folder:
find -maxdepth 1 -name '*.fits' -exec ./script.sh 'PG 1104+243' {} \;
If you are going to copy/move/alter or do something with the files you find, you might be better off, in terms of complexity and ease of quoting, using a loop like this:
#!/bin/bash
find . -name \*.fits -print0 | while read -d '' -r file; do
echo Checking file: $file
imhead "$file" | grep -q 'PG 1104+243'
if [ $? -eq 0 ]; then
echo Object matches: $file
fi
done
I have this simple bash script:
I run ns simulator on each file passed in argument where last argument is some text string to search for.
#!/bin/bash
nsloc="/home/ashish/ns-allinone-2.35/ns-2.35/ns"
temp="temp12345ashish.temp"
j=1
for file in "$#"
do
if [ $j -lt $# ]
then
let j=$j+1
`$nsloc $file > $temp 2>&1`
if grep -l ${BASH_ARGV[0]} $temp
then
echo "$file Successful"
fi
fi
done
I expected:
file1.tcl Successful
I am getting:
temp12345ashish.temp
file1.tcl Successful
When i run the simulator command myself on the terminal i do not get the file name to which output is directed.
I am not getting from where this first line of output is getting printed.
Please explain it.
Thanks in advance.
See man grep, and see specifically the explanation of the -l option.
In your script (above), you are using -l, so grep is telling you (as instructed) the filename where the match occurred.
If you don't want to see the filename, don't use -l, or use -q with it also. Eg:
grep -ql ${BASH_ARGV[0]} $temp
Just silence the grep:
if grep -l ${BASH_ARGV[0]} $temp &> /dev/null